CHAPTER 13 SOLVING SIMULTANEOUS EQUATIONS
EXERCISE 51 Page 105
1. Solve the simultaneous equations 2 x − y =
6
x+y=6
(1) + (2) gives:
2x − y =
6
(1)
x+y=6
(2)
3x = 12
From (1):
8–y=6
i.e.
8–6=y
from which,
x=
12
=4
3
from which, y = 2
2. Solve the simultaneous equations 2x – y = 2
x – 3y = –9
3 × (1) gives:
(3) – (2) gives:
2x – y = 2
(1)
x – 3y = –9
(2)
6x – 3y = 6
5x = 15
From (1):
6–y=2
i.e.
6–2=y
(3)
from which,
x=
15
=3
5
from which, y = 4
3. Solve the simultaneous equations x – 4y = – 4
5x – 2y = 7
x – 4y = –4
2 × (2) gives:
(3) – (1) gives:
(1)
5x – 2y = 7
(2)
10x – 4y = 14
(3)
9x = 18 from which,
180
x=
18
=2
9
© 2014, John Bird
From (1):
2 – 4y = – 4
i.e.
2 + 4 = 4y
i.e.
6 = 4y
from which, y =
6
= 1.5
4
4. Solve the simultaneous equations 3x – 2y = 10
5x + y = 21
2 × (2) gives:
(1) + (3) gives:
3x – 2y = 10
(1)
5x + y = 21
(2)
10x + 2y = 42
(3)
13x = 52 from which,
From (1):
12 – 2y = 10
i.e.
12 – 10 = 2y
i.e.
2 = 2y
x=
from which, y =
52
=4
13
2
=1
2
5. Solve the simultaneous equations 5p + 4q = 6
2p – 3q = 7
5p + 4q = 6
(1)
2p – 3q = 7
(2)
3 × (1) gives:
15p + 12q = 18
(3)
4 × (2) gives:
8p – 12q = 28
(4)
(3) + (4) gives:
From (1):
23p = 46 from which,
p=
46
=2
23
10 + 4q = 6
i.e.
4q = 6 – 10
i.e.
4q = –4
181
from which, q
=
−4
= –1
4
© 2014, John Bird
6. Solve the simultaneous equations 7x + 2y = 11
3x – 5y = –7
5 × (1) gives:
2 × (2) gives:
(3) + (4) gives:
From (1):
7x + 2y = 11
(1)
3x – 5y = –7
(2)
35x + 10y = 55
(3)
6x – 10y = –14
41x = 41
(4)
from which,
x=
41
=1
41
y=
4
=2
2
y=
58
=2
29
x=
6
=3
2
7 + 2y = 11
i.e.
2y = 11 – 7
i.e.
2y = 4
from which,
7. Solve the simultaneous equations 2x – 7y = –8
3x + 4y = 17
2x – 7y = –8
(1)
3x + 4y = 17
(2)
3 × (1) gives:
6x – 21y = –24
(3)
2 × (2) gives:
6x + 8y = 34
(4)
(4) – (3) gives:
From (1):
29y = 58 from which,
2x – 14 = –8
i.e.
2x = 14 – 8
i.e.
2x = 6
from which,
8. Solve the simultaneous equations a + 2b = 8
b – 3a = –3
182
© 2014, John Bird
Rearranging gives:
2 × (2) gives:
a + 2b = 8
(1)
– 3a + b = –3
(2)
– 6a + 2b = –6
(3)
(1) – (3) gives:
From (1):
7a = 14 from which,
a=
14
=2
7
2 + 2b = 8
i.e.
2b = 8 – 2
i.e.
2b = 6
from which,
b=
6
=3
2
9. Solve the simultaneous equations a + b = 7
a–b=3
a+b=7
(1)
a–b=3
(2)
(1) + (2) gives:
2a = 10
From (1):
from which,
a=
10
=5
2
5+b=7
i.e.
b=7–5
from which,
b=2
10. Solve the simultaneous equations 2x + 5y = 7
x + 3y = 4
2 × equation (2) gives:
(3) – (1) gives:
Substituting in (1) gives:
2x + 5y = 7
(1)
x + 3y = 4
(2)
2x + 6y = 8
(3)
y=1
2x + 5 = 7
i.e.
