Chapter(5(
! 5.! − 3( z − 8) + 10 = − 5!! !
Chapter(5(Opener(
− 3 z + 24 + 10 = − 5!!
Try$It$Yourself$(p.$201)$
− 3 z + 34 =
! 1.! 5 + 4 z − 2 z = 5 + ( 4 − 2) z !!
− 34
− 3 z = − 39
= 2z + 5
− 3z
− 39
=
−3
−3
= 5c + 1c + 40 + 3
− !8
− !34
= 5 + 2z
( 2.! 5(c + 8) + c + 3 = 5c + 40 + c + 3 !
! 3.! − 5 x + 8 =
−5
z = 13!!!
= (5 + 1)c + 40 + 3
! ! The!solution!is! z = 13. !
= 6c + 43
! ! Check:! − 3( z − 8) + 10 = − 5 !
−7 !
?
− 3(13 − 8) + 10 = − 5
!
− !8
?
− 3(5) + 10 = − 5
− 5 x = −15
?
−15 + 10 = − 5
− 5x
−15
=
−5
−5
−5 = −5
x = 3!!!!
( ( Check:! − 5 x + 8 = − 7
! 6.! !!!!2 = 10c − 4( 2c − 9) !
!
!!!!2 = 10c − 8c + 36
?
− 5(3) + 8 = − 7
!!!!!2 = 2c + 36
− !36
?
−15 + 8 = − 7
−7 = −7

2c
− 34
!!!!!!!!!!!!!!!!!
=
2
2
−17 = c !!!!!!!!!!!!!!!!!!!!!!!!
! 4.! 7 w + w − 15 = 17!!! !
+ !15
17
! ! The!solution!is! c = −17. !
+ !15
8w =
!!!!! − !36
− 34 = 2c !!!!!!!!!
! ! The!solution!is! x = 3. !
8w − 15 =
!
! ! Check:! 2 = 10c − 4( 2c − 9)
32
?
2 = 10( −17) − 4 2( −17) − 9
8w
32
=
!!
8
8
w = 4!!!!
?
2 = −170 − 4( − 34 − 9)
?
2 = −170 − 4( − 43)
! ! The!solution!is! w = 4. !
! ! Check:! 7 w + w − 15 = 17
2 = −170 + 172
?
7( 4) + 4 − 15 = 17
?
28 + 4 − 15 = 17
?
32 − 15 = 17
17 = 17 
!
?
!
2 = 2
Section(5.1(
5.1$Activity$(pp.$202–203)$
( 1.( a.! C = 10 x + 500 !
!
b.! R = 60 x !
!
c.! C = 10 x + 500 !
R = 60 x
!
Copyright*©*Big*Ideas*Learning,*LLC*
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!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
139
Chapter(5(
! 2.( a.( (
!
!
!
! !
x"
0!
1!
2!
3!
C!
500!
510!
520!
530! 540! 550! 560!
R"
0!
60!
120!
180! 240! 300! 360!
x"
7!
8!
9!
C!
570!
580!
590!
600! 610!
R"
420!
480!
540!
600! 660!
10!
4!
5!
6!
y
600
(
−4
500
! The!solution!is! ( − 0.6, −1.38). A!graphing!calculator!
!
was!used!because!of!the!decimals.!
!
b.! !
!
0!
1!
2!
3!
4!
y = x"
0!
1!
2!
3!
4!
y = − 2x + 9
9!
7!
5!
3!
1!
c.! !
y
1
−3 −2 −1
−3
(−1.5, −3.5)
200
100
0 1 2 3 4 5 6 7 8 9 10 11 12 x
! The!solution!is! ( −1.5, − 3.5). A!graph!was!sketched!
!
because!the!value!of!x!in!the!solution!was!not!an!
integer.!
!
c.( The!graphs!intersect!at! (10, 600). !You!need!to!rent!!
5.1$On$Your$Own$(pp.$204 –205)$
! 4.( a.! Sample-answer:! Xmin = 0, ! Xmax = 16, !
! ! Ymin = 0, Ymax = 800 !
! 1.! y = x − 1 !
!
! ! !
b.( Equation!2!has!a!greater!slope!and!passes!through!the!
origin.!
( !
800
y = −x + 3
4
y y=x−1
3
y = 10x + 500
2
(2, 1)
1
−4 −3 −2 −1
y = 60x
0
0
!
c.( (10, 600) !!
!
d.! (10, 600) !
−3
−4
16
! 5.! Use!a!table!to!determine!when!the!equations!have!the!
same!value,!or!graph!both!equations!and!find!the!point!of!
intersection.!Check!your!solution!by!substituting!it!into!
each!equation!and!making!sure!both!are!satisfied.!
140* Big(Ideas(Math(Blue*
*
y = −x − 5
y = 3x + 1
the!bedroom!for!10!nights!before!you!break!even.!This!
is!the!same!breakHeven!point!as!in!Activity!2.!
(
1 2 3 x
R = 60x
300
!
x"
! The!solution!is! (3, 3). A!table!was!used!because!of!the!
!
equation! y = x. !
400
0
4
(−0.6, −1.38) −4
!
C = 10x + 500
4
a.! !
11!
!
b.! The!cost!equals!the!revenue!when! x = 10. !You!need!
to!rent!the!bedroom!for!10!nights!before!you!break!
even.!
! 3.( a–b.! !
! 6.( Sample-answers:(
Worked;Out*Solutions*
1 2 3 4 x
y = −x + 3
! ! Check:! ( 2, 1) !
! ! y = x −1
?
y = −x + 3 !
?
1 = 2 −1
1 = −2 + 3
1 = 1
1 = 1
! ! So,!the!solution!is! ( 2, 1). !
!
!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
! 2.! y = − 5 x + 14 !
! 4.! y = − 4 x − 7 !
y = x − 10
x + y = 2
! ! !
! ! !
y
14
12
4
x+y=2
y = −5x + 14
10
y
(−3, 5)
3
2
8
6
1
y = x − 10
4
−4 −3 −2
2
O
2 4 6 8
−1
−2
12 x
−4
y = −4x − 7
(4, −6)
−6
−8
x
1 2
−7
−10
−8
! ! Check:! ( 4, − 6) !
! ! Check:! ( − 3, 5) !
y = − 5 x + 14
! !
y = x − 10 !
?
! ! y = − 4x − 7
?
− 6 = − 5( 4) + 14
−6 = −6 
− 6 = − 20 + 14
−3 + 5 = 2
2 = 2
?
5 = 12 − 7
−6 = −6 
! ! So,!the!solution!is! ( − 3, 5). !
! 3.! y = x
! 5.! x − y = 5
!
y = 2x + 1
!
− 3 x + y = −1
! ! !
y = 2x + 1
4
3
y
1
2
(−1, −1)
y
2
3
1

5 = 5
! ! So,!the!solution!is! ( 4, − 6). !
! ! !
!
?
5 = − 4( − 3) − 7
− 6 = 4 − 10
?
x + y = 2
?
−4 −3 −2
y=x
−1
1 2 x
−3x + y = −1
1 2 x
−2
−3
(−2, −7)
y = x
−1 = −1 
x−y=5
−7
! ! Check:! ( −1, −1) !
! !
−5
−8
y = 2x + 1
!
?
! ! Check:! ( − 2, − 7) !
?
! !
−1 = 2( −1) + 1
−1 = − 2 + 1
x − y = 5
− 3x + y = −1
?
− 3( − 2) + ( − 7) = −1
?
6 + ( − 7) = −1
− 2 − ( − 7) = 5
−1 = −1 
?
−2 + 7 = 5
! ! So,!the!solution!is! ( −1, −1). !
5 = 5
!
?

−1 = −1

! ! So,!the!solution!is! ( − 2, − 7). !
!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
141
Chapter(5(
! 6.!
! 2.! Solving!a!system!of!linear!equations!means!finding!the!
ordered!pair! ( x , y ) !that!represents!the!point!of!
1
x + y = −6 !
2
6x + 2 y = 8
! ! !
intersection!of!the!graphs!of!the!equations!in!the!system.!
! 3.! By!substituting!the!point! (3, 4) in!each!equation,!you!can!
y
4
verify!that!the!point!is!a!solution!of!each!equation!in!the!
system.!
6x + 2y = 8
3
2
1
Practice$and$Problem$Solving$
1 2 3 4 5 x
O
! 4.! C = 15 x + 150 !
1
x + y = −6
2
−2
−3
R = 45 x
−4
! ! !
−5
−7
−8
(4, −8)
−9
! ! Check:! ( 4, − 8) !
6x + 2 y = 8
!
6( 4) + 2( − 8) = 8
24 − 16 = 8
2 + ( − 8) = − 6
8 = 8


5
4
3
2
1
O
! ! !
! !
1 2 3 4 x
165!
180!
195!
210!
225!
240!
R"
0!
45!
90!
135!
180!
225!
270!
R = 45 x
!
?
225 = 45(5)
?
225 = 225 
?
x"
0!
1!
2!
3!
4!
5!
6!
C"
80!
104!
128!
152!
176!
200!
224!
R"
0!
44!
88!
132!
176!
220!
264!
C = 24 x + 80
x + 3 y = 17
176 = 44( 4)
?
176 = 176 
176 = 96 + 80
!
!
?
176 = 176 
?
2 + 3(5) = 17
! ! So,!the!breakHeven!point!is! ( 4, 176). !
?
2 + 15 = 17
17 = 17
R = 44 x
?
176 = 24( 4) + 80
! ! Check:! ( 2, 5) !

150!
! ! Check:! ( 4, 176) !
x+y=7
7 = 7
C"
R = 44 x
(2, 5)
?
6!
225 = 225 
x + 3y = 17
2+5 = 7
5!
! 5.! C = 24 x + 80 !
!
! ! x + y = 7
4!
! ! So,!the!breakHeven!point!is! (5, 225). !
x + 3 y = 17
y
3!
225 = 75 + 150
! ! So,!the!solution!is! ( 4, − 8). !
! ! !
2!
225 = 15(5) + 150
?
?
! 7.! x + y = 7
1!
C = 15 x + 150
! !
?
?
1
(4) + (− 8) = − 6
2
−6 = −6
0!
! ! Check:! (5, 225) !
1
x + y = −6
2
! !
x"

!
!
! ! The!solution!is! ( 2, 5). So,!the!kicker!made!2!extra!points!
!
and!5!field!goals.!
5.1$Exercises$(pp.$206 –207)$
Vocabulary$and$Concept$Check$
( 1.! yes;!Any!set!of!two!or!more!linear!equations!in!the!same!
variables!is!a!system!of!linear!equations.!
142* Big(Ideas(Math(Blue*
*
Worked;Out*Solutions*
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
!10.! y = 2 x + 9 !
! 6.! C = 36 x + 200 !
y = 6 − x
R = 76 x
! ! !
x"
0!
1!
2!
3!
4!
5!
6!
C"
200!
236!
272!
308!
344!
380!
416!
R"
0!
76!
152!
228!
304!
380!
456!
! ! !
y=6−x
y
10
(−1, 7)
4
2
! ! Check:! (5, 380) !
! !
O
C = 36 x + 200
R = 76 x
!
?
?
380 = 76(5)
?
380 = 380 
380 = 36(5) + 200
380 = 180 + 200
! ! Check:! ( −1, 7) !
! ! y = 2x + 9
! ! Check:! (6, 7) !
?
7 = 9− 2
7 = 6+1
7 = 7
7 = 7
!11.! y = − x − 4 !
y = − x + 13
y =
?
7 = − 6 + 13
3
x + 4
5
! ! !
7 = 7
! ! 7 = 7
! ! So,!the!solution!is! (6, 7 ). !
3
y= x+4
5
!
(−5, 1)
−6 −5 −4
−2
6.5 = 6.5 
?
6.5 = 3( 2.5) − 1
−1 = 2 − 3
y = −x − 4
?
6.5 = 7.5 − 1
?
!
! !
1 = − ( − 5) − 4
?
1 = 5− 4
1 = 1
1 x
3
x + 4
5
? 3
1 = ( − 5) + 4 !
5
y =
?
1 = −3 + 4
1 = 1
! ! So,!the!solution!is! ( − 5, 1). !
!
! ! Check:! (3, −1) !
?
−1
! ! Check:! ( − 5, 1) !
y = 3x − 1
2
! 9.! C;! y = x − 3 !
3
=
−
y
2x + 5
2
x −3
3
? 2
−1 = (3) − 3
3
1
−5
6.5 = 6.5 
! !
! ! So,!the!solution!is! ( 2.5, 6.5). !
y =
3
y = −x − 4
! ! Check:! ( 2.5, 6.5) !
6.5 = 2.5 + 4
y
−2
y = 3x − 1
?
5
2
! 8.! A;! y = x + 4 !
y = x + 4
?
7 = −2 + 9
! ! So,!the!solution!is! ( −1, 7). !
y = − x + 13
?
7 = 6 − ( −1)
?
! 7.! B;! y = 1.5 x − 2 !
!
?
7 = 2( −1) + 9
! ! So,!the!breakHeven!point!is! (5, 380). !
7 = 1.5(6) − 2
y = 6− x
?
380 = 380 
y = 1.5 x − 2
2 4 x
y = 2x + 9
!
y = − 2x + 5
?
−1 = − 2(3) + 5
?
−1 = − 6 + 5
−1 = −1 
−
=
−
1
1

! !
! ! So,!the!solution!is! (3, −1). !
Copyright*©*Big*Ideas*Learning,*LLC*
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!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
143
Chapter(5(
!12.! y = 2 x + 5 !
!14.! y − x = 17 !
y = 4x + 2
1
y = x −1
2
! ! !
! ! !y − x = 17
y
6
5
y = 2x + 5
3
2
25
20
y=
1
x−1
2
(5, 22)
10
5
1
−5 −4
y
y = 4x + 2
−5
−1
2 3 x
5 10 15 20 x
! ! Check:! (5, 22) !
−2
(−4, −3)
y − x = 17
! !
−4
y = 4x + 2 !
?
! ! Check:! ( − 4, − 3) !
y = 2x + 5
1
x −1
2
? 1
− 3 = ( − 4) − 1 !
2
! ! − 3 = 2( − 4) + 5
?
22 = 20 + 2
−3 = − 2 − 1
−3 = −3 
−3 = −3 
22 = 22 
! ! So,!the!solution!is! (5, 22). !
!15.! x − y = 7 !
0.5 x + y = 5
?
− 3 = −8 + 5
! ! !
! ! So,!the!solution!is! ( − 4, − 3). !
8
y
0.5x + y = 5
4
!13.! x + y = 27 !
2
y = x + 3
O
(8, 1)
2 4
8 x
−4
y=x+3
40
?

17 = 17
y =
?
! ! !
?
22 = 4(5) + 2
22 − 5 = 17
−6
y
x−y=7
−8
20
10
(12, 15)
10 20
! ! Check:! (8, 1) !
x
! ! x − y = 7
x + y = 27
x + y = 27
y = x +3
?
15 = 12 + 3

15 = 15
! ! So,!the!solution!is! (12, 15). !
?