2x = 7 – 5 = 2
and x =
2
=1
2
Thus, x = 1 and y = 1 and may be checked by substituting into both of the original equations
183
© 2014, John Bird
11. Solve the simultaneous equations 3s + 2t = 12
4s – t = 5
2 × equation (2) gives:
(1) + (3) gives:
Substituting in (1) gives:
3s + 2t = 12
(1)
4s – t = 5
(2)
8s – 2t = 10
(3)
11s = 22
6 + 2t = 12
from which,
i.e.
s=
22
=2
11
2t = 12 – 6 = 6
and
t=
6
=3
2
Thus, s = 2 and t = 3 and may be checked by substituting into both of the original equations
12. Solve the simultaneous equations 3x – 2y = 13
2x + 5y = –4
3x – 2y = 13
(1)
2x + 5y = –4
(2)
2 × equation (1) gives:
6x – 4y = 26
(3)
3 × equation (2) gives:
6x + 15y = –12
(4)
(3) – (4) gives:
Substituting in (1) gives:
–19y = 38
3x + 4 = 13
i.e.
from which,
y=
3x = 13 – 4 = 9
38
= –2
−19
and
x=
9
=3
3
Thus, x = 3 and y = –2 and may be checked by substituting into both of the original equations
13. Solve the simultaneous equations 5m – 3n = 11
3m + n = 8
3 × equation (2) gives:
5m – 3n = 11
(1)
3m + n = 8
(2)
9m + 3n = 24
(3)
184
© 2014, John Bird
(1) + (3) gives:
14m = 35
Substituting in (1) gives:
12.5 – 3n = 11
i.e.
12.5 – 11 = 3n
i.e.
1.5 = 3n
from which,
and
m=
n=
35
= 2.5
14
1.5
= 0.5
3
14. Solve the simultaneous equations 8a – 3b = 51
3a + 4b = 14
8a – 3b = 51
(1)
3a + 4b = 14
(2)
4 × (1) gives:
32a – 12b = 204
(3)
3 × (2) gives:
9a + 12b = 42
(4)
(3) + (4) gives:
41a = 246
From (1):
48 – 3b = 51
i.e.
48 – 51 = 3b
i.e.
–3 = 3b
from which,
a=
246
=6
41
from which,
b=
−3
= –1
3
15. Solve the simultaneous equations 5x = 2y
3x + 7y = 41
5x – 2y = 0
(1)
3x + 7y = 41
(2)
3 × equation (1) gives:
15x – 6y = 0
(3)
5 × equation (2) gives:
15x + 35y = 205
(4)
(4) – (3) gives:
Substituting in (1) gives:
41y = 205
5x – 10 = 0
i.e.
from which,
5x = 10
185
y=
and
205
=5
41
x=
10
=2
5
© 2014, John Bird
16. Solve the simultaneous equations 5c = 1 – 3d
2d + c + 4 = 0
Rearranging gives:
5 × equation (2) gives:
(3) – (1) gives:
Substituting in (1) gives:
5c + 3d = 1
(1)
c + 2d = –4
(2)
5c + 10d = –20
(3)
7d = –21
5c – 9 = 1
i.e.
from which,
5c = 10
186
and
d=
−21
= –3
7
c=
10
=2
5
© 2014, John Bird
EXERCISE 52 Page 107
1. Solve the simultaneous equations 7p + 11 + 2q = 0
–1 = 3q – 5p
Rearranging gives:
7p + 2q = –11
(1)
5p – 3q = 1
(2)
3 × equation (1) gives:
21p + 6q = –33
(3)
2 × equation (2) gives:
10p – 6q = 2
(4)
(3) + (4) gives:
31p = –31
Substituting in (1) gives:
–7 + 2q = –11
2. Solve the simultaneous equations
from which,
i.e.
p=–1
2q = –11 + 7 = –4
and
q=
x
y
+
=4
2
3
y
x
–
=0
9
6
x
y
Rearranging gives: (6) + (6) =
(6)(4)
2
3
i.e.
3x + 2y = 24
(1)
x
y
(18) − (18) =
(18)(0)
6
9
i.e.
3x – 2y = 0
(2)
and
(1) – (2) gives:
4y = 24
Substituting in (1) gives:
3x + 12 = 24
3. Solve the simultaneous equations
Rearranging gives:
−4
= –2
2
i.e.
from which,
3x = 24 – 12 = 12
and
y=6
x=
12
=4
3
a
– 7 = –2b
2
2
12 = 5a + b
3
a
(2) − (2)7 =
−(2)(2b)
2
i.e.