4+1 = 5
5 = 5

!16.( a.! R = 35 x !
!
!
144* Big(Ideas(Math(Blue*
Worked;Out*Solutions*

! ! So,!the!solution!is! (8, 1). !
!
*
0.5(8) + 1 = 5
?
12 + 15 = 27
27 = 27
7 = 7
!
!
?
8−1 = 7
! ! Check:! (12, 15) !
! !
0.5 x + y = 5
?
b.! !
x"
0!
50!
100!
150!
C"
2000!
2750!
3500!
4250!
R"
0!
1750!
3500!
5250!
! So,!100!rides!are!needed!to!break!even.!
!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
!17.! 2.2 x + y = 12.5 !
18.( 2.1x + 4.2 y = 14.7
1.4 x − 4 y = 1
! ! !
!
− 5.7 x − 1.9 y = −11.4
5
!23.( a.! x = the!time!you!will!catch!up!to!your!friend !
!
! 3.4 x =
− 3x
7
3x + 0.5 !
− !3 x !!!!!!!!!!
0.4 x = 0.5 !!!!!!!!!!
−2
0.4 x
0.5
=
!!!!!!!!!!!
0.4
0.4
x = 1.25!!!!!!!!!!!
10
−4 Intersection
Intersection
X=5
Y=1.5
X=1
−3
! ! The!solution!is! (5, 1.5). !
8
Y=3
−1
! The!solution!is! (1, 3). !
!19.! −1.1x − 5.5 y = − 4.4 !
!
!
0.8 x − 3.2 y = −11.2
! ! !
6
(
−10
2
Intersection
X=-6
Y=2
−2
! ! The!solution!is! ( − 6, 2). !
!20.! The!solution!does!not!give!a!point,!only!an!xHcoordinate.!
So,!the!solution!is! ( 4, 3). !
!21.! no;!Linear!equations!either!intersect!at!exactly!one!point,!
do!not!intersect!at!all,!or!intersect!at!infinitely!many!
points.!
!22.! x = math!problems
! So,!you!will!catch!up!to!your!friend!in!1.25!hours,!or!!
1!hour,!15!minutes.!
b.! 3.4 x = 8.5
3.4 x
8.5
=
3.4
3.4
x = 2.5 !
( You!will!cross!the!finish!line!after!2.5!hours.!Find!how!
far!your!friend!has!traveled!after!2.5!hours.!
!
! 3 x + 0.5 = 3( 2.5) + 0.5 = 8 !
!
! When!you!finish!the!race,!your!friend!will!have!rowed!
8!miles.!So,!you!will!be!0.5!mile!ahead!of!your!friend!
when!you!finish!the!race.!
!24.( a.! x = month
y = hair!length
! Your!friend:!
!
! m =
!
! y − y1 = m( x − x1 ) !
!
y = science!problems
y + 10 = x
y − 4 = 0.5 x − 1.5
y = 0.5 x + 2.5
! ! !y
40
x + y = 42
35
y + 10 = x
30
25
20
(26, 16)
15
!
! Her!cousin:!
!
! m =
!
! y − y1 = m( x − x1 ) !
10
y − 7 = 0.4 x − 1.2
0 5 10 15 20 25 30 35 40 45 x
y = 0.4 x + 5.8
! ! Check:! ( 26, 16) !
! !
y2 − y1
9−7
2
=
=
= 0.4 !
x2 − x1
8−3
5
y − 7 = 0.4( x − 3)
5
0
y2 − y1
6.5 − 4
2.5
=
=
= 0.5 !
x2 − x1
8−3
5
y − 4 = 0.5( x − 3)
! ! x + y = 42 !
45
!
!
x + y = 42
y + 10 = x
?
!
?
26 + 16 = 42
42 = 42
!
16 + 10 = 26

26 = 26

!
! Your!friend’s!hair!length!is!modeled!by!
y = 0.5 x + 2.5 and!her!cousin’s!hair!length!is!
modeled!by! y = 0.4 x + 5.8. !
!
! ! So,!you!have!26!math!problems!and!16!science!problems.!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
145
Chapter(5(
!
!
!27.( 6 x − 3( x + 8) = 9!!!!! !
b.! Yes,!it!is!growing!at!a!faster!rate.!
! !y
25
6 x − 3 x − 24 = 9!!!!!
y = 0.4x + 5.8
3 x − 24 =
(33, 19)
20
15
+ 24
10
0
0 5 10 15 20 25 30 35 x
!
! Check:! (33, 19) !
!
!
y = 0.5 x + 2.5
?
19 = 0.5(33) + 2.5
?
!
+ !24
3x =
y = 0.5x + 2.5
5
9
33
3x
33
=
!
3
3
x = 11!!!
y = 0.4 x + 5.8
!
! ! The!solution!is! x = 11. !
?
19 = 0.4(33) + 5.8
! ! Check:
?
19 = 16.5 + 2.5
19 = 13.2 + 5.8
19 = 19 
19 = 19 
6 x − 3( x + 8) = 9
6(11) − 3(11 + 8) = 9
?
66 − 3(19) = 9
! So,!your!friend’s!hair!will!be!the!same!length!as!your!
cousin’s!hair!in!month!33.!
?
66 − 57 = 9
9 = 9
Fair$Game$Review$
25.!
!
?
3
1
3
1
c − c +3 = 7
c − c + 3 = 7!!!! !Check:
4
4
4
4
?
1
3
1
c +3 =
7
(8) − (8) + 3 = 7
2
4
4
?
− !3
− !3
6− 2+3 = 7
1
c =
4
7 = 7
2
1 
2 c  = 2( 4)
2 
!
(28.! C;! m =
=
y2 − y1
x2 − x1

!!
−2 − ( −1)
−2 − 3
−1
=
−5
1
=
5

!
c = 8!!!!
(
! ! The!solution!is! c = 8. !
!26.! 5( 2 − y ) + y = − 6!! !!Check:! 5( 2 − y ) + y = − 6
10 − 5 y + y = − 6!!
10 − 4 y =
−6
5( 2 − 4) + 4 = − 6
?
− !10
5( − 2) + 4 = − 6
− 4 y = −16
−10 + 4 = − 6
− !10!!!!!!!!!
−4y
−16
=
−4
−4
!
?
?
−6 = −6

y = 4!!!!!
! ! The!solution!is! y = 4. !
146* Big(Ideas(Math(Blue*
*
Worked;Out*Solutions*
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
Section(5.2(
!
! Substitute! 6 x − 11 for!y.!
2x + 3 y = 7
5.2$Activity$(pp.$208 –209)$
2 x + 3(6 x − 11) = 7
! 1.( a.! 6 x − y = 11 !
2 x + 18 x − 33 = 7
2x + 3y = 7
!
! Method!1:!Solve!for!x.!
!
! 6 x − y = 11
+y
20 x − 33 =
!
!
!
20 x =
+ y !!!!!!
2x + 3y = 7
2( 2) + 3 y = 7
4 + 3y =
11 1
+ y for!x.!
6
6
!
2x + 3 y = 7
!
 11 1 
2 + y  + 3 y = 7
6 
6
11 1
+ y + 3y = 7
3
3
11 10
+
y = 7
3
3
!
11
11
−
−
3
3
!
!
!
!
−3
2x =
! The!solution!is! ( 2, 1). !
! So,!the!solution!is!the!same!using!both!methods.!
b.! 2 x − 3 y = −1 !
!
!
!
!
!
4
! Substitute! y + 1 !for!x.!
2 x − 3 y = −1
2( y + 1) − 3 y = −1
2 y + 2 − 3 y = −1
− y + 2 = −1
!
−2
−2
!
! Method!2:!Solve!for!y.!
6 x − y = 11 !!!!! !!!!!!
!
!!!!!! !
− y = − 6 x + 11
(−1)(− y)
!
y = 3
! The!solution!is! ( 2, 1). !
!
+ y !! !
− y = −3
(−1)(− y ) = (−1)(− 3)
!
− 6x
+y
x = y +1
2x
4
=
2
2
x = 2
− 6x
!
3
! Method!1:!Solve!for!x.!
x − y = 1!!!!!!!
7
−3
3y =
!
2 x + 3(1) = 7
!
7
−4
x − y =1
2x + 3y = 7
!
!
−4
3y
3
=
3
3
y = 1
10
10
y =
3
3
3  10 
3  10 
 y =
 
10  3 
10  3 
y = 1
2x + 3 =
40
20 x
40
=
20
20
x = 2
6x
11 y
=
+
6
6
6
11 1
x =
+ y
6
6
! Substitute!
+ 33 !
+ 33
6 x = 11 + y
!
7
!
!
!
!
! The!solution!is! ( 4, 3). !
!
= ( −1)( − 6 x + 11)
y = 6 x − 11 !!
!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
2 x − 3 y = −1
2 x − 3(3) = −1
2 x − 9 = −1
+9
+9
2x = 8
2x
8
=
2
2
x = 4
!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
147
Chapter(5(
!
!
5x − 4 y = − 3
! Method!2:!Solve!for!y.!
x − y = 1
−x
−x
!
!
− y = −x + 1
= ( −1)( − x + 1)
y = x −1
(−1)( − y )
!
!
! Substitute! x −
2x − 3y
2 x − 3( x − 1)
2 x − 3x + 3
−x + 3
!
−3
1 for!y.!
= −1
= −1
= −1
= −1
−3
!
!
5
 1
5 − y +  −
3
3

5
25
− y +
−
3
3
17
− y +
3
17
34
y = −
3
3
3  17 
3  34 
− − y = − − 
17  3 
17  3 
y = 2
!
= ( −1)( − 4)
3x + y = 5
3x + 2 = 5
2 x − 3 y = −1
−2
2( 4) − 3 y = −1
!
−8
−3y = −9
!
!
! The!solution!is! (1, 2). !
!
! Method!2:!Solve!for!y.!
3x + y = 5
! − !3 x
!
− !3 x
y = − 3x + 5
!
−3y
−9
=
−3
−3
y = 3
!
!
!
! The!solution!is! ( 4, 3). !
!
! So,!the!solution!is!the!same!using!both!methods.!
c.! 3 x + y = 5
!
!
! Substitute! − 3x + 5 for!y.!
!
5x − 4 y
5 x − 4( − 3 x + 5)
5 x + 12 x − 20
17 x − 20
!
+ 20
5x − 4 y = − 3
!
! Method!1:!Solve!for!x.!
3x + y = 5
−y
!
!
!
−y
3x = − y + 5
−2
3x = 3 !
3x
3
=
3
3
x = 1
!
8 − 3 y = −1
!
3x
5 !
y
= − +
3
3
3
1
5
x = − y +
3
3
1
5
! Substitute! − y + for!x.3
3
3x + y = 5
3(1) + y = 5
!
3+ y =
−3
Worked;Out*Solutions*
5!
−3
y =
148* Big(Ideas(Math(Blue*
= −3
= −3
= −3
= −3
+ 20 !
17 x = 17
17 x
17
=
17
17
x = 1
!
*
!
−
x = 4
− 8!!!!!!!!
4y = −3
25
= −3
3
25
25
−
−
3
3
−x = −4
(−1)(− x)
4y = −3
2
!
! The!solution!is! (1, 2). !
!
!
! So,!the!solution!is!the!same!using!both!methods.!
!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
!
d.! 5 x − y = 2
!
!
3 x − 6 y = 12
!
! Method!1:!Solve!for!x.!
5x − y = 2
+y
+y
!
5x = y + 2
!
!
!
1
2
! Substitute! y + for!x.!
5
5
− 27 x =
0
− 27 x
0
=
− 27
− 27
x = 0
5x
y
2 !
=
+
5
5
5
1
2
x = y +
5
5
3 x − 6 y = 12
!
! Substitute! 5 x − 2 for!y.!
3 x − 6 y = 12
3 x − 6(5 x − 2) = 12
3 x − 30 x + 12 = 12
− 27 x + 12 =
12
−
12
−
12 !
!
2
1
3 y +  − 6 y = 12
5
5
3
6
y + − 6 y = 12
5
5
27
6
− y +
= 12
5
5
!
6
6
−
−
5
5
27
54
− y =
5
5
5  27 
5  54 
− − y = −  
27  5 
27  5 
y = −2
!
!
5x − y =
5(0) − y =
0− y =
!
−y =
1
−
−
( )( y ) =
y =
!
! The!solution!is! (0, − 2). !
!
!
! So,!the!solution!is!the!same!using!both!methods.!
!
e.! x + y = −1
5 x + y = −13
!
! Method!1:!Solve!for!x.!
x + y = −1
!
!
!
!
!
5( − y − 1) + y = −13
− 5 y − 5 + y = −13
− !2
0
− 4 y − 5 = −13
!
!
!
!
! The!solution!is! (0, − 2). !
!
! Method!2:!Solve!for!y.!
5x − y = 2
− 5x
− 5 x !!!! !!!!
!
− y = − 5 x + 2!!!!!!!!! !
−
−
1
( )( y ) = (−1)(− 5 x + 2)
y = 5x − 2
+5!
+5
−4y =
5x
0
=
5
5
x = 0
!
!
5 x + y = −13
2
5x =
−y
! Substitute! − y − 1 for!x.!
5 x − ( − 2) = 2
− !2
−y
x = −y − 1
5x − y = 2
5x + 2 =
2
2
2
!
2
(−1)(2)
−2
−8
−4y
−8
=
−4
−4
y = 2
x + y = −1
!
!
x + 2 = −1
−2
−2
!
x = −3
!
! The!solution!is! ( − 3, 2). !
!
! Method!2:!Solve!for!y.!
x + y = −1
! − x !!!!!!!
!
−x
!
y = −x − 1
!
Copyright*©*Big*Ideas*Learning,*LLC*
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!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
149
Chapter(5(
!
! Substitute! − x − 1 for!y.!
!
5x + y
5 x + ( − x − 1)
5x − x − 1
4x − 1
!
+1
=
=
=
=
−13
−13
−13
−13
+1!
4 x = −12
4x
−12
=
4
4
x = −3
!
!
x + y = −1
− 3 + y = −1
+ 3!!!!!!!
+3
!
2x − 6 y = − 6
− !2 x
!
!
!
!
!
! So,!the!solution!is!the!same!using!both!methods.!
!
f.! 2 x − 6 y = − 6 !
7x − 8y = 5
! Method!1:!Solve!for!x.!
2x − 6 y = − 6
+ 6y
+ 6y
!
!
!
− 8y
− 8y
− 8y
− 21
+ 21
= 5
= 5
= 5
5
=
+ 21 !
2x − 6 y = −6
2 x − 6( 2) = − 6
2 x − 12 = − 6
!
!
+ 12
+ 12
13
x = 13 !
3
3  13 
3
(13)
 x =
13  3 
13
x = 3
2x − 6 y = − 6
2(3) − 6 y = − 6
2x =
6
2x
6
=
2
2
x = 3
6 − 6y =
!
!
−6
−6
−6
− 6 y = −12
!
−6y
−12
=
−6
−6
y = 2
!
! The!solution!is! (3, 2). !
!
! So,!the!solution!is!the!same!using!both!methods.!
! 2.( a.! Sample-answer: ( 2, 6) !
!
b.! Sample-answer:! 6 x − y = 6, 2 x + 3 y = 22 !
!
c.! Answer-should-include,-but-is-not-limited-to:!Students!
should!solve!the!system!of!linear!equations!by!solving!
for!either!x!or!y,!then!substituting!back!into!the!
equations!to!verify!the!ordered!pairs!shown!in!part!(a).!
!
!
!
! The!solution!is! (3, 2). !
150* Big(Ideas(Math(Blue*
*
1

7 x − 8 x + 1 = 5
3

8
7x − x − 8 = 5
3
13
x−8 = 5!
!
!
3
+8
+8 !
! Substitute! 3 y − 3 for!x.!
13 y = 26
13 y
26
=
13
13
y = 2
!
1
! Substitute! x + 1 for!y.!
3
2x = 6 y − 6 !
2x
6y
6
=
−
2
2
2
x = 3y − 3
!
1
x +1
3
y =
! The!solution!is! ( − 3, 2). !
7x
7(3 y − 3)
21 y − 21
13 y
−6y
− 2x
6 !
=
−
−6
−6
−6
!
7x − 8y = 5
!
!
− !2 x
− 6 y = − 2x − 6
y = 2
!
! Method!2:!Solve!for!y.!
Worked;Out*Solutions*
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
! 3.! ( A,!C) x + y = − 3 !
! ! (G,!H) x + y = 0
x − y = −3
x + y = −3
! ! !!
−y
x + y = 0
!
−y
−y
x − y = −16
x − y = −3
(− y)
− 3) − y = − 3
−2y − 3 = −3
!
+3
!
+3
−2y =
0
!
!
!
!
!
x + y = −3
x = −3
!
!
=
!
(A,!C)
!
(− 3, 0)
( x, y )
= (G,!H)
= ( − 8, 8)
− 2 y = 10
10
−2y
=
−2
−2
− 4 y − 18 − y = −13
− 5 y − 18 = −13 !!!
!
+ 18
x =
!
!
( x, y )
+ 18 !!!
−5 y =
5 !!
−5 y
5
=
−5
−5
y = −1
x + ( − 5) = 0
+5
!
2( − 2 y − 9) − y = −13
− y = 10
x−5 =
!
2 x − y = −13
! ! !!
x+ y = 0
!
!
−2y
y = −5
!
!
x = −2y − 9
−y
x − y = 10
!
!
−2y
x = −y
!
0
− !8
x + 2y = −9
x + y = 0!! !
(− y)
−8
2 x − y = −13
x − y = 10
−y
x +8 =
! ! ( I,!L) x + 2 y = − 9
! ! ( D,!E) x + y = 0 !
! ! !!
!
x = −8
! x + 0 = −3 !
=
− 2 y = −16
−2y
−16
=
−2
−2
x + y = 0!!
y = 0
( x, y )
− y = −16
y = 8
0
−2y
=
−2
−2
!
−y
x = −y
x = −y − 3
(− y
!
x − y = −16
0
x + 2 y = −9
!
x + 2( −1) = − 9
+5
!
5
= ( D,!E)
= (5, − 5)
x − 2 = −9 !
+2
x = −7
!
!
!
Copyright*©*Big*Ideas*Learning,*LLC*
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+2
( x, y )
= ( I,!L)
= ( − 7, −1)
!
!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
151
Chapter(5(
! ! ( M,!N) x + 2 y = 4
! ! ( R,!S) 2 x + y = 21 !
!
2 x − y = −12
x− y = 6
x + 2y = 4
− 2y
!! 2 x + y = 21
! ! !
− 2y
−y
x = −2y + 4
!
2 x = − y + 21
2 x − y = −12 !
!
2x
−y
21
=
+
2
2
2
1
21
x = − y +
2
2
2( − 2 y + 4) − y = −12
− 4 y + 8 − y = −12
− 5 y + 8 = −12
−8
x − y = 6
−8
21 
 1
− y +
 − y = 6
2
 2
− 5 y = − 20
−5 y
− 20
=
−5
−5
3
21
− y +
= 6
2
2
21
21
−!
−!
2
2
y = 4
x + 2y = 4
x + 2( 4) = 4
!
x+8 =
!
−8
4
!
( x, y )
−8
= ( M,!N)
= ( − 4, 4)
!
!
!
! ! (O,!P) x + 2 y = − 2
2 x + 3 = 21!
!
−3
!
x + 2y = −2
!
2x − y = 6
!
2( − 2 y − 2) − y = 6
!
−4y − 4 − y = 6
−5 y − 4 =
+4
−3 !
2 x = 18 ! !
2x
18
=
2
2
x = 9
− 2y
x = −2y − 2
! ! !!
!
2 x + y = 21
2x − y = 6
− 2y
3
9
− y = −
2
2
2 3 
2 9 
− − y = − − 
3 2 
3 2 
y = 3
!
x = −4
!
!
−y
6 !!
+4
!
!
( x, y )
= ( R,!S)
= (9, 3)
!
!
!!
− 5 y = 10 !!
−5 y
10
=
−5
−5
y = −2
x + 2y = −2
x + 2( − 2) = − 2
!
!
x − 4 = −2 !
+4
+4
x =
!
!
( x, y )
= (O,!P)
= ( 2, − 2)
2
!
152* Big(Ideas(Math(Blue*
*
Worked;Out*Solutions*
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
! ! (T,!U) 2 x + y = − 7 !
x − y = 1
x − y = 10
2x + y = −7
! ! !!
−y
 1

 − y + 10  − y = 1
 2

3
1
− y + 10 =
2
− 10
− 10
!
−y
2x = − y − 7
2x
−y 7
=
−
2
2
2
1
7
x = − y −
2
2
x − y = 10
! ! !
!
7
 1
 − y −  − y = 10
2
 2
2 x + ( − 9) = − 7
2x − 9 = −7
+9
2x =
2
!
!
!
= (T, U)
= (1, − 9)
!
!
!!
( x, y )
= ( V,!W )
= (7, 6)
!
5.2$On$Your$Own$(pp.$210 –211)$
! 1.!
y = 2x + 3
y = 2x + 3
5x =
− 2x
2 x + y = 20
! Check :!! (1,!5)
y = 2x + 3 !
y = 5x
!
x− y =1
−y
2 x = 14 !
2x
14
=
2
2
x = 7
! ! ( V,!W) 2 x + y = 20 !
! ! !!
!!
−6
! 4.! Solve!for!a!variable!in!one!equation.!Substitute!the!
expression!for!that!variable!into!the!other!equation!and!
solve!the!equation.!Substitute!the!variable!value!that!you!
know!into!one!of!the!equations!to!find!the!value!of!the!
other!variable.!
2x
2
=
2
2
x = 1
( x, y )
!
! ! So,!the!quote!is!“GIVE!ME!A!PLACE!TO!STAND,!AND!
I!WILL!MOVE!THE!EARTH.”!
2x + y = − 7
+9
2
(− 9)
3
y = 6
−6
y = −9
!
−
2 x + 6 = 20
3
27
− y =
2
2
2 3 
2  27 
− − y = −   !
3 2 
3 2 
!
−9
2 x + y = 20
3
7
− y −
= 10
2
2
7
7
+
+
2
2
! ! !
3
− y =
2
2 3 
− − y =
3 2 
3x =
!
−y
?
?
5 = 5
5 = 5
3
3x
3
=
3
3
x = 1
2 x = − y + 20
2x
−y
20
=
+
2
2
2
1
x = − y + 10
2
5 = 5(1)
5 = 2 +3
− 2x
y = 5x
?
5 = 2(1) + 3
2x + 3
!
y = 5x
! !
= 5(1) !
= 5
! ! The!solution!is! (1, 5). !
!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
153
Chapter(5(
! 2.! 4 x + 2 y = 0 !
y =
! 3.! x = 5 y + 3
2 x + 4 y = −1
1
x −5
2
4x + 2 y = 0
1

4 x + 2 x − 5  = 0
2

4 x + x − 10 = 0
! !
5 x − 10 =
! !
+ 10
0
+ 10
5x =
!
10
5x
10
=
5
5
x = 2
= −4
! ! The!solution!is! ( 2, − 4). !
1
x −5
2
? 1
− 4 = ( 2) − 5
2
y =
?
4( 2) + 2( − 4) = 0
?
0 = 0
14 y = − 7
14 y
−7
=
14
14
1
y = −
2
x = 5y + 3
!
−4 = −4 
2 x + 4 y = −1
1
 1
= 5 −  + 3
2
 2
1 ? 5
= − +3
2
2
1
1
= 
2
2
−4 = 1 − 5

−1
−1
−1
−1
−6
!
!
1
 1
2  + 4 −  = −1
 2
 2
?
?
8−8 = 0
=
=
=
=
1 1
! ! Check:!  , −  !
 2 2
! ! x = 5y + 3
! ! Check:! ( 2, − 4) !
4x + 2 y = 0
2x + 4 y
2(5 y + 3) + 4 y
10 y + 6 + 4 y
14 y + 6
−6
 1
= 5 −  + 3
 2
! !
!
5
= − +3
2
1
=
2
1 1
! ! The!solution!is!  , − . !
2 2
1
y =
x −5
2
1
= ( 2) − 5 !
! !
2
= 1−5
! !
!
?
?
1 + ( − 2) = −1
−1 = − 1

! 4.! x = number!of!cups!of!lemonade !
y = number!of!cups!of!orange!juice
! !
! !
x + y = 100
!
2 x + 3 y = 240
x + y = 100
−y
−y
!
x = − y + 100
2x + 3 y
2( − y + 100) + 3 y
− 2 y + 200 + 3 y
! !
y + 200
− 200
= 240
= 240
= 240
!
= 240
− 200
y =
! !
x + y = 100
x + 40 = 100
− 40
− 40
40
!
x = 60
! ! The!juice!stand!sold!60!cups!of!lemonade!and!40!cups!of!
orange!juice.!
$
154* Big(Ideas(Math(Blue*
*
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Copyright*©*Big*Ideas*Learning,*LLC*
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Chapter(5(
5.2$Exercises$(pp.$212–213)$
! 5.! Sample-answer:!
Vocabulary$and$Concept$Check$
1
x + 5y = 6
!
! ! 4
x + 6 y = 10
! 1.! Solve!one!equation!for!one!of!the!variables.!Substitute!!
the!expression!into!the!other!equation!and!solve!for!the!
second!variable.!Substitute!the!value!of!the!second!
variable!into!one!of!the!original!equations!and!solve!!
for!the!first!variable.!
x + 6 y = 10
! !
− 6y
− 6y
x = − 6 y + 10
! 2.! Solve!for!a!variable!with!a!coefficient!of!1!or! −1, or!else!
1
x + 5y = 6
4
solve!for!the!variable!that!is!easiest.!
1
(− 6 y + 10) + 5 y = 6
4
3
5
− y + + 5y = 6
2
2
7
5
y +
= 6
2
2
5
5
−
−
2
2
! 3.! sometimes;!A!solution!obtained!by!graphing!may!not!be!
exact.!
Practice$and$Problem$Solving$
! 4.! Sample-answer:!
2 x + 3 y = 13
4x − y = 5
4x − y = 5
! ! − !4 x !!!!!!
− !4 x
7
7
y =
2
2
2 7 
2 7 
 y =  
7 2 
7 2
y = 1
!
− y = − 4x + 5
(−1)(− y )
(−1)(− 4 x
=
+ 5)
y = 4x − 5
2 x + 3 y = 13
! !
14 x − 15 =
! !
+ 15
x + 6(1) = 10
! !
13
28
14 x
28
=
14
14
x = 2
x + 6 = 10 !
−6
+ 15 !
14 x =
!
x + 6 y = 10
2 x + 3( 4 x − 5) = 13
2 x + 12 x − 15 = 13
!
x =
−6
4
! ! So,!the!solution!is! ( 4, 1). !
!
!
4x − y = 5
4( 2) − y = 5
8− y =
! !
−8
5
−8
−y =
(−1)(− y )
!
−3
= ( −1)( − 3)
y = 3
! ! So,!the!solution!is! ( 2, 3). !
Copyright*©*Big*Ideas*Learning,*LLC*
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Big(Ideas(Math(Blue(
Worked;Out*Solutions*
155
Chapter(5(
!10.! y = x − 4
! 6.! Sample-answer:!
5 x − 9 y = −15
! !
2 x + 10 y = 62
! !
− 10 y
x−4 =
2 x = −10 y + 62
!
+4
− 3x =
− 155
y = x − 4
! !
−15
− 155 !
− 34 y = −170
− 34 y
−170
=
− 34
− 34
! ! So,!the!solution!is! ( 2, − 2). !
! ! Check:! ( 2, − 2) !
y = x − 4
− 50
2x =
y = 4 x − 10
?
! !
?
− 2 = 4( 2) − 10
−2 = 2 − 4
− 2 = −2
2 x + 10(5) = 62
! !
= 2 − 4!
= −2
y = 5
2 x + 10 y = 62
− 6!!!!!!!!!
x = 2
5( − 5 y + 31) − 9 y = −15
− 34 y + 155 =
+ 4!!!!!!!!!
− 3x
−6
=
−3
−3
5 x − 9 y = −15
− 25 y + 155 − 9 y = −15
− 4x
− 3 x − 4 = −10!!!!!!!!!
2x
−10 y
62
=
+
2
2
2
x = − 5 y + 31
2 x + 50 =
4 x − 10 !
− 4 x !!!!!!
− 10 y
! !
!
y = 4 x − 10
2 x + 10 y = 62
?

−2 = −2 
62
− 50
12
!
− 2 = 8 − 10
!
2x
12
=
2
2
x = 6
! ! So,!the!solution!is! (6, 5). !
!11.! y = 2 x + 5 !
y = 3x − 1
3x − 1 =
− 2x
! !
x −1 =
5
+1
+1
x =
! 7.! 4 x − y = 3; The!coefficient!of!y!is! −1. !
! 8.! x + 6 y = 0; The!coefficient!of!x!is!1,!and!there!is!no!
constant.!
! 9.! 2 x + 10 y = 14; Dividing!by!2!to!solve!for!x!yields!
integers.!
2x + 5
− 2x
!
6
y = 3x − 1
! !
= 3(6) − 1
= 18 − 1
!
= 17
! ! So,!the!solution!is! (6, 17). !
! ! Check:! (6, 17 ) !
y = 2x + 5
?
! !
!
156* Big(Ideas(Math(Blue*
*
Worked;Out*Solutions*
17 = 2(6) + 5
?
y = 3x − 1
?
17 = 3(6) − 1
?
17 = 12 + 5
17 = 18 − 1
17 = 17 
17 = 17 
!
!
Copyright*©*Big*Ideas*Learning,*LLC*
All*rights*reserved.*
Chapter(5(
!13.! 4 x − 2 y = 14 !
!12.! x = 2 y + 7 !
3x − 2 y = 3
y =
3( 2 y + 7) − 2 y = 3
1

4 x − 2 x − 1 = 14
2


4 x − x + 2 = 14
6 y + 21 − 2 y = 3
4 y + 21 =
− 21
! !
1
x −1
2
3
− 21
3 x + 2 = 14
4 y = −18 !
! !
−18
4y
=
4
4
9
y = −
2
3x = 12
3x
12
=
3
3
x = 4
x = 2y + 7
! !
 9
= 2 −  + 7
!
 2
= −9 + 7
1
x −1
2
1
= ( 4) − 1 !
2
= 2 −1
y =
! !
= −2
9

! ! So,!the!solution!is!  − 2, − . !
2

= 1
! ! So,!the!solution!is! ( 4, 1). !
9

! ! Check:!  − 2, −  !
2

! ! Check:! ( 4, 1) !
x = 2y + 7
?
 9
− 2 = 2 −  + 7
 2
! !
−2!
−2
3x − 2 y = 3
 9 ?
3( − 2) − 2 −  = 3
 2
?
− 2 = −9 + 7
−2 = −2 
y =
?
!
?
! ! 4( 4) − 2(1) = 14
−6 + 9 = 3
3 = 3
1
x −1
2
? 1
1 = ( 4) − 1 !
2
4 x − 2 y = 14
?
?
16 − 2 = 14

14 = 14
1 = 2 −1

1 = 1
!14.! 2 x = y − 10 !
x + 7 = y
2 x = ( x + 7) − 10
! !
2x =
−x
x −3
−x
!
x = −3
y = x + 7
! !
= ( − 3) + 7 !
= 4
! ! So,!the!solution!is! ( − 3, 4). !
! ! Check:! ( − 3, 4) !
2 x = y − 10
?
! ! 2( − 3) = 4 − 10
−6 = −6 
!
Copyright*©*Big*Ideas*Learning,*LLC*
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x + 7 = y
?
−3 + 7 = 4
4 = 4
!

!
Big(Ideas(Math(Blue(
Worked;Out*Solutions*
157
Chapter(5(
!17.( a.! x = price!of!an!adult!ticket !
1
y = 0!
3
12 x + 3 = y
!15.! 8 x −
y = price!of!a!student!ticket
1
8 x − (12 x + 3) = 0
3
8x − 4x − 1 = 0
4x − 1 =
0
+1
+1
! !
4x =
1
!
!
!
b.!
260 y
1040
=
260
260
y = 4
x = 2y
!
!
!
! So,!the!price!of!an!adult!ticket!is!$8,!and!the!price!of!a!
student!ticket!is!$4.!
!18.! y − x = 0
( ( y − x = 0
+x
12 x + 3 = y
2x − 5 y = 9
2x − 5x = 9
?
1
12  + 3 = 6
 4
?
!
! !
3+3 = 6

6 = 6
!16.( a.! x = number!of!students!in!drama!club

y − x = 0
!
y − ( − 3) = 0
! !
!
+ 10) + y = 64
2 y + 10 =
− 10
2y =
64
− 10
! ! So,!the!solution!is! ( − 3, − 3). !
! ! Check:! ( − 3, − 3) !
y − x = 0
2x − 5 y = 9
?
! !
?
− 3 − ( − 3) = 0
2( − 3) − 5( − 3) = 9
?
0 = 0
!
!
?
−3 + 3 = 0
! x = y + 10 !
= 27 + 10
−3
y = −3
54
2y
54
=
2
2
y = 27
!
y +3 = 0 !
−3
x + y = 64 !
(y
!
x = −3
y = number!of!students!in!yearbook!club
b.!
− 3x = 9
− 3x
9
=
−3
−3
?
2 − 2 = 0
!
!
+x
y = x
1
8x − y = 0
3
?
1 1
8  − (6) = 0
! !  4 3
0 = 0
!
2x − 5 y = 9
1 
! ! Check:!  , 6  !
4 
x = y + 10
= 2( 4) !
!
= 8
1 
! ! So,!the!solution!is!  , 6 . !
4 
x + y = 64
64 x + 132 y = 1040 !
260 y = 1040
= 6
!
!
128 y + 132 y = 1040
!
1
= 12  + 3
 4
= 3+ 3
!
x = 2y
64( 2 y ) + 132 y = 1040
4x
1
=
4
4
1
x =
4
! ! y = 12 x + 3
64 x + 132 y = 1040
− 6 + 15 = 9

9 = 9

!
= 37
!
! So,!there!are!37!students!in!the!drama!club!and!!
27!students!in!the!yearbook!club.!
158* Big(Ideas(Math(Blue*
*
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Chapter(5(
!19.! x + 4 y = 14 !
!20.! − 2 x − 5 y = 3 !
3 x + 7 y = 22
3x + 8 y = − 6
! ! x + 4 y = 14
− 4y
! ! − 2x − 5 y = 3
!
+ 5y
− 4y
− 2x
5y
3
=
+
−2
− 2 −2
3 x + 7 y = 22
3( − 4 y + 14) + 7 y = 22
5
3
x = − y −
2
2
−12 y + 42 + 7 y = 22
− 5 y + 42 =
− 42
22
3
 5
3 − y −  + 8 y = − 6!!!
2
 2
15
9
− y − + 8 y = − 6!!!
2
2
1
9
y −
= − 6!!!
2
2
9
9 !
! !
+
+ !!!!
2
2
− 42 !
− 5 y = − 20
−5y
− 20
=
−5
−5
y = 4
x + 4 y = 14
x + 4( 4) = 14
! !
x + 16 =
1
3
y = − !!!!
2
2
1 
 3
2 y  = 2 − 
2 
 2
y = − 3!!!!!
14 !
− 16
− 16
x =
+ 5y
−2 x = 5 y + 3
x = − 4 y + 14
! !
!
−2
! ! So,!the!solution!is! ( − 2, 4). !
! ! Check:! ( − 2, 4) !
− 2x − 5 y = 3
x + 4 y = 14
− 2 x − 5( − 3) = 3
3x + 7 y = 22
?
! !
3( − 2) + 7( 4) = 22
?
!
?
− 2 + 16 = 14
14 = 14
− 2 x + 15 =
?
− 2 + 4( 4) = 14
− 6 + 28 = 22

22 = 22
! !