187
a + 4b = 14
(1)
© 2014, John Bird
2
(3)(12) = (3)(5a) + (3) b
3
and
i.e.
2 × equation (2) gives:
15a + 2b = 36
(2)
30a + 4b = 72
(3)
(3) – (1) gives:
29a = 58
Substituting in (1) gives:
2 + 4b = 14
4. Solve the simultaneous equations
i.e.
4b = 14 – 2 = 12
from which,
and
b=
a=2
12
=3
4
3
s – 2t = 8
2
s
+ 3t = –2
4
Rearranging gives:
3
(2) s − (2)2t =
(2)(8)
2
i.e.
3s – 4t = 16
(1)
and
s
(4) + (4)3t =(4)(−2)
4
i.e.
s + 12t = –8
(2)
3s + 36t = –24
(3)
3 × equation (2) gives:
(1) – (3) gives:
– 40t = 40
from which,
t=
Substituting in (1) gives:
3s + 4 = 16
5. Solve the simultaneous equations
i.e.
(14)
3s = 16 – 4 = 12
i.e. t = –1
and
s=
12
=4
3
x
2y
49
+
=
5
3
15
y
3x
5
–
+ =0
2
7
7
x
2y
49
Rearranging gives: (15) + (15)
=
(15)
5
3
15
and
− 40
40
i.e. 3x + 10y = 49
3x
5
y
i.e.
0
− (14) + (14) =
7
2
7
2 × equation (1) gives:
6x – 7y = –10
6x + 20y = 98
188
(1)
(2)
(3)
© 2014, John Bird
(3) – (2) gives:
27y = 108
Substituting in (1) gives:
3x + 40 = 49
i.e.
from which, y =
3x = 49 – 40 = 9
and
x=
108
=4
27
9
=3
3
u
12
v
25
u+ –
=0
4
2
6. Solve the simultaneous equations v – 1 =
u
Rearranging gives: (12)v − (12)(1) =
(12)
12
i.e.
12v – u = 12
(1)
v
25
i.e.
(4)u + (4) − (4)
0
=
4
2
v + 4u = 50
(2)
4 × equation (1) gives:
48v – 4u = 48
(3)
(2) + (3) gives:
49v = 98
and
Substituting in (1) gives:
24 – u = 12
i.e.
24 – 12 = u
from which, v =
and
98
=2
49
u = 12
7. Solve the simultaneous equations 1.5x – 2.2y = –18
2.4x + 0.6y = 33
Multiplying both equations by 10 gives: 15x – 22y = –180
6 × equation (1) gives:
22 × equation (2) gives:
(1)
24x + 6y = 330
(2)
90x – 132y = –1080
(3)
528x + 132y = 7260
(3) + (4) gives:
Substituting in (1) gives: 150 – 22y = –180
618x = 6180
(4)
from which,
x = 10
i.e. –22y = –180 – 150 = –330 and y =
−330
= 15
−22
Thus, x = 10 and y = 15 and may be checked by substituting into both of the original equations.
189
© 2014, John Bird
8. Solve the simultaneous equations 3b – 2.5a = 0.45
1.6a + 0.8b = 0.8
Multiplying equation (1) by 100 gives: – 250a +300b = 45
(1)
Multiplying equation (1) by 10 gives:
(2)
4 × equation (1) gives:
150 × equation (2) gives:
16a +8b = 8
–1000a + 1200b = 180
(3)
2400a + 1200b = 1200
(4) – (3) gives:
3400a = 1020
from which,
Substituting in (1) gives: –75 + 300b = 45
(4)
a=
1020
= 0.3
3400
i.e. 300b = 45 + 75 = 120 and b =
190
120
= 0.4
300
© 2014, John Bird
EXERCISE 53 Page 109
1. Solve the simultaneous equations
2
3
+
= 14
y
x
3
5
–
= –2
y
x
Let
1
= a and
x
1
=b
y
Thus, the equations become:
3a + 2b = 14
(1)
and
5a – 3b = –2
(2)
3 × equation (1) gives:
9a + 6b = 42
(3)
2 × equation (2) gives:
10a – 6b = –4
(4)
(3) + (4) gives:
Substituting in (1) gives:
Since
19a = 38
6 + 2b = 14
1
1 1
= a then x = =
a 2
x
Thus, x =
i.e.