3
− 15
− 15
− 2 x = −12
!
− 2x
−12
=
−2
−2
x = 6
! ! So,!the!solution!is! (6, − 3). !
! ! Check:! (6, − 3) !
− 2x − 5 y = 3
3x + 8 y = − 6
?
! !
?
− 2(6) − 5( − 3) = 3
3(6) + 8( − 3) = − 6
?
−12 + 15 = 3
3 = 3
!
Copyright*©*Big*Ideas*Learning,*LLC*
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!
?
18 − 24 = − 6

−6 = −6

!
Big(Ideas(Math(Blue(
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159
Chapter(5(
!21.! The!expression!for!y!was!not!substituted!back!into!the!
other!equation.!Instead,!the!expression!was!substituted!
back!into!the!same!equation.!
3x − 2 y = 4
!23.! x = number!of!cats !
y = number!of!dogs
x + y = 65
! !
3 x − 2( − 2 x + 5) = 4
3 x + 4 x − 10 = 4
7 x − 10 =
! !
+ 10
7x =
y =
x + y = 65
4
7
x = 65
6
13
x = 65
!
! !
6
6  13 
6
(65)
 x =
13  6 
13
x +
+ 10 !
14
7x
14
=
7
7
x = 2
x = 30
2x + y = 5
2( 2) + y = 5
! !
4+ y =
−4
7
y = x
6
7
! !
= (30) !
6
= 35
5!
−4
y =
1
! ! So,!the!solution!is! ( 2, 1). !
!22.! x = !measure!of!a!base!angle
! ! So,!there!are!30!cats!and!35!dogs!in!the!shelter.!
!
!24.! x = the!tens!place!digit !
y = measure!of!the!obtuse!angle
! ! y = 2.5 x !
! ! 2 x + y = 180 !
! !
2 x + y = 180 !
y = the!ones!place!digit
! ! x + y = 8
! ! x + y = 8
−y
!
−y
x = −y + 8
4.5 x = 180
10 y + x = 10 x + y + 36
4.5 x
180
=
4.5
4.5
x = 40
10 y + ( − y + 8) = 10( − y + 8) + y + 36
10 y − y + 8 = −10 y + 80 + y + 36
y = 2.5 x
9 y + 8 = − 9 y + 116
= 2.5( 40) !
= 100
!
10 y + x = 10 x + y + 36
2 x + 2.5 x = 180
! !
!
7
x
6
−8
! !
−8
!
9 y = − 9 y + 108
! ! So,!the!base!angles!are! 40° and!the!obtuse!angle!is!100°. !
+ 9y
+ 9y
18 y = 108
18 y
108
=
18
18
y = 6
x + y = 8
! !
x + 6 =
−6
x =
8
−6
!
2
! ! So,!the!number!is!26.!
!
160* Big(Ideas(Math(Blue*
*
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!
Copyright*©*Big*Ideas*Learning,*LLC*
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Chapter(5(
!25.! d = number!of!dance!songs !
!29.! B;!
! ! ∠1 is!also!107°. !
r = number!of!rock!songs
c = number!of!country!songs
∠1 + ∠2 =
! ! d + r + c = 1075 !
! !
d = 3r
107° + ∠2 =
− 107°!!!!!!!!!!
d + r + c = 1075
− 105
− 105
5r =
Available!at!BigIdeasMath.com.!
!
970
5r
970
=
5
5
r = 194
! ! d = 3r
73°
Study(Help((
5r + 105 = 1075
! !
!
! ! So,!the!measure!of!angle!2!is! 73°. !
+ r + ( r + 105) = 1075
3r + r + r + 105 = 1075
Quiz(5.1–5.2((
! 1.! B;! y = x − 2
!
y = − 2x + 1
! ! Check:! (1, −1) !
y = x−2
!
y = − 2x + 1
?
?
= 3(194)
−1 = 1 − 2
−1 = − 2(1) + 1
= 582
−1 = −1 
−1 = − 2 + 1
c = r + 105
= 299
! ! So,!there!are!582!dance!songs,!194!rock!songs,!and!!
299!country!songs!on!the!system.!
Fair$Game$Review$
3x − 9 =
!26.!
− 7y
7y !
3x − 7 y − 9 =
y = x−3
?
0 = 3−3
9
8 − 5 y = − 2x !
+ 2x
+ 2x
8 − 5 y + 2x =
−8
0
0 = 0
−5 y + 2x = −8
2x − 5 y = −8
! ! Check:! ( − 2, − 3) !
1
x−2
2
? 1
− 3 = ( − 2) − 2
2
y =
−y
( ( The!equation!in!standard!form!is( 6 x − y = 3. !
!
1
x − 2!
2
y = 4x + 5
y + 3!
6x − y = 3
?
0 = −1 + 1
! 3.! A;! y =
( ( The!equation!in!standard!form!is( 2 x − 5 y = − 8. !
−y
1
y = − x +1
3
?
1
0 = − (3) + 1
3
0 = 0
! !
! ! So,!the!solution!is! (3,!0). !
−8
6x =
!
! ! Check:! (3, 0) !
+9
( ( The!equation!in!standard!form!is( 3x − 7 y = 9. !
!
1
y = − x +1
3
0
3x − 7 y =
!27.!
! 2.! C;! y = x − 3
− 7y
+9
?
−1 = −1 
! !
! ! So,!the!solution!is! (1, −1). !
= 194 + 105
!28.!
180°
− !107°
∠2 =
c = r + 105
(3r )
180°
?
− 3 = −1 − 2
y = 4x + 5
?
− 3 = 4( − 2) + 5
?
− 3 = −8 + 5
−3 = −3 
−3 = −3 
! !
! ! So,!the!solution!is! ( − 2, − 3). !
!
!
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Big(Ideas(Math(Blue(
Worked;Out*Solutions*
161
Chapter 5
4. y = 2 x − 3
6. 4 x + 2 y = 2
y = −x + 9
3x = 4 − y
y= −x+ 9
4x + 2y = 2
y
y
4
3
8
6
(4, 5)
4
2
1
x
O
3x = 4 − y
O
2 4 6 8
2 3 4 5 x
−2
−3
−4
y = 2x − 3
(3, − 5)
−5
Check: ( 4, 5)
y = 2x − 3
y = −x + 9
?
?
5 = 2( 4) − 3
5 = −4 + 9
Check: (3, − 5)
4x + 2 y = 2
3x = 4 − y
?
?
5 = 8−3
?
4(3) + 2( − 5) = 2
5 = 5
3(3) = 4 − ( − 5)
?
5 = 5
?
12 − 10 = 2
So, the solution is ( 4, 5).
2 = 2
y = −3x + 1
7. y = x − 8
y = 2 x − 14
y = − 3x + 1
(− 1, 4)
y
−1
y =
2 x − 14
x−8 =
2 x − 14
− 2x
6x + y = − 2
− 4− 3− 2
9 = 9
So, the solution is (3, − 5).
5. 6 x + y = − 2
5
4
9 = 4+5
− x − 8 = −14
1 2 x
−2
−3
?
6( −1) + 4 = − 2
?
−6 + 4 = − 2
−2 = −2
+8
+8
−x =
−6
(−1)(− x) = (−1)(− 6)
Check: ( −1, 4)
6x + y = −2
− 2x
x = 6
y = −3x + 1
?
4 = − 3( −1) + 1
?
4 = 3 +1
4 = 4
So, the solution is ( −1, 4).
y = x−8
= 6−8
= −2
So, the solution is (6, − 2).
Check: (6, − 2)
y = x−8
?
y = 2 x − 14
?
−2 = 6 − 8
− 2 = 2(6) − 14
−2 = −2
− 2 = 12 − 14
?
−2 = −2
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Chapter 5
Check: ( − 4, −1)
8. x = 2 y + 2
2x − 5 y = 1
x − 5y = 1
2x − 5 y = 1
8 − 9 = −1
1 =1
−1 = −1
10. a. x = number of movies rented
−4
y = amount owed
= ( −1)( −3)
y = 3
x = 2y + 2
y = 2 x + 15
Members
y = 3x
Nonmembers
b.
= 2(3) + 2
= 6+2
=8
So, the solution is (8, 3).
Check: (8, 3)
x = 2y + 2
2x − 5 y = 1
?
?
−4 + 5 = 1
1
− y = −3
( −1)( − y)
− 2( − 4) + 9( −1) = −1
?
4y + 4 − 5y = 1
−4
?
− 4 − 5( −1) = 1
2( 2 y + 2) − 5 y = 1
−y + 4 =
− 2 x + 9 y = −1
?
y = 2x + 15
y
45
40
35
30
25
20
15
10
5
0
(15, 45)
y = 3x
0 3 6 9 12 15 18 21 x
?
8 = 2(3) + 2
2(8) − 5(3) = 1
?
Check: (15, 45)
?
8 = 6+2
16 − 15 = 1
8 = 8
1 = 1
y = 2 x + 15
y = 3x
?
?
45 = 2(15) + 15
45 = 3(15)
?
9. x − 5 y = 1
45 = 30 + 15
45 = 45
− 2 x + 9 y = −1
45 = 45
x − 5y = 1
So, it is beneficial to have a membership when you
plan to rent more than 15 new release movies per year.
+ 5y
+ 5y
x = 5y + 1
11. x = the first number
− 2 x + 9 y = −1
y = the second number
− 2(5 y + 1) + 9 y = −1
x + y = 38
−10 y − 2 + 9 y = −1
y = x+8
− y − 2 = −1
+2
+2
−y =
(−1)(− y)
1
= ( −1)(1)
x + y = 38
x + ( x + 8) = 38
2 x + 8 = 38
−8
y = −1
x − 5y = 1
2x
30
=
2
2
x = 15
x − 5( −1) = 1
x+5 =
−5
1
−5
x = −4
So, the solution is ( − 4, −1).
−8
2 x = 30
y = x+8
= 15 + 8
= 23
So, the numbers are 15 and 23.
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163
Chapter 5
12.
Section 5.3
= length
w = width
5.3 Activity (pp. 216–217)
= 2w
1. a. 2 x + y = 4
2 + 2 w = 180
2 + 2w = 180
2( 2w) + 2w = 180
2x − y = 0
Method 1:
2x + y = 4
4 w + 2w = 180
−2x + y = 0
6w = 180
2y = 4
6w 180
=
6
6
w = 30
The result is 2 y = 4.
= 2w
You can solve for y and then substitute the answer
back into one of the equations to find x.
2y = 4
= 2(30)
= 60
So, the length is 60 feet and the width is 30 feet.
13. n = number of nurses
d = number of doctors
2y
4
=
2
2
y = 2
2x + y = 4
2x + 2 =
4
n + d = 77
−2
9d = 2 n
2x =
n + d = 77
2x
2
=
2
2
x = 1
−d
−d
n = −d + 77
−2
2
9d = 2n
The solution is (1, 2).
9d = 2( −d + 77)
Method 2:
9d = − 2d + 154
+ 2d
+ 2d
2x + y = 4
+ 2x − y = 0
= 4
4x
11d = 154
11d
154
=
11
11
d = 14
The result is 4 x = 4.
You can solve for x and then substitute the answer
back into one of the equations to find y.
n + d = 77
4x = 4
n + 14 =
4x
4
=
4
4
x =1
− 14
n =
77
− 14
63
So, there are 63 nurses and 14 doctors employed at the
hospital.
2x + y = 4
2(1) + y = 4
2 + y =
−2
4
−2
y =
2
The solution is (1, 2).
So, the solution is the same using both methods.
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Chapter 5
b. 3 x − y = 4
c. x + 2 y = 7
3x + y = 2
x − 2 y = −5
Method 1:
Method 1:
3x − y =
4
x + 2y = 7
− 3x − y = − 2
−x + 2y = 5
−2 y =
4 y = 12
2
The result is − 2 y = 2.
The result is 4 y = 12.
You can solve for y and then substitute the answer
back into one of the equations to find x.
You can solve for y and then substitute the answer
back into one of the equations to find x.
−2 y = 2
4 y = 12
−2 y
2
=
−2
−2
4y
12
=
4
4
y = 3
y = −1
x + 2(3) = 7
3x − y = 4
x+6 =
3 x − ( −1) = 4
3x + 1 =
−1
3x =
3
The solution is (1, −1).
Method 2:
−6
x =
−1
3x
3
=
3
3
x = 1
1
The solution is (1, 3).
Method 2:
x + 2y =
7
+ x − 2 y = −5
=
2x
2
The result is 2 x = 2.
3x − y = 4
You can solve for x and then substitute the answer
back into one of the equations to find y.
+ 3x + y = 2
= 6
6x
−6
4
7
2x = 2
The result is 6 x = 6.
You can solve for x and then substitute the answer
back into one of the equations to find y.
2x
2
=
2
2
x =1
6x = 6
x + 2y = 7
6x
6
=
6
6
x =1
1 + 2y =
−1
3(1) − y = 4
3− y =
4
−3
−y =
−1
2y =
3x − y = 4
−3
7
1
(−1)(− y) = (−1)(1)
6
2y
6
=
2
2
y = 3
The solution is (1, 3).
So, the solution is the same using both methods.
y = −1
The solution is (1, −1).
So, the solution is the same using both methods.
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165
Chapter 5
2. 2 x + y = 2
3. Box 1:
x + 5y = 1
2x + y = 0
a. no; You must first multiply one of the equations
x− y = 3
so that either x or y is eliminated when adding or
subtracting.
b. You can multiply each term by 5 so that the
y-coefficients are the same.
c. You can multiply each term by 2 so that the
Add the two equations.
2x + y = 0
+x− y = 3
= 3
3x
3x
3
=
3
3
x = 1
x-coefficients are the same.
d. To use the method in part (b), multiply Equation 1 by
5 and subtract Equation 2 from Equation 1.
10 x + 5 y = 10
2x + y = 0
− x − 5 y = −1
2(1) + y = 0
= 9
9x
2 + y =
9x
9
=
9
9
x = 1
−2
y = −2
The solution is (1, − 2), which corresponds to H.
2x + y = 2
Box 2:
2(1) + y = 2
2 + y =
−2
x+ y = 2
2
2x − 2 y = 4
−2
y =
Multiply the first equation by 2 and then subtract
Equation 2 from Equation 1.
0
The solution is (1, 0).
2x + 2 y =
To use the method in part (c), multiply Equation 2 by
2 and subtract Equation 2 from Equation 1.
2x +
0
−2
y =
2
−2x + 2 y = −4
4y =
0
4y
0
=
4
4
y = 0
− 2 x − 10 y = − 2
−9 y =
4
0
−9 y
0
=
−9
−9
x+ y = 2
x+0 = 2
y = 0
x = 2
2x + y = 2
The solution is ( 2, 0), which corresponds to Y.
2x + 0 = 2
2x = 2
2x
2
=
2
2
x =1
The solution is (1, 0).
So, the solution is the same using both methods.
e.
4
0
3
−2
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Chapter 5
Box 3:
Box 5:
3x + 3 y = 0
x+ y = 5
2 x − 2 y = −8
x− y =1
Multiply the second equation by
3
and then subtract
2
Equation 2 from Equation 1.
3
(2 x − 2 y = −8)
2
3x + 3 y =
3 x − 3 y = −12
0
−3 x + 3 y = 12
Subtract Equation 2 from Equation 1.
x+ y = 5
− x + y = −1
2y = 4
2y
4
=
2
2
y = 2
6 y = 12
x+ y = 5
6y
12
=
6
6
y = 2
x+ 2 =
3x + 3 y = 0
0
−6
−2
x =
3
The solution is (3, 2), which corresponds to T.
3x + 3( 2) = 0
3x + 6 =
−2
5
−6
Box 6:
x+ y = 4
3x = − 6
x − y = −2
−6
3x
=
3
3
x = −2
Subtract Equation 2 from Equation 1.
The solution is ( − 2, 2), which corresponds to P.
2 x + y = −3
x− y = 0
Add the two equations.
2 x + y = −3
0
= −3
3x
−x + y = 2
2y = 6
2y
6
=
2
2
y = 3
Box 4:
+x − y =
x+ y = 4
3x
−3
=
3
3
x = −1
x+ y = 4
x+3 =
−3
x =
4
−3
1
The solution is (1, 3), which corresponds to I.
x− y = 0
−1 − y =
+1
0
+1
−y =
(−1)(− y)
1
= ( −1)(1)
y = −1
The solution is ( −1, −1), which corresponds to A.
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167
Chapter 5
Box 7:
5.3 On Your Own (pp. 218–220)
x + 2y = 5
1. 2 x − y = 9
2 x − y = −5
4 x + y = 21
Multiply the first equation by 2 and then subtract
Equation 2 from Equation 1.
Add the two equations.
2 x + 4 y = 10
−2x + y =
5
2x − y =
x + 2y = 5
−6
6x
30
=
6
6
x = 5
2x − y = 9
2(5) − y = 9
x + 2(3) = 5
x+6 =
= 30
6x
5 y = 15
5y
15
=
5
5
y = 3
9
+ 4 x + y = 21
10 − y =
5
−6
x = −1
The solution is ( −1, 3), which corresponds to A.
So, the answer to the puzzle is HYPATIA.
4. You can use elimination to solve a system of linear
equations by adding or subtracting the two equations so
that you eliminate one of the variables. You may have to
multiply one or both equations by a constant first. After
eliminating a variable, solve for the remaining variable,
then substitute its value into one of the equations and
solve for the other variable.
− 10
9
− 10
−y =
(−1)(− y)
−1
= ( −1)( −1)
y = 1
So, the solution is (5, 1).
Check: (5, 1)
2x − y = 9
?
2(5) − 1 = 9
?
4 x + y = 21
?
4(5) + 1 = 21
?
10 − 1 = 9
20 + 1 = 21
9 = 9
21 = 21
5. You can add equations when the coefficient of one of the
variables in one equation is the negative of the coefficient
of that variable in the other equation.
Example: 5 x − 2 y = 4
x + 2y = 2