and since
from which,
2b = 14 – 6 = 8
a=
and
38
=2
19
b=
8
=4
2
1
1 1
= b then y = =
y
b 4
1
1
and y =
and may be checked by substituting into both of the original equations
4
2
2. Solve the simultaneous equations
4 3
– = 18
a b
2
5
+ = –4
a
b
Let
1
1
= x and
=y
a
b
Thus, the equations become:
4x – 3y = 18
(1)
and
2x + 5y = –4
(2)
4x + 10y = –8
(3)
2 × equation (2) gives:
191
© 2014, John Bird
(1) – (3) gives:
–13y = 26
Substituting in (1) gives:
Since
4x + 6 = 18
1
1 1
= x then a = =
a
x 3
Thus, a =
i.e.
from which,
4x = 18 – 6 = 12
26
= –2
−13
y=
and
x=
12
=3
4
1
1
1
1
= y then b = =
= −
b
y −2
2
and since
1
1
and b = − and may be checked by substituting into both of the original equations
2
3
3. Solve the simultaneous equations
1
3
+
=5
5q
2p
5
1
35
–
=
p
2q
2
Let
1
1
= a and
=b
p
q
Thus, the equations become:
1
3
a+ b=
5
2
5
1
35
5a − b =
2
2
Rearranging gives:
1
3
(10) a + (10) b =
(10)(5)
2
5
i.e.
5a + 6b = 50
1
35
(2)(5a ) − (2) b =
(2)
2
2
i.e.
10a – b = 35
Thus,
5a + 6b = 50
(1)
and
10a – b = 35
(2)
10a + 12b = 100
(3)
2 × equation (1) gives:
(3) – (2) gives:
Substituting in (1) gives:
Since
13b = 65
5a + 30 = 50
1
1 1
= a then p = =
p
a 4
Thus, p =
and since
i.e.
from which,
5a = 50 – 30 = 20
b=
65
=5
13
and
a=
20
=4
5
1
1 1
= b then q = =
q
b 5
1
1
and q = and may be checked by substituting into both of the original equations
4
5
192
© 2014, John Bird
4. Solve the simultaneous equations
3
5
+
= 1.1
y
x
3 7
–
= –1.1
y
x
Let
1
= a and
x
1
=b
y
Thus, the equations become:
5a + 3b = 1.1
(1)
and
3a – 7b = –1.1
(2)
3 × equation (1) gives:
15a + 9b = 3.3
(3)
5 × equation (2) gives:
15a – 35b = –5.5
(4)
(3) – (4) gives:
44b = 8.8
Substituting in (1) gives:
Since
5a + 0.6 = 1.1
1
1
= a then x = = 10
a
x
and since
i.e.
from which,
b=
5a = 1.1 – 0.6 = 0.5
8.8 1
=
44 5
and
a=
0.5 1
=
5 10
1
1
= b then y = = 5
y
b
Thus, x = 10 and y = 5 and may be checked by substituting into both of the original equations
5. Solve the simultaneous equations
c +1 d + 2
–
+1=0
3
4
13
1− c 3 − d
+
+
=0
5
4
20
Rearranging gives:
and
(12)
(20)
c +1
d +2
− (12)
+ (12)(1) =
0 i.e. 3(c + 1) – 4(d + 2) + 12 = 0
4
3
1− c
3− d
13
+ (20)
+ (20)
=
0 i.e. 4(1 – c) + 5(3 – d) + 13 = 0
5
4
20
Since 3(c + 1) – 4(d + 2) + 12 = 0
then
3c + 3 – 4d – 8 + 12 = 0
i.e. 3c – 4d = –7
and
then
4 – 4c + 15 – 5d + 13 = 0
i.e. 4c + 5d = 32
Thus,
4(1 – c) + 5(3 – d) + 13 = 0
3c – 4d = –7
(1)
4c + 5d = 32
(2)
193
© 2014, John Bird
4 × equation (1) gives:
12c – 16d = –28
(3)
3 × equation (2) gives:
12c + 15d = 96
(4)
(4) – (3) gives:
31d = 124
Substituting in (1) gives:
3c – 16 = –7
from which,
i.e.
d=
124
=4
31
3c = 16 – 7 = 9
and
c=
9
=3
3
Thus, c = 3 and d = 4 and may be checked by substituting into both of the original equations
3r + 2 2 s − 1 11
–
=
5
5
4
6. Solve the simultaneous equations
3 + 2r
5 − s 15
+
=
3
4
4
Rearranging gives:
(20)
and
3r + 2
2s − 1
11
i.e. 4(3r + 2) – 5(2s – 1) = 44
− (20)
=
(20)
5
4
5
(12)
3 + 2r
5− s
15
i.e.