You can subtract equations when one variable has the
same coefficient in both equations.
Example: 2 x + 3 y = − 5
6x + 3y = 3
You have to multiply first when neither of the above is
true.
Example: 3 x + 4 y = −8
2 x − 2 y = 18
6. The Multiplication Property of Equality states that
multiplying both sides of an equation by the same
constant produces an equivalent equation.
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Chapter 5
Check: ( − 2, 0)
2. − 5 x + 2 y = 13
5 x + y = −1
3x + 4 y = − 6
Add the two equations.
−6 = −6
−14 = −14
Multiply the second equation by
1
(6 x + 3 y = 24)
3
2x + y = 8
3 x + y = 11
− 2 x − y = −8
x
Check: ( −1, 4)
5 x + y = −1
?
5( −1) + 4 = −1
− 5 + 4 = −1
13 = 13
−1 = −1
3. 3 x + 4 y = − 6
=
3
3 x + y = 11
3(3) + y = 11
9 + y = 11
−9
−9
y =
So, the solution is (3, 2).
Check: (3, 2)
3 x + y = 11
7 x + 4 y = −14
Subtract Equation 2 from Equation 1.
2
?
5 + 8 = 13
= −6
= 14
=
8
8
=
−4
= −2
1
and then subtract
3
Equation 2 from Equation 1.
So, the solution is ( −1, 4).
3x + 4 y
−7x − 4 y
− 4x
− 4x
−4
x
−14 + 0 = −14
6 x + 3 y = 24
−5x = 5
5
−5x
=
−5
−5
x = −1
?
?
−6 + 0 = −6
4. 3 x + y = 11
− 5 x + 2 y = 13
− 5 x + 2( 4) = 13
− 5 x + 8 = 13
−8
−8
− 5( −1) + 2( 4) = 13
7( − 2) + 4(0) = −14
?
3 y = 12
3y
12
=
3
3
y = 4
?
?
3( − 2) + 4(0) = − 6
−5 x + 2 y = 13
+ 5 x + y = −1
− 5 x + 2 y = 13
7 x + 4 y = −14
?
?
3(3) + 2 = 11
?
6 x + 3 y = 24
?
6(3) + 3( 2) = 24
?
9 + 2 = 11
18 + 6 = 24
11 = 11
24 = 24
3x + 4 y = −6
3( − 2) + 4 y = − 6
−6 + 4 y = −6
+6
+6
4y = 0
4y
0
=
4
4
y = 0
So, the solution is ( − 2, 0).
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169
Chapter 5
Check: (0, 3)
5. 4 x − 5 y = −19
−x − 2 y = 8
5 y = 15 − 5 x
Multiply the second equation by 4 and then add the two
equations.
4x − 5 y
−4x − 8 y
−13 y
−13 y
−13
y
15 = 15
3 = 3
5 x + 6 y = 185
To find the cost x of each peony, eliminate the y-terms.
Multiply the first equation by 2. Multiply the second
equation by 3. Then subtract Equation 2 from Equation 1.
8 x + 18 y =
380
−15 x − 18 y = −555
4x
− 24
=
4
4
x = −6
−7 x
= −175
−7 x
−175
=
−7
−7
So, the solution is ( − 6, −1).
x = 25
Check: ( − 6, −1)
4 x − 5 y = −19
−x − 2y = 8
?
5.3 Exercises (pp. 221–223)
?
Vocabulary and Concept Check
6+2 = 8
−19 = −19
8 = 8
6. 5 y = 15 − 5 x
y = −2x + 3
Multiply the second equation by 5 and then subtract
Equation 2 from Equation 1.
= 15 − 5 x
= −15 + 10 x
5x
=
5x
=
5
= x
So, the cost of each peony is $25.
?
− ( − 6) − 2( −1) = 8
− 24 + 5 = −19
=
=
=
=
3 = 0 +3
4 x + 9 y = 190
4 x = − 24
5y
5y
5y
5y
?
15 = 15 − 0
y = cost of a geranium
4 x − 5(−1) = −19
4 x + 5 = −19
−5
−5
5y
−5 y
0
0
5
0
3 = − 2(0) + 3
7. x = cost of a peony
4 x − 5 y = −19
?
?
5(3) = 15 − 5(0)
?
= −19
= 32
= 13
13
=
−13
= −1
4( − 6) − 5( −1) = −19
y = −2x + 3
?
1. If necessary, multiply one or both equations by a constant
so at least one pair of like terms has the same or opposite
coordinates. Add or subtract the equations to eliminate
one of the variables. Solve the resulting equation for the
remaining variable. Substitute the value back into one of
the original equations and solve for the other variable.
2. You should use multiplication when it is not possible to
eliminate a variable by addition or subtraction.
3. 2 x + 3 y = 11
3 x − 2 y = 10
This system is the only one that requires multiplication
by a constant in order to find the solution.
15 − 5 x
15 − 5(0)
15 − 0
15
5y
15
=
5
5
y = 3
So, the solution is (0, 3).
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Chapter 5
Practice and Problem Solving
6. 3 x + 2 y = 3
3x − 2 y = −9
4. x + y = 3
x− y = 1
Subtract Equation 2 from Equation 1.
Add the two equations.
x+ y = 3
3x + 2 y =
3
−3x + 2 y =
9
+x− y = 1
4 y = 12
= 4
4y
12
=
4
4
y = 3
2x
2x
4
=
2
2
x = 2
3x + 2 y = 3
x+ y = 3
2+ y =
3x + 2(3) = 3
3
−2
3x + 6 =
−2
y =
−6
1
−3
3x
=
3
3
x = −1
5. − x + 3 y = 0
x + 3 y = 12
Add the two equations.
0
− x − y = −3
6 y = 12
Add the two equations.
6y
12
=
6
6
y = 2
x + 3y =
2y =
− x + 3( 2) = 0
0
−6
− x = −6
(−1)(− x)
5
− x − y = −3
− x + 3y = 0
−6
So, the solution is ( −1, 3).
7. x + 3 y = 5
+ x + 3 y = 12
−x + 6 =
−6
3x = −3
So, the solution is ( 2, 1).
− x + 3y =
3
= ( −1)( − 6)
x = 6
So, the solution is (6, 2).
2
2y
2
=
2
2
y = 1
x + 3y = 5
x + 3(1) = 5
x+3 =
−3
5
−3
x =
2
So, the solution is ( 2, 1).
Check:
x + 3y = 5
?
2 + 3(1) = 5
?
2 +3 = 5
− x − y = −3
?
− 2 − 1 = −3
−3 = −3
5 = 5
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171
Chapter 5
Check: (1, − 3)
8. x − 2 y = − 7
3x + 2 y = 3
4 x + 3 y = −5
Add the two equations.
4x
−1 + 3( − 3) = −10
?
?
3
4 − 9 = −5
−1 − 9 = −10
= −4
−5 = −5
−10 = −10
4x
−4
=
4
4
x = −1
10. 2 x + 7 y = 1
2 x − 4 y = 12
Subtract Equation 2 from Equation 1.
x − 2 y = −7
2x + 7 y =
−1 − 2 y = − 7
1
+1
− 2 x + 4 y = −12
− 2 y = −6
11y = −11
−2 y
−6
=
−2
−2
−11
11y
=
11
11
y = −1
+1
y = 3
2x + 7 y = 1
So, the solution is ( −1, 3).
2 x + 7( −1) = 1
Check: ( −1, 3)
2x − 7 =
x − 2 y = −7
?
−1 − 2(3) = − 7
3x + 2 y = 3
?
3( −1) + 2(3) = 3
?
?
−1 − 6 = − 7
−3 + 6 = 3
−7 = −7
3 = 3
9. 4 x + 3 y = − 5
− x + 3 y = −10
Subtract Equation 2 from Equation 1.
4x + 3 y =
−5
x − 3y =
10
=
5
5x
?
4(1) + 3( − 3) = − 5
x − 2 y = −7
+ 3x + 2 y =
− x + 3 y = −10
?
5x
5
=
5
5
x = 1
+7
1
+7
2x =
8
2x
8
=
2
2
x = 4
So, the solution is ( 4, −1).
Check: ( 4, −1)
2x + 7 y = 1
?
2( 4) + 7( −1) = 1
?
2 x − 4 y = 12
?
2( 4) − 4( −1) = 12
?
8−7 =1
8 + 4 = 12
1 =1
12 = 12
4 x + 3 y = −5
4(1) + 3 y = − 5
4 + 3 y = −5
−4
−4
3 y = −9
−9
3y
=
3
3
y = −3
So, the solution is (1, − 3).
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Chapter 5
Check: ( − 2, − 5)
11. 2 x + 5 y = 16
3x − 5 y = −1
3x − 2 y = 4
Add the two equations.
6x − 2 y = − 2
?
?
3( − 2) − 2( − 5) = 4
2 x + 5 y = 16
6( − 2) − 2( − 5) = − 2
?
?
+ 3 x − 5 y = −1
− 6 + 10 = 4
−12 + 10 = − 2
= 15
4 = 4
−2 = −2
5x
5x
15
=
5
5
x = 3
13. The student subtracted the x-terms and the constant
terms, but added the y-terms.
Add the two equations.
2 x + 5 y = 16
5x + 2 y = 9
2(3) + 5 y = 16
+ 3x − 2 y = −1
6 + 5 y = 16
−6
= 8
8x
−6
8x
8
=
8
8
x = 1
5 y = 10
5y
10
=
5
5
y = 2
5x + 2 y = 9
5(1) + 2 y = 9
So, the solution is (3, 2).
5 + 2y =
Check: (3, 2)
−5
2 x + 5 y = 16
?
2(3) + 5( 2) = 16
?
?
9 − 10 = −1
16 = 16
−1 = −1
So, the solution is (1, 2).
14. a. x = number of raffle tickets you sell
12. 3 x − 2 y = 4
y = number of raffle tickets your friend sells
6x − 2 y = −2
Subtract Equation 2 from Equation 1.
x − y = 14
x + y = 58
3x − 2 y = 4
−6x + 2 y = 2
b. Add the two equations.
= 6
x − y = 14
− 3x
6
=
−3
−3
+ x + y = 58
x = −2
2x
72
=
2
2
x = 36
= 72
2x
3x − 2 y = 4
3( − 2) − 2 y = 4
−6 − 2 y =
+6
4
2y
4
=
2
2
y = 2
3(3) − 5( 2) = −1
?
−3x
2y =
3 x − 5 y = −1
6 + 10 = 16
9
−5
4
+6
− 2 y = 10
−2 y
10
=
−2
−2
y = −5
x + y = 58
36 + y =
− 36
58
− 36
y =
22
So, you sell 36 raffle tickets and your friend sells
22 raffle tickets.
So, the solution is ( − 2, − 5).
Copyright*©*Big*Ideas*Learning,*LLC*
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Big Ideas Math Blue
Worked;Out*Solutions*
173
Chapter 5
15. a. x = time it takes to jog around the block
17. x + 4 y = 1
y = time it takes to jog around the park
3 x + 5 y = 10
2 x + y = 10
Multiply the first equation by − 3 and then add the two
2 x + 3 y = 22
equations.
b. To find the time y it takes to jog around the park,
eliminate the x-terms. Subtract Equation 2 from
Equation 1.
2x + y =
10
− 3 x − 12 y = − 3
+ 3 x + 5 y = 10
−7 y =
7
− 2 x − 3 y = − 22
−7 y
7
=
−7
−7
− 2 y = −12
y = −1
−2 y
−12
=
−2
−2
x + 4y = 1
x + 4( −1) = 1
y = 6
So, it takes 6 minutes for you to jog around the park.
x−4 =
+4
1
+4
x =
16. 2 x − y = 0
5
3x − 2 y = −3
So, the solution is (5, −1).
Multiply the first equation by − 2 and then add the two
Check: (5, −1)
equations.
−4x + 2 y =
x + 4y = 1
0
?
5 + 4( −1) = 1
+ 3x − 2 y = −3
−x
= −3
(−1)(− x)
= ( −1)( −3)
?
3 x + 5 y = 10
?
3(5) + 5( −1) = 10
?
5−4 = 1
15 − 5 = 10
1 = 1
10 = 10
x = 3
2x − y = 0
2(3) − y = 0
6− y =
−6
0
−6
− y = −6
(−1)(− y)
= ( −1)( − 6)
y = 6
So, the solution is (3, 6).
Check: (3, 6)
2x − y = 0
?
2(3) − 6 = 0
?
3x − 2 y = −3
?
3(3) − 2(6) = − 3
9 − 12 = − 3
0 = 0
−3 = −3
174* Big Ideas Math Blue*
*
?
6 −6 = 0
Worked;Out*Solutions*
Copyright*©*Big*Ideas*Learning,*LLC*
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Chapter 5
Check: ( − 2, −1)
18. − 2 x + 3 y = 7
5x + 8 y = − 2
3x + 3 = 3 y
Multiply the first equation by 5. Multiply the second
equation by 2. Then add the two equations.
2x − 6 y = 2
?
?
3( − 2) + 3 = 3( −1)
2( − 2) − 6( −1) = 2
?
?
−10 x + 15 y = 35
− 6 + 3 = −3
−4 + 6 = 2
+ 10 x + 16 y = − 4
−3 = −3
2 = 2
31 y = 31
20. 2 x − 6 = 4 y
31 y
31
=
31
31
y = 1
7 y = −3x + 9
Rewrite both equations.
−2x + 3y = 7
2x − 4 y = 6
− 2 x + 3(1) = 7
3x + 7 y = 9
−2x + 3 =
−3
7
Multiply the first equation by 3. Multiply the second
equation by − 2. Then add the two equations.
−3
−2x =
4
6 x − 12 y =
18
−2x
4
=
−2
−2
− 6 x − 14 y = −18
x = −2
− 26 y
0
=
− 26
− 26
− 26 y =
So, the solution is ( − 2, 1).
0
y = 0
Check: ( − 2, 1)
−2x + 3 y = 7
?
− 2( − 2) + 3(1) = 7
5x + 8 y = − 2
?
5( − 2) + 8(1) = − 2
?
?
4 +3 = 7
−10 + 8 = − 2
7 = 7
−2 = −2
19. 3 x + 3 = 3 y
2x − 6 y = 2
2x − 6 = 4 y
2 x − 6 = 4(0)
2x − 6 =
0
+6
+6
2x =
6
2x
6
=
2
2
x = 3
Rewrite the first equation.
So, the solution is (3, 0).
3x − 3 y = − 3
Check: (3, 0)
Multiply the first equation by − 2 and then add the two
equations.
2x − 6 = 4 y
?
−6x + 6 y = 6
2(3) − 6 = 4(0)
+ 2x − 6 y = 2
6 −6 = 0
− 4x
= 8
?
0 = 0
7 y = −3x + 9
?
7(0) = − 3(3) + 9
?
0 = −9 + 9
0 = 0
− 4x
8
=
−4
−4
x = −2
3x + 3 = 3 y
3( − 2) + 3 = 3 y
−6 + 3 = 3 y
−3 = 3 y
−3
3y
=
3
3
−1 = y
So, the solution is ( − 2, −1).
Copyright*©*Big*Ideas*Learning,*LLC*
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Big Ideas Math Blue
Worked;Out*Solutions*
175
Chapter 5
21. 5 x = 4 y + 8
23. a. 4 x − y = 3
3 y = 3x − 3
ax + 10 y = 6
Rewrite both equations.
To solve the system by elimination, the value of a
should be the same or the opposite of the x-coefficient
in the first equation, 4. So, the value of a should be
− 4 or 4.
5x − 4 y = 8
−3x + 3 y = −3
Multiply the first equation by 3. Multiply the second
equation by 4. Then add the two equations.
15 x − 12 y =
24
=
12
3x
12
=
3
3
x = 4
24. Line 1: ( − 2, 1), ( 2, 7 )
5x = 4 y + 8
m =
5( 4) = 4 y + 8
20 = 4 y + 8
−8
− 6 x + by = 9
To solve the system by elimination, the value of b
should be the same or opposite of the y-coefficient in
the first equation, 7. So, the value of b should be
− 7 or 7.
−12 x + 12 y = −12
3x
b. x − 7 y = 6
y2 − y1
7 −1
6
3
=
=
=
x2 − x1
2 − ( − 2)
4
2
Line 2: ( − 4, −1), (0, 5)
−8
5 − ( −1)
y2 − y1
6
3
=
=
=
x2 − x1
0 − ( − 4)
4
2
12 = 4 y
m =
12
4y
=
4
4
3 = y
The lines do not intersect because the lines are parallel.
25. Line 1: (3, − 2), (7, −1)
So, the solution is ( 4, 3).
m =
Check: ( 4, 3)
5x = 4 y + 8
?
5( 4) = 4(3) + 8
?
3 y = 3x − 3
Line 2: (5, 2), (6, − 2)
?
3(3) = 3( 4) − 3
?
20 = 12 + 8
9 = 12 − 3
20 = 20
9 = 9
22. The y-term was multiplied by 5 instead of −5.
x+ y =1
−1 − ( − 2)
y2 − y1
1
=
=
x2 − x1
7 −3
4
m =
y2 − y1
−2 − 2
−4
=
=
= −4
x2 − x1
6−5
1
Because the product of the slopes of the lines is
1
(− 4) = −1, the lines are perpendicular. The lines
4
intersect because perpendicular lines intersect at right
angles.
5 x + 3 y = −3
Multiply the first equation by − 5 and then add the two
equations.
−5 x − 5 y = −5
+ 5 x + 3 y = −3
− 2 y = −8
−2 y
−8
=
−2
−2
y = 4
x+ y = 1
x+ 4 =
−4
1
−4
x = −3
So, the solution is ( − 3, 4).
176* Big Ideas Math Blue*
*
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Chapter 5
26. Blue plane: ( 2, 4), (6, 12)
m =
y2 − y1
12 − 4
8
=
=
= 2
x2 − x1
6−2
4
y − y1 = m( x − x1 )
y − 4 = 2( x − 2)
y − 4 = 2x − 4
+4
+4
27. a.
x = number of points a multiple choice question is worth
y = number of points a short response question is worth
23 x + 10 y = 86
28 x + 5 y = 76
b. Multiply the second equation by − 2 and then add the
two equations.
23 x + 10 y =
y = 2x
Red plane: (15, 9), (6, 12)
− 56 x − 10 y = −152
− 33 x
y − y1
12 − 9
3
1
m = 2
=
=
= −
6 − 15
3
x2 − x1
−9
m( x − x1 )
1
y − 9 = − ( x − 15)
3
1
y−9 = − x+5
3
+9
+9
The system is: y = 2 x
1
y = − x + 14
3
Subtract Equation 2 from Equation 1.
y = 2x + 0
1
− y = x − 14
3
7
0 = x − 14
3
+ 14
+ 14
= − 66
− 33 x
− 66
=
− 33
− 33
y − y1 =
1
y = − x + 14
3
86
x = 2
23x + 10 y = 86
23( 2) + 10 y = 86
46 + 10 y =
− 46
86
− 46
10 y =
40
10 y
40
=
10
10
y = 4
So, multiple choice questions are worth 2 points each
and short response questions are worth 4 points each.
28. no; If x = − 6, then there would be a negative number of
adult tickets sold, which is impossible.
7
x
3
3
3 7
x
(14) =
7
7 3
14 =
6 = x
y = 2x
= 2(6)
= 12
So, the solution is (6, 12), which is the location of the
airport.
Copyright*©*Big*Ideas*Learning,*LLC*
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Big Ideas Math Blue
Worked;Out*Solutions*
177
Chapter 5
29. x = cost per hour to go parasailing
30. Sample answer: You could choose coefficients for x
and y. Substitute 2 for x and − 4 for y, and add to get
y = cost per hour to go horseback riding
2 x + 5 y = 205
a constant.
3 x + 3 y = 240
Example: 2( 2) + 3( − 4) = − 8
Multiply the first equation by 3. Multiply the second
equation by − 2. Then add the two equations.
So, the second equation could be 2 x + 3 y = −8.
6 x + 15 y
−6 x − 6 y
9y
9y
9
y
= 615
= − 480
= 135
135
=
9
= 15
2x + y = 0
2 x + 3 y = −8
Subtract Equation 2 from Equation 1.
2x + y = 0
−2x − 3y = 8
−2 y = 8
−2 y
8
=
−2
−2
y = −4
2 x + 5 y = 205
2 x + 5(15) = 205
2 x + 75 = 205
− 75
− 75
2 x = 130
2x
130
=
2
2
x = 65
2x + y = 0
2 x + ( − 4) = 0
2x − 4 = 0
+4
+4
2x = 4
2x
4
=
2
2
x = 2
You want to parasail for 1 hour and horseback ride for
2 hours.
x + 2y
65 + 2(15)
65 + 30
95
=
=
=
=
c
c
c
c
So, the cost is $95.
The solution is ( 2, − 4), so it is a system of linear
equations.
31. x = the amount in grams of the 90% gold alloy
to be used
y = the amount in grams of the 50% gold alloy
to be used
x+ y =8
0.9 x + 0.5 y = 0.75(8) = 6
Multiply the second equation by − 2 and then add the
equations.
x+ y =
8
−1.8 x − y = −12
− 0.8 x
= −4
− 0.8 x
−4
=
− 0.8
− 0.8
x = 5
x+ y = 8
5+ y =
−5
8
−5
y =
3
So, 5 grams of the 90% gold alloy and 3 grams of the
50% alloy should be used.
178* Big Ideas Math Blue*
*
Worked;Out*Solutions*
Copyright*©*Big*Ideas*Learning,*LLC*
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Chapter 5
33. 2 x − y + 3 z = −1
32. x = speed of the power boat
x + 2 y − 4 z = −1
y = speed of the current
y − 2z = 0
Use the distance formula, d = rt , to write the system of
equations.
Multiply the second equation by − 2 and then add
Downstream:
Equations 1 and 2.
d = 10 mi, r = ( x + y) mi/h, t = 30 min =
1
h
2
Upstream:
d = 10 mi, r = ( x − y) mi/h, t = 50 min =
1
( x + y) = 10
2
5
( x − y) = 10
6
1
x+
2
5
x−
6
5
h
6
1
y = 10
2
5
y = 10
6
3
and then add the two
Multiply the second equation by
5
equations.
3 5
5
x − y = 10
5 6
6
1
x+
2
1
+ x−
2
x
1
1
x− y = 6
2
2
1
y = 10
2
1
y = 6
2
= 16
1
1
x + y = 10
2
2
1
1
(16) + y = 10
2
2
1
8 + y = 10
2
−8
−8
1
y = 2
2
1
2 y = 2( 2)
2
y = 4
So, the speed of the current is 4 miles per hour.
2x −
y + 3 z = −1
− 2 x − 4 y + 8z = 2
− 5 y + 11z =
1
Multiply the third equation by 5 and then add Equation 3
and the result from above.
5 y − 10 z = 0
−5 y + 11z = 1
z =1
y − 2z = 0
y − 2(1) = 0
y−2 =
0
+2 = +2
y =
2
2 x − y + 3 z = −1
2 x − 2 + 3(1) = −1
2 x − 2 + 3 = −1
2 x + 1 = −1
−1
−1
2x = −2
2x
−2
=
2
2
x = −1
So, the solution is ( −1, 2, 1).
Fair Game Review
34. 4 n + 1 = n − 8
3n = − 9
4n + 1 =
−n
n −8
−n
3n + 1 = −8
−1
−1
3n = − 9
−9
3n
=
3
3
n = −3
3n = −9
−9
3n
=
3
3
n = −3
So, the equations are equivalent.
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Big Ideas Math Blue
Worked;Out*Solutions*
179
Chapter 5
35. 2 a + 6 = 12
c. no; This means that your cousin will always be
three years older than you.
a +3 = 6
2a + 6 = 12
−6
a +3 =
−6
2a =
6
−3
−3
a =
6
2. a. C = 15 x + 500
R = 15 x
3
2a
6
=
2
2
a = 3
So, the equations are equivalent.
3
= 5
2
14v − 3 = 15
36. 7v −
3
= 5
2
3
3
+
+
2
2
7v −
0
1
2
3
4
5
C
500
515
530
545
560
575
R
0
15
30
45
60
75
x
6
7
8
9
10
C
590
605
620
635
650
R
90
105
120
135
150
14v − 3 = 15
+3
+3
b. You will never break even. Because you are selling
the backpacks for the same amount it costs you to
make them, you will never recover your investment.
You need to sell each backpack for more than $15.
14v = 18
14v
18
=
14
14
9
v =
7
13
2
1
1 13
(7v) =
7
7 2
7v =
v =
x
3. a.
y
5
4
y = 2x + 4
13
14
2
1
So, the equations are not equivalent.
−5 −4 −3
−1
−1
3y − 6x = 12
37. D; 2 y − x = 7
+x
−4
−5
+x
2y = x + 7
b. yes; The two lines intersect at every point because
2y
x 7
=
+
2
2
2
1
7
y = x+
2
2
they are the same line.
c. Any point on the line y = 2 x + 4 is a solution of the
puzzle.
d. y = 2 x + 4
1
.
2
So, the slope is
1 2 3 4 5 x
Section 5.4
x
0
1
2
3
4
5
y
4
6
8
10
12
14
x
6
7
8
9
10
y
16
18
20
22
24
5.4 Activity (pp. 224–225)
1. a.
y
12
y=t +3
10
e. Yes, because the equations represent the same line.
y=t
8
f. There are many solutions. The solution of the system
6
is all points on the line y = 2 x + 4.
4
2
0
0
2
4
6
8
10
12 t
b. The vertical distance is 3. It represents the difference
in age between you and your cousin.
180* Big Ideas Math Blue*
*
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Chapter 5
4. A system of linear equations has no solution when the
two lines are parallel because parallel lines do not
intersect. A system of linear equations has many
solutions when the equations are the same when written
in slope-intercept form.
Sample answer: The system of equations consisting of
y = x + 4 and y = x + 6 has no solution because the
lines are parallel, as you can see from the graph.
y
12
11 y = x + 6
10
2. y = − 5 x − 2
5x + y = 0
5x + y = 0
5 x + ( − 5 x − 2) = 0
5x − 5x − 2 = 0
−2 = 0
The equation − 2 = 0 is never true. So, the system of
linear equations has no solution.
Check:
9
8
7
6
4
y
y=x+4
5
4
3
2
1
0
y = −5x − 2
−2 −1
1 2 x
−3
5x + y = 0
0 1 2 3 4 5 6 7 8 9 x
The system of equations consisting of y = 2 x + 7 and
2 y − 4 x = 14 has many solutions because they are the
same line, as you can see from the graph.
2 x + 3 y = −1
5
4
3
2( 2 y + 10) + 3 y = −1
4 y + 20 + 3 y = −1
2
1
−6 −5
3. x = 2 y + 10
2 x + 3 y = −1
y
y = 2x + 7 7
2y − 4x = 14
The lines are parallel.
7 y + 20 =
−3 −2 −1
−1
1 2 x
− 20
−1
− 20
7 y = − 21
7y
− 21
=
7
7
y = −3
5.4 On Your Own (pp. 226–227)
1. y = − x + 3
y = −x + 5
x = 2 y + 10
y = −x + 3
x = 2( − 3) + 10
−x + 5 = −x + 3
+x
x = − 6 + 10
+x
x = 4
5 = 3
So, the solution is ( 4, − 3).
The equation 5 = 3 is never true. So, the system of
linear equations has no solution.
Check: ( 4, − 3)
Check:
x = 2 y + 10
y = −x + 5
6
?
y
4 = 2( − 3) + 10
?
4
4 = − 6 + 10
4 = 4
y = −x + 3
2 x + 3 y = −1
?
2( 4) + 3( − 3) = −1
?
8 − 9 = −1
−1 = −1
1
−2 −1
1 2 x
The lines are parallel.
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Big Ideas Math Blue
Worked;Out*Solutions*
181
Chapter 5
4. x + y = 3
6. 2 x − 4 y = 10
x − y = −3
−12 x + 24 y = − 60
Add the two equations.
Multiply the first equation by 6 and add the two
equations.
x+ y =
3
12 x − 24 y =
+ x − y = −3
=
2x
0
0 =
2x
0
=
2
2
x = 0
0
x+ y = 3
The equation 0 = 0 is always true. So, the solutions are
all points on the line 2 x − 4 y = 10. The system of
linear equations has infinitely many solutions.
0 + y = 3
Check:
y = 3
2x − 4y = 10
1
So, the solution is (0, 3).
y
O
1 2 3
5 x
−2
Check: (0, 3)
x+ y = 3
x − y = −3
?
−12x + 24y = −60
?
0 +3 = 3
0 − 3 = −3
3 = 3
−3 = −3
4x + 2 y = 0
Multiply the first equation by − 2 and add the two
equations.
12 x + 24 y = 162
The equation 0 = 54 is never true. So, the system of
linear equations has no solution.
The equation 0 = −10 is never true. So, the system of
linear equations has no solution.
5.4 Exercises (pp. 228–229)
Vocabulary and Concept Check
y
2x + y = 5
1. The graph of a system of linear equations with no
solution is two parallel lines, and the graph of a system of
linear equations with infinitely many solutions is one line
because the equations in the system have the same
y-intercept and slope.
3
2
O
Multiply Equation 1 by 3 and subtract Equation 2 from
Equation 1.
0 = 54
0
0 = −10
5
4
7. 4 x + 8 y = 54
−12 x − 24 y = −108
− 4 x − 2 y = −10
+ 4x + 2 y =
Check:
Both equations are the same line.
12 x + 24 y = 108
5. 2 x + y = 5
1 2
4 x
2. When solving a system of linear equations algebraically,
−2
you know the system has no solution when you reach an
invalid statement such as − 7 = 2.
−3
−4
4x + 2y = 0
The lines are parallel.
182* Big Ideas Math Blue*
*
60
−12 x + 24 y = − 60
Worked;Out*Solutions*
When solving a system of linear equations algebraically,
you know the system has infinitely many solutions when
you reach a valid statement such as 2 = 2.
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Chapter 5
Practice and Problem Solving
Subtract Equation 2 from Equation 1.
y =
3 y − 12 x = 1
3 4x +
8. y = 2 x − 2
y = 2x + 9
1
3. y = 4 x +
3
3 y − 12 x = 1
1
− 12 x = 1
3
12 x + 1 − 12 x = 1
1 =1
2x − 2
− y = − 2x − 9
0 =
−11
The equation 0 = −11 is never true. So, the system of
linear equations has no solution.
Check:
The equation 1 = 1 is always true. So, the solutions are
1
all points on the line y = 4 x + . The system of linear
3
equations has infinitely many solutions.
4.
y
10
9
y = 2x + 9
5
4
1
x+3 = y
2
x = 2y + 6
3
2
1
−4 −3 −2 −1
x = 2y + 6
−6
7
6
2 x
y = 2x − 2
−6
x − 6 = 2y
The lines are parallel.
x 6
2y
− =
2
2
2
1
x−3= y
2
9. y = 3 x + 1
− x + 2 y = −3
The system of linear equations has no solution because the
lines have the same slope but different y-intercepts.
− x + 2 y = −3
− x + 2(3x + 1) = −3
− x + 6 x + 2 = −3
5. y = 5 x − 9
5 x + 2 = −3
y = 5x + 9
−2
The system of linear equations has no solution because
the lines have the same slope but different y-intercepts.
5 x = −5
5x
−5
=
5
5
x = −1
6. y = 6 x + 2
y = 3x + 1
The system of linear equations has one solution because
the lines have different slopes.
y = 3x + 1
= 3( −1) + 1
= −3 + 1
7. y = 8 x − 2
y − 8x = − 2
−2
y = 8x − 2
The system of linear equations has infinitely many
solutions because both equations are the same line.
= −2
So, the solution is ( −1, − 2).
Check: ( −1, − 2)
y = 3x + 1
?
− 2 = 3( −1) + 1
?
− 2 = −3 + 1
−2 = −2
Copyright*©*Big*Ideas*Learning,*LLC*
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− x + 2 y = −3
?
−( −1) + 2( − 2) = − 3
?
1 − 4 = −3
−3 = −3
Big Ideas Math Blue
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183
Chapter 5
π
x+π
3
−π x + 3 y = − 6π
10. y =
12.
−π x + 3 y = − 6π
−π x + 3
π
3
x+π
= − 6π
1
x+ y =1
3
2x + 6 y = 6
Multiply the first equation by −6 and add the two
equations.
− 2x − 6 y = −6
−π x + π x + 3π = − 6π
+ 2x + 6 y =
6
3π = − 6π
0 =
0
The equation 3π = − 6π is never true. So, the system of
linear equations has no solution.
Check:
p
3
y
4
y = x +p
The equation 0 = 0 is always true. So, the solutions are
1
all points on the line x + y = 1. The system of linear
3
equations has infinitely many solutions.
Check:
2x + 6y = 6
2
2
1
−5 −4
−2 −1
y
x
1 2 3 4 5 6 7 x
O
−2
−3
1 2 3
1
x+y=1
3
−4
−5
− p x + 3y = −6p
−6
The equations represent the same line.
The lines are parallel.
1
6
x + 6 y = 30
11. y = − x + 5
x + 6 y = 30
1
x + 6 − x + 5 = 30
6
x − x + 30 = 30
30 = 30
The equation 30 = 30 is always true. So, the solutions
1
are all points on the line y = − x + 5. The system of
6
linear equations has infinitely many solutions.
Check:
1
6
y=− x+5
6
y
4
3 x + 6y = 30
2
1
−1
1 2 3 4 x
The equations represent the same line.
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Chapter 5
13. − 2 x + y = 1.3
17. When two equations have the same slope and the same
y-intercept, they represent the same line, meaning they
have infinitely many solutions.
2(0.5 x − y) = 4.6
Rewrite the second equation.
x − 2 y = 4.6
Multiply the second equation by 2 and add the two
equations.
−2x +
y =
1.3
+ 2x − 4 y =
9.2
8 x + 16 y = 128
−3 y
10.5
=
−3
−3
a.
y = − 3.5
− 2 x + y = 1.3
32 + 16 y = 128
− 32
48
8y
48
=
8
8
y = 6
− 2x
4.8
=
−2
−2
− 32
16 y =
96
16 y
96
=
16
16
y = 6
So, you and your friend work 6 hours at the second
job.
x = − 2.4
b. Because Equation 2 is double Equation 1, if you and
So, the solution is ( − 2.4, − 3.5).
Check: ( − 2.4, − 3.5)
2(0.5 x − y) = 4.6
− 2 x + y = 1.3
− 2( − 2.4) + ( − 3.5) = 1.3
64
− 16
8y =
4.8
?
8 x + 16 y = 128
8( 4) + 16 y = 128
− 16
1.3
+ 3.5
−2x =
4 x + 8 y = 64
4( 4) + 8 y = 64
16 + 8 y =
− 2 x + ( − 3.5) = 1.3
+ 3.5
When two equations have different slopes, they
will always intersect exactly once, no matter their
y-intercepts.
18. 4 x + 8 y = 64
−3 y = 10.5
− 2 x − 3.5 =
When two equations have the same slope but different
y-intercepts, they represent parallel lines and have no
solution because the lines will never intersect.
?
2 0.5( − 2.4) − ( − 3.5) = 4.6
?
4.8 − 3.5 = 1.3
1.3 = 1.3
?
2( −1.2 + 3.5) = 4.6
?
2( 2.3) = 4.6
4.6 = 4.6
your friend each work the same number of hours at the
second job, then you will also each work the same
number of hours at the first job.
19. x = number of songs downloaded
y = total cost of the music
y = 0.99 x + 10
y = 0.99 x
no; Because you paid $10 before buying the same
number of songs at the same price you spend $10 more.
14. The student correctly identified the lines having the same
slope, but failed to recognize that they have different
y-intercepts. This indicates the lines are parallel and there
is no solution.
15. y = 2 x + 3
y = 2x
Sample answer:
no; The pigs are running at the same speed (the slopes of
the lines are the same), so your pig will always be
3 feet ahead of your friend’s pig (the y-intercepts are
different).
16. The system of linear equations will have exactly one
solution because the lines have different slopes and will
intersect in one point.
Copyright*©*Big*Ideas*Learning,*LLC*
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20. y = ax + 1
y = bx + 4
always; When a = b, the system of linear equations
always has no solution because the lines are parallel.
The slopes are the same, but the y-intercepts are different.
sometimes; When a ≥ b, the system of linear equations
sometimes has no solution. There is no solution when
a = b because the lines are parallel. When a > b, there
is one solution because the equations have different slopes
and will intersect.
never; When a < b, the system of linear equations will
never have no solution. There will always be one solution
because the equations have different slopes and will
intersect.
Big Ideas Math Blue
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185
Chapter 5
21. x = cost of a lift ticket
Check: ( 2, 3) when a = 2, b = 2:
y = cost of a ski rental
36 x + 18 y = 684
12 x − 2by = 12
?
12( 2) − 2( 2)(3) = 12
24 x + 12 y = 456
?
24 − 12 = 12
Multiply Equation 1 by 2. Multiply Equation 2 by 3.
Then subtract Equation 2 from Equation 1.
12 = 12
72 x + 36 y = 1368
3ax − by = 6
− 72 x − 36 y = −1368
?
3( 2)( 2) − ( 2)(3) = 6
0 = 0
?
The equation 0 = 0 is always true. The system of linear
equations has infinitely many solutions. So, it is not
possible to determine how much each lift ticket costs.
22. 12 x − 2by = 12
12 − 6 = 6
6 = 6
So, the solution is ( 2, 3) when a = 2 and b = 2.
To determine if the system has any other solutions,
substitute a = 2 and b = 2 into the system and solve.
3ax − by = 6
Substitute x = 2 and y = 3 in the first equation to find
the value of b.
12 x − 4 y = 12
6x − 2 y = 6
12 x − 2by =
12
12( 2) − 2b(3) =
12
24 − 6b =
12
12 x − 4 y = 12
− 24
− 12 x + 4 y = 12
− 24
Multiply Equation 2 by 2 and then subtract Equation 2
from Equation 1.
0 = 0
− 6b = −12
− 6b
−12
=
−6
−6
b = 2
Substitute x = 2, y = 3, and b = 2 into the second
equation to find the value of a.
The equation 0 = 0 is always true. So, the system has
infinitely many solutions.
Fair Game Review
23. m =
y2 − y1
6−0
6
=
=
= 3
x2 − x1
2−0
2
3ax − by = 6
y − y1 = m( x − x1 )
3a ( 2) − 2(3) = 6
6a − 6 =
+6
6
y − 0 = 3( x − 0)
y = 3x
+6
6a = 12
6a
12
=
6
6
a = 2
24. m =
3 − ( −3)
y2 − y1
6
=
=
= 2
x2 − x1
3−0
3
y − y1 = m( x − x1 )
y − ( −3) = 2( x − 0)
y+3 =
2x
−3
−3
y = 2x − 3
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Chapter 5
25. m =
y2 − y1
2 −5
−3
1
=
=
= −
x2 − x1
0 − ( −6)
6
2
m( x − x1 )
y − y1 =
1
x − ( − 6)
2
1
y−5 = − x−3
2
+5
+5
2. 2 x = x − 3
y = 2x
y = x−3
3
y−5 = −
y
2
1
−4 −3 −2 −1
x
1 2
y = 2x
−3
1
y = − x+2
2
y=x−3
−5
(−3, −6)
26. B;
− 2( y + 5) ≤ 16
The graphs intersect at ( − 3, − 6). So, the solution is
− 2 y − 10 ≤
x = −3.
+ 10
16
+ 10
Check: x = −3
26
2x = x − 3
−2 y ≤
−2 y
26
≥
−2
−2
y ≥ −13
5.4 Extension (pp. 230–231)
?
2( − 3) = − 3 − 3
−6 = −6
3. 3x + 1 = 3 x + 2
y = 3x + 1
Practice
1. 2 x + 3 = 4
y = 2x + 3
y = 3x + 2
y = 3x + 2
y = 4
5
3
1
O
6
5
y=4
y
y
4
3
(0.5, 4)
2
y = 2x + 3
1 2 x
y = 3x + 1
−2
1 2 x
The lines are parallel. So, there is no solution.
The graphs intersect at (0.5, 4). So, the solution is
x = 0.5.
Check:
3x + 1 =
− 3x
Check: x = 0.5
3x + 2
− 3x
1 ≠ 2
2x + 3 = 4
?
2(0.5) + 3 = 4
?
1+3 = 4
4 = 4
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Big Ideas Math Blue
Worked;Out*Solutions*
187
Chapter 5
4.
6. 3 − 2 x = − 2 x + 3
1
x = x+8
3
1
y = x
3
y = x+8
y = 3 − 2x
y = −2x + 3
y = −2x + 3
10
8
y=x+8
5
y
1
3
y= x
4
2
−12
y
3
2
1
1
O
−6
2 4 x
x
y = 3 − 2x
−4
(−12, −4)
The lines are the same. So there are infinitely many
solutions.
−6
The graphs intersect at ( −12, − 4). So, the solution is
Check: 3 − 2 x = − 2 x + 3
+ 2x
x = −12.
+ 2x
3 = 3
Check: x = −12
1
x = x+8
3
7. Sample answer: The equation −3x = −3x − 2 has no
?
1
( −12) = −12 + 8
3
−4 = −4
5. 1.5 x + 2 = 11 − 3 x
y = 1.5 x + 2
solution because adding 3x to each side of the equation
produces the equation 0 = − 2, which is never true. To
change the equation so that it has infinitely many
solutions, subtract 2 from the left side of the equation.
So, adding 3x to each side of the equation
− 3x − 2 = −3x − 2 produces the equation − 2 = − 2,
which is always true.
y = 11 − 3x
y = 1.5x + 2
y = 11 − 3x
y
6
5
4
8. 6 x − 2 = x + 11
y = 6x − 2
y = x + 11
(2, 5)
20
y = x + 11
3
2
y = 6x − 2
1 2 3
x
−2 Intersection
X=2.6
The graphs intersect at ( 2, 5). So, the solution is x = 2.
Check: x = 2
1.5 x + 2 = 11 − 3 x
?
1.5( 2) + 2 = 11 − 3( 2)
?
3 + 2 = 11 − 6
5 = 5
7
Y=13.6
−4
The graphs intersect at ( 2.6, 13.6). So, the solution is
x = 2.6.
Check: x = 2.6
6 x − 2 = x + 11
?
6( 2.6) − 2 = 2.6 + 11
?
15.6 − 2 = 13.6
13.6 = 13.6
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Chapter 5
9.
11. 0.5 x + 12 = x + 9
4
2
x −1 = x + 6
3
3
4
y = x −1
3
2
y = x+6
3
y = 0.5 x + 12
y = x+9
y=x+9
16
(6, 15)
14
18
12
10
2
y= x+6
3
8
Intersection
X=10.5
Y=13
4
2
21
0
y = 0.5x + 12
6
4
y= x−1
3
−6
y
18
0
0 2 4 6 8 10 x
The graphs intersect at (10.5, 13). So, the solution is
The graphs intersect at (6, 15). So, the plants are the
x = 10.5.
same height after 6 months.
Check: x = 10.5 =
21
2
Quiz 5.3–5.4
1. x + 2 y = 4
4
2
x −1 = x + 6
3
3
? 2 21
4 21
−1 =
+6
3 2
3 2
−x − y = 2
Add the two equations.
x + 2y = 4
−x −
?
14 − 1 = 7 + 6
y = 2
y = 6
13 = 13
x + 2y = 4
10. 1.75 x = 2.25 x + 10.25
x + 2(6) = 4
y = 1.75 x
x + 12 =
y = 2.25 x + 10.25
4
− 12
y = 2.25x + 10.25
0
−28
2
− 12
x = −8
So, the solution is ( − 8, 6).
y = 1.75x
Intersection
X=-20.5
Y=-35.875
−50
The graphs intersect at ( − 20.5, − 35.875). So, the solution
is x = − 20.5.
Check: x = − 20.5
Check: ( −8, 6)
x + 2y = 4
?
− 8 + 2(6) = 4
?
−x − y = 2
?
−( − 8) − 6 = 2
?
− 8 + 12 = 4
8−6 = 2
4 = 4
2 = 2
1.75 x = 2.25 x + 10.25
?
1.75( − 20.5) = 2.25( − 20.5) + 10.25
?
− 35.875 = − 46.125 + 10.25
− 35.875 = − 35.875
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Big Ideas Math Blue
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189
Chapter 5
Check: ( 2, 1)
2. 2 x − y = 1
x + 3y − 4 = 0
3x = − 4 y + 10
Rewrite the second equation.
x + 3y = 4
4 x + 3 y = 11
?
?
3( 2) = − 4(1) + 10
4( 2) + 3(1) = 11
?
Multiply the second equation by − 2 and then add the
6 = − 4 + 10
two equations.
2x − y = 1
6 = 6
− 2 x − 6 y = −8
4. 2 x + 5 y = 60
?
8 + 3 = 11
11 = 11
−7 y = −7
2 x − 5 y = − 20
−7 y
−7
=
−7
−7
Add the two equations.
2x + 5 y =
y = 1
+ 2 x − 5 y = − 20
2x − y = 1
2x − 1 =
+1
=
40
4x
=
4
x =
40
4
10
2x + 5 y =
60
2(10) + 5 y =
60
4x
1
+1
2x =
60
2
2x
2
=
2
2
x = 1
20 + 5 y =
So, the solution is (1, 1).
Check: (1, 1)
2x − y = 1
x + 3y − 4 = 0
?
?
2(1) − 1 = 1
1 + 3(1) − 4 = 0
?
60
− 20
?
2 −1 = 1
1+3−4 = 0
1 = 1
0 = 0
− 20
5y =
40
5y
=
5
y =
40
5
8
So, the solution is (10, 8).
Check: (10, 8)
2 x + 5 y = 60
3. 3 x = − 4 y + 10
?
2(10) + 5(8) = 60
4 x + 3 y = 11
Rewrite the first equation.
3x + 4 y = 10
?
2 x − 5 y = − 20
?
2(10) − 5(8) = − 20
?
20 + 40 = 60
20 − 40 = − 20
60 = 60
− 20 = − 20
Multiply the first equation by 4. Multiply the second
equation by −3. Then add the two equations.
12 x + 16 y =
40
−12 x − 9 y = −33
7y =
7
7y
7
=
7
7
y = 1
3x = − 4 y + 10
3x = − 4(1) + 10
3x = − 4 + 10
3x = 6
3x
6
=
3
3
x = 2
So, the solution is ( 2, 1).
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Chapter 5
5. 3 x − 2 y = 16
7. − 2 x + y = − 2
6 x − 4 y = 32
3x + y = 3
Multiply the first equation by − 2 and then add the two
Subtract Equation 2 from Equation 1.
equations.
−2x + y = −2
− 6 x + 4 y = − 32
−3x − y = −3
+ 6x − 4 y =
32
−5 x
0 =
0
The equation 0 = 0 is always true. So, the solutions are
all points on the line 3x − 2 y = 16. The system of
linear equations has infinitely many solutions.
Check:
6x − 4y = 32
4
y
x = 1
−2x + y = −2
− 2(1) + y = − 2
+2
2 4 6 x
−2
−4
+2
y =
0
So, the solution is (1, 0).
−6
−8
3x − 2y = 16
Both equations are the same line.
6. 4 y = x − 8
Check: (1, 0)
−2x + y = −2
3x + y = 3
?
1
− x + y = −1
4
Rewrite the first equation.
− x + 4 y = −8
Multiply the second equation by − 4 and then add the
− x + 4 y = −8
+ x − 4y =
?
− 2(1) + 0 = − 2
3(1) + 0 = 3
−2 = −2
3 = 3
1
y+ 2
3
9x − y = −6
8. 3 x =
Rewrite the first equation.
two equations.
3x −
4
1
y = 2
3
Multiply Equation 1 by − 3 and then add the two
0 = −4
The equation 0 = − 4 is never true. So, the system of
linear equations has no solution.
Check:
equations.
−9x + y = −6
+ 9x − y = −6
0 = −12
1
4
y
− x + y = −1
The equation 0 = −12 is never true. So, the system of
linear equations has no solution.
4 x
−4 −3 −2 −1
−3
−4
−5 x
−5
=
−5
−5
−2 + y = −2
2
1
= −5
4y = x − 8
The lines are parallel.
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Big Ideas Math Blue
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191
Chapter 5
9. 4 x − 1 = 2 x
11. 1 − 3 x = − 3x + 2
y = 4x − 1
y = −3x + 1
y = 2x
y = −3x + 2
y = 4x − 1
y = −3x + 1
y
4
y
3
y = 2x
2
1
(0.5, 1)
−4 −3 −2 −1
1 2 3 4 x
1
−1
2 x
−2
−3
y = −3x + 2
The graphs intersect at (0.5, 1). So, the solution is
x = 0.5.
The lines are parallel. So, there is no solution.
Check: 1 − 3x = − 3x + 2
+ 3x
Check: x = 0.5
1 ≠ 2
4x − 1 = 2x
?
4(0.5) − 1 = 2(0.5)
?
2 −1 = 1
+ 3x
12. 1 − 5 x = 3 − 7 x
y = −5x + 1
y = −7 x + 3
1 = 1
y = −7x + 3
1
1
2
10. − x + 1 = − x + 1
−4 −3 −2
−3
−4
(1, −4)
−5
−6
y = −5x + 1
−7
1
2
y=− x+1
−1
1 2 3 4 x
−2
1
y = − x +1
2
y = −x + 1
2
O
y
The graph intersects at (1, − 4). So, the solution is
(0, 1)
1 2
4 x
x = 1.
Check: x = 1
−2
y = −x + 1
1 − 5x = 3 − 7 x
?
The graphs intersect at (0, 1). So, the solution is x = 0.
1 − 5(1) = 3 − 7(1)
?
Check: x = 0
1
− x + 1 = −x + 1
2
?
1
− ( 0) + 1 = − 0 + 1
2
1−5 = 3−7
−4 = −4
?
0 +1 = 1
1 = 1
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Chapter 5
2. y = − x + 4
13. x = cost of a boat tour
x + 3y = 0
y = cost of a walking tour
x + 2 y = 19
1.5 x + 3 y = 28.50
Multiply the first equation by 3. Multiply the second
equation by 2. Then subtract Equation 2 from Equation 1.
3x + 6 y =
y = −x + 4
2
1
57
O
− 3x − 6 y = − 57
−2
0 =
y
5
4
3
3 4 5 6 7 x
(6, −2)
x + 3y = 0
−3
0
The equation 0 = 0 is always true. The system of linear
equations has infinitely many solutions. So, it is not
possible to determine how much each tour costs.
Check: (6, − 2)
y = −x + 4
x + 3y = 0
?
?
6 + 3( − 2) = 0
− 2 = −6 + 4
14. a. x = number of bicycle rentals
?
−2 = −2
y = number of in-line skate rentals
6−6 = 0
0 = 0
x + y = 20
So, the solution is (6, − 2).
25 x + 20 y = 455
b. Multiply Equation 1 by − 20 and then add the two
equations.
2x − 3y = −2
− 20 x − 20 y = − 400
+ 25 x + 20 y =
455
5x =
55
11 + y =
y
1
−5 −4 −3
O
1 x
−2
−3
(−4, −2)
2x − 3y = −2
20
Check: ( − 4, − 2)
− 11
y =
3
x − y = −2
5x
55
=
5
5
x = 11
x + y = 20
− 11
3. x − y = − 2
x − y = −2
9
2x − 3y = −2
?
So, the business had 11 bicycle rentals and 9 in-line
skate rentals today.
Chapter 5 Review
1. y = 2 x − 3
6
(5, 7)
y = 2x − 3
?
7 = 2(5) − 3
−2
2 4 6 8 x
y = 2x − 3
−8
?
?
−4 + 2 = −2
−8 + 6 = − 2
−2 = −2
−2 = −2
Check: (5, 7 )
y
y=x+2 4
−8 −6 −4
2( − 4) − 3( − 2) = − 2
So, the solution is ( − 4, − 2).
y = x+ 2
8
?
− 4 − ( − 2) = − 2
?
7 = 10 − 3
y = x+2
?
7 = 5+2
7 = 7
7 = 7
So, the solution is (5, 7 ).
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193
Chapter 5
Check: ( −8, 0)
4. y = − 3 x − 7
y = x+9
1
x + y = −4
2
x + 9 = −3x − 7
+ 3x
?
1
(− 8) + 0 = − 4
2
+ 3x
4 x + 9 = −7
−9
?
0 = 2( − 8) + 16
?
−9
−4 + 0 = −4
4 x = −16
−4 = −4
4x
− 16
=
4
4
x = −4
?
0 = −16 + 16
0 = 0
6. − x + 5 y = 28
x + 3 y = 20
x + 3 y = 20
y = x+9
− 3y
= −4 + 9
− 3y
x = −3 y + 20
= 5
So, the solution is ( − 4, 5).
− x + 5 y = 28
−( −3 y + 20) + 5 y = 28
Check: ( − 4, 5)
y = − 3x − 7
y = x+9
?
5 = − 3( − 4) − 7
?
5 = 12 − 7
3 y − 20 + 5 y = 28
8 y − 20 =
?
5 = −4 + 9
+ 20
1
x + y = −4
2
y = 2 x + 16
28
+ 20
8y =
5 = 5
48
8y
48
=
8
8
y = 6
5 = 5
5.
y = 2 x + 16
x + 3 y = 20
x + 3(6) = 20
1
x + y = −4
2
1
x + ( 2 x + 16) = − 4
2
5
x + 16 = − 4
2
− 16
− 16
5
x = − 20
2
2 5
2
x = ( − 20)
5 2
5
x = −8
x + 18 =
20
− 18
− 18
x =
2
So, the solution is ( 2, 6).
Check: ( 2, 6)
− x + 5 y = 28
?
− 2 + 5(6) = 28
?
x + 3 y = 20
?
2 + 3(6) = 20
?
− 2 + 30 = 28
2 + 18 = 20
28 = 28
20 = 20
y = 2 x + 16
= 2( − 8) + 16
= −16 + 16
= 0
So, the solution is ( − 8, 0).
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Chapter 5
7. x = number of jars of jam
9. 3 x − 2 y = 1
y = number of packages of bread mix
9x − 6 y = 3
x+ y = 8
Multiply Equation 1 by − 3 and then add the two
6 x + 5 y = 45
equations.
Multiply the first equation by −6 and then add the two
−9x + 6 y = −3
equations.
+ 9x − 6 y = 3
− 6 x − 6 y = − 48
+ 6x + 5 y =
45
−y =
−3
(−1)(− y)
0 = 0
The equation 0 = 0 is always true. So, the solutions are
all points on the line 3x − 2 y = 1. The system of linear
equations has infinitely many solutions.
= ( −1)( − 3)
Check:
y = 3
3x − 2y = 1
x+ y =
8
4
x+3 =
8
3
2
−3
1
5
O
−3
x =
So, there are 5 jars of jam and 3 packages of bread mix in
the gift basket.
8. x + 2 y = − 5
y
9x − 6y = 3
1 2 3 x
Both equations are the same line.
10. 8 x − 2 y = 16
x − 2 y = −5
−4x + y = 8
Subtract Equation 2 from Equation 1.
Multiply Equation 2 by 2 and then add the two equations.
x + 2 y = −5
8 x − 2 y = 16
− x + 2y = 5
− 8 x + 2 y = 16
4y = 0
0 = 32
4y
0
=
4
4
y = 0
The equation 0 = 32 is never true. So, the system of
linear equations has no solution.
Check:
x + 2 y = −5
−4x + y = 8
y
10
8
x + 2(0) = − 5
x + 0 = −5
x = −5
2
So, the solution is ( − 5, 0).
−4
x + 2 y = −5
?
4 6 x
−4
−6
Check: ( − 5, 0)
− 5 + 2(0) = − 5
O
x − 2 y = −5
?
− 5 − 2(0) = − 5
?
?
−5 + 0 = −5
−5 − 0 = −5
−5 = −5
−5 = −5
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−8 8x − 2y = 16
The lines are parallel.
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195
Chapter 5
11. 2 x − 9 = 7 x + 11
1
x + 10
2
y = 4x − 4
2. y =
y = 2x − 9
y = 7 x + 11
1
y
12
y = 2 x + 10
10
y
14
y = 7x + 11
12
(4, 12)
8
−6 −4
2
6
4
2
6 x
−6
−8
−2
−4
y = 2x − 9
−16
−18
2 4 6 x
y = 4x − 4
Check: ( 4, 12)
x = − 4.
1
x + 10
2
? 1
12 = ( 4) + 10
2
Check: x = − 4
12 = 2 + 10
12 = 16 − 4
12 = 12
12 = 12
(−4, −17)
y =
The graphs intersect at ( − 4, −17 ). So, the solution is
y = 4x − 4
?
12 = 4( 4) − 4
?
2 x − 9 = 7 x + 11
?
2( − 4) − 9 = 7( − 4) + 11
?
− 8 − 9 = − 28 + 11
−17 = −17
?
So, the solution is ( 4, 12).
3. y + x = 0
3 y + 6 x = −9
y+x=0
Chapter 5 Test
4
1. y = 4 − x
y
O
1 x
3y + 6x = −9
(4, 0)
1 2 3 4
1
−4 −3 −2
y=x−4
1
3
2
(−3, 3)
y = x− 4
y
−3
−4
6 x
−2
Check: ( − 3, 3)
y=4−x
y+ x = 0
Check: ( 4, 0)
y = 4− x
?
y = x−4
?
0 = 4−4
0 = 4−4
0 = 0
0 = 0
So, the solution is ( 4, 0).
196* Big Ideas Math Blue*
*
3 y + 6 x = −9
?
Worked;Out*Solutions*
?
3 + ( − 3) = 0
3(3) + 6( − 3) = − 9
?
?
3−3 = 0
9 − 18 = − 9
0 = 0
−9 = −9
So, the solution is ( − 3, 3).
Copyright*©*Big*Ideas*Learning,*LLC*
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Chapter 5
Check: (7, 13)
4. − 3 x + y = 2
−x + y − 4 = 0
x + y = 20
−3x + y = 2
+ 3x
y = 2x − 1
?
?
13 = 2(7) − 1
7 + 13 = 20
+ 3x
?
20 = 20
y = 3x + 2
13 = 14 − 1
13 = 13
−x + y − 4 = 0
− x + ( 3 x + 2) − 4 = 0
2x − 2 =
+2
6. x − y = 3
0
x + 2 y = −6
+2
2x =
x− y = 3
+y
+y
2
2x
2
=
2
2
x = 1
x = y+3
x + 2 y = −6
+ 2 y = −6
3y + 3 = −6
−3
−3
( y + 3)
− 3x + y = 2
− 3(1) + y = 2
−3 + y =
+3
2
3 y = −9
−9
3y
=
3
3
y = −3
+3
y =
5
So, the solution is (1, 5).
Check: (1, 5)
− 3x + y = 2
?
− 3(1) + 5 = 2
−x + y − 4 = 0
?
−1 + 5 − 4 = 0
?
−3 + 5 = 2
2 = 2
5. x + y = 20
y = 2x − 1
x + y = 20
x + ( 2 x − 1) = 20
3 x − 1 = 20
+1
+1
0 = 0
x− y = 3
x − ( − 3) = 3
x+3 = 3
−3
−3
x =
0
So, the solution is (0, − 3).
Check: (0, − 3)
x− y = 3
?
0 − ( − 3) = 3
?
x + 2 y = −6
?
0 + 2( − 3) = − 6
?
0+3 = 3
0 − 6 = −6
3 = 3
−6 = −6
3x = 21
3x
21
=
3
3
x = 7
y = 2x − 1
= 2(7) − 1
= 14 − 1
= 13
So, the solution is (7, 13).
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197
Chapter 5
7. 2 x + y = 3
9. − 2 x + y + 3 = 0
x− y = 3
3 x + 4 y = −1
Add the two equations.
Rewrite the first equation.
2x + y = 3
− 2 x + y = −3
+x− y = 3
Multiply the first equation by − 4 and then add the two
= 6
3x
equations.
3x
6
=
3
3
x = 2
8 x − 4 y = 12
+ 3x + 4 y = −1
11x
2x + y = 3
11x
11
=
11
11
x =1
2( 2) + y = 3
4+ y =
3
−4
= 11
−4
−2x + y + 3 = 0
y = −1
− 2(1) + y + 3 = 0
−2 + y + 3 = 0
So, the solution is ( 2, −1).
Check: ( 2, −1)
2x + y = 3
y +1 =
0
−1
−1
x− y = 3
?
y = −1
?
2( 2) + ( −1) = 3
2 − ( −1) = 3
?
So, the solution is (1, −1).
?
4 −1 = 3
2 +1 = 3
3 = 3
3 = 3
Check: (1, −1)
−2x + y + 3 = 0
?
8. x + y = 12
− 2(1) + ( −1) + 3 = 0
3x = 2 y + 6
?
Rewrite the second equation.
2 x + 2 y = 24
+ 3x − 2 y =
0 = 0
−1 = −1
y = 5x + 1
y = 8x − 1
Divide both sides of the first equation by 2.
2y
16 x 2
=
−
2
2
2
y = 8x − 1
x + y = 12
6 + y = 12
−6
y =
10. y = 4 x + 8
11. 2 y = 16 x − 2
5x
30
=
5
5
x = 6
−6
?
3 − 4 = −1
The system of linear equations has one solution because
the lines have different slopes.
6
= 30
5x
?
3(1) + 4( −1) = −1
−2 − 1 + 3 = 0
3x − 2 y = 6
Multiply the first equation by 2 and then add the two
equations.
3 x + 4 y = −1
The system of linear equations has infinitely many
solutions because both equations are the same line.
6
So, the solution is (6, 6).
Check: (6, 6)
x + y = 12
?
6 + 6 = 12
12 = 12
3x = 2 y + 6
?
3(6) = 2(6) + 6
?
18 = 12 + 6
18 = 18
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Chapter 5
12. y = − 3 x + 2
14. 8 x − 14 = − 2 x − 4
6 x + 2 y = 10
y = 8 x − 14
Solve the second equation for y.
y = − 2x − 4
6 x + 2 y = 10
− 6x
y = 8x − 14
− 6x
4
2 y = − 6 x + 10
2y
− 6 x 10
=
+
2
2
2
y = − 3x + 5
The system of linear equations has no solution
because the lines have the same slope but different
y-intercepts.
13.
1
x−4 =
4
1
y = x−
4
3
y = x+
4
3
x+2
4
3
4
4
2
4 6 x
−4
(1, −6)
−6
−8
y = −2x − 4
The graphs intersect at (1, − 6). So, the solution is
x = 1.
Check: x = 1
8 x − 14 = − 2 x − 4
4
?
8(1) − 14 = − 2(1) − 4
2
?
8 − 14 = − 2 − 4
y= x+2
−12
y
−8
−2
y
−6 = −6
2 x
15. x = cost of the pears
y = cost of the apples
−6
(−12, −7)
−8
1
4
y= x−4
The graphs intersect at ( −12, − 7). So, the solution is
x = −12.
Check: x = −12
1
3
x−4 = x+ 2
4
4
? 3
1
(−12) − 4 = ( −12) + 2
4
4
2 x + 6 y = 14
3 x + 9 y = 21
Multiply the first equation by 3. Multiply the second
equation by − 2. Then add the two equations.
6 x + 18 y =
42
− 6 x − 18 y = − 42
0 =
0
The equation 0 = 0 is always true. The system of linear
equations has infinitely many solutions. So, it is not
possible to determine how much each pear and apple
costs.
?
−3 − 4 = −9 + 2
−7 = −7
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199
Chapter 5
16. x = number of lilies in the bouquet
y = number of tulips in the bouquet
x + y = 12
Chapter 5 Standards Assessment
2
3
4x + 6 y = −6
1. D; y = − x − 1
3 x + 2 y = 32
4x + 6 y = −6
Multiply the first equation by −3 and then add the two
2
4x + 6 − x − 1 = −6
3
4x − 4x − 6 = −6
−6 = −6
equations.
− 3x − 3 y = − 36
+ 3x + 2 y =
32
−y =
−4
(−1)(− y) =
( −1)(− 4)
y = 4
x + 4 = 12
The slope of the line is − 0.25. The slope of a line that is
−4
x =
8
So, there are 8 lilies and 4 tulips in the bouquet.
17. x = cost of a special
y = cost of a glass of milk
4 x + 2 y = 28
3x + 4 y = 26.25
perpendicular to this line is the negative reciprocal of
1
− 0.25. So, because − 0.25 = − , the slope of a line
4
4
perpendicular to the line y = − 0.25x + 3 is m =
= 4.
1
3. G
4. A; x + 3 y = 10
Multiply the first equation by − 2 and then add the two
equations.
−8 x − 4 y = − 56
+ 3x + 4 y = 26.25
−5 x
linear equations has infinitely many solutions.
2. 4; y = − 0.25 x + 3
x + y = 12
−4
The equation − 6 = − 6 is always true. So, the system of
x = 2y − 5
x + 3 y = 10
(2 y − 5) + 3 y = 10
5 y − 5 = 10
+5
+5
= − 29.75
5 y = 15
5y
15
=
5
5
y = 3
−5 x
− 29.75
=
−5
−5
x = 5.95
4 x + 2 y = 28
4(5.95) + 2 y =
23.8 + 2 y =
− 23.8
28
28
− 23.8
2y =
4.2
2y
4.2
=
2
2
y = 2.1
Find the cost for 2 specials and 2 glasses of milk.
2 x + 2 y = 2(5.95) + 2( 2.1)
= 11.9 + 4.2
= 16.1
So, it costs $16.10 for 2 specials and 2 glasses of milk.
x = 2y − 5
= 2(3) − 5
= 6−5
=1
So, the solution is (1, 3).
5. F; The lines are parallel and will never intersect. So, the
system has no solution.
6. D; x = price of an adult ticket
y = price of a child ticket
2 x + 2 y = 62
x + 4 y = 70
7. I; y = − 7 x + 5
The slope is − 7. The y-intercept is 5.
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Chapter 5
8. −11; 7 w − 3w = 2(3w + 11)
4w =
− 6w
6w + 22
− 6w
− 2w
22
=
−2
−2
w = −11
So, −11 is the value of w that makes the equation true.
y2 − y1
to find the slope
x2 − x1
of the line passing through ( −1, 5) and ( 4, 7 ).
9. C; Use the slope formula m =
y − y1
7 −5
2
m = 2
=
=
x2 − x1
4 − ( −1)
5
2
is parallel to the line passing
5
through ( −1, 5) and ( 4, 7 ). So, the equation
A line having a slope of
2
x + 1 is parallel to the line passing through
5
(−1, 5) and ( 4, 7).
y =
10. $7.50; x = cost of a T-shirt
y = cost of a pair of shorts
3 x + 2 y = 42.5
12 y + 60 =
− 12 y
32 y
− 12 y
60 =
20 y
60
20 y
=
20
20
3 = y
So, the value of y is 3.
12. A; A system of linear equations that has the same slope
and the same y-intercept has infinitely many solutions
because they are the same line.
13. G; x = the number
1
x + 10 = 13
3
− 10
− 10
1
x =
3
3
1
3 x = 3(3)
3
x = 9
So, the number is 9.
5 x + 3 y = 67.5
14. B; 4 x + 7 y = 16
3 x + 2 y = 42.5
− 7y
− 3x
2 y = − 3 x + 42.5
− 3 x 42.5
2y
=
+
2
2
2
3
y = − x + 21.25
2
5 x + 3 y = 67.5
3
5 x + 3 − x + 21.25 = 67.5
2
5x −
rectangles, you can use the equation 12( y + 5) = 32 y
and solve for y.
12( y + 5) = 32 y
− 2w = 22
− 3x
11. H; Because the figures have the same area and are both
− 7y
4 x = − 7 y + 16
− 7 y 16
4x
=
+
4
4
4
7
x = − y+4
4
7
7
So, x = − y + 4, or x = 4 − y.
4
4
9
x + 63.75 = 67.5
2
1
x + 63.75 = 67.5
2
− 63.75
− 63.75
1
3.75
x =
2
1
2 x = 2(3.75)
2
x = 7.5
So, the cost of each T-shirt is $7.50. The linear equations
modeling the situation are solved for x, the cost of a
T-shirt.
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201
*