+ (12)
=
(12)
4
3
4
Since 4(3r + 2) – 5(2s – 1) = 44 then
and
3(3 + 2r) + 4(5 – s) = 45
Thus,
2 × equation (2) gives:
(3) – (1) gives:
Thus, r = 3 and s =
12r + 8 – 10s + 5 = 44
then
9 + 6r + 20 – 4s = 45
i.e. 12r – 10s = 31
i.e.
12r – 10s = 31
(1)
6r – 4s = 16
(2)
12r – 8s = 32
(3)
2s = 1
Substituting in (1) gives:
3(3 + 2r) + 4(5 – s) = 45
12r – 5 = 31
from which,
i.e.
s=
6r – 4s = 16
1
2
12r = 31 + 5 = 36
and
r=
36
=3
12
1
and may be checked by substituting into both of the original equations
2
194
© 2014, John Bird
5
20
=
x+ y
27
7. Solve the simultaneous equations
4
16
=
2x − y
33
Multiplying both sides of the first equation by 27(x + y) gives: i.e.
27( x + y )
5
20
= 27( x + y )
x+ y
27
i.e. 135 = 20x + 20y
i.e. 20x + 20y = 135
(1)
Multiplying both sides of the second equation by 33(2x – y) gives: i.e.
33(2 x − y )
4
16
= 33(2 x − y )
2x − y
33
i.e. 132 = 32x – 16y
i.e. 32x – 16y = 132
(2)
4 × equation (1) gives:
80x + 80y = 540
(3)
5 × equation (2) gives:
160x – 80y = 660
(4)
(3) + (4) gives:
240x = 1200
Substituting in (1) gives:
8. If 5x –
If 5x –
i.e.
x=
20y = 135 – 100 = 35
1200
=5
240
and
y=
35
3
=1
20
4
4
3
xy + 1
5
= 1 and x +
=
find the value of
y
y
y
2
3
=1
y
and if x +
100 + 20y = 135
from which,
4 5
=
y 2
then
5x =
3
+1
y
then x = −
and
x=
3 1
+
5y 5
4 5
+
y 2
3 1
4 5
+ =− +
y 2
5y 5
i.e.
i.e.
3 + 20 25 − 2
=
5y
10
i.e.
23 23
=
5 y 10
and
(23)(10) = (5y)(23)
i.e.
230 = 115y
Equating x values gives:
from which,
y=
3 4 5 1
+ = −
5y y 2 5
230
=2
115
195
© 2014, John Bird
Substituting into the first equation gives: 5x –
Thus,
3
=1
2
i.e.
5x = 2.5
and x =
2.5 1
=
5
2
1
( 2) + 1 1 + 1
xy + 1  2 
=
=1
=
y
2
2
196
© 2014, John Bird
EXERCISE 54 Page 112
1. In a system of pulleys, the effort P required to raise a load W is given by P = aW + b, where a and
b are constants. If W = 40 when P = 12 and W = 90 when P = 22, find the values of a and b.
P = aW + b, hence if W = 40 when P = 12, then:
12 = 40a + b
(1)
and
22 = 90a + b
(2)
if W = 90 when P = 22, then:
Equation (2) – equation (1) gives:
10 = 50a
from which,
a=
10 1
or 0.2
=
50 5
1
12 = 40   + b
5
Substituting in (1) gives:
from which,
i.e. 12 = 8 + b
b=4
Thus, a = 0.2 and b = 4 and may be checked by substituting into both of the original equations
2. Applying Kirchhoff's laws to an electrical circuit produces the following equations:
5 = 0.2I1 + 2(I1 – I 2 )
12 = 3I 2 + 0.4I 2 – 2(I 1 – I 2 )
Determine the values of currents I 1 and I 2
Rearranging 5 = 0.2 I1 + 2 ( I1 − I 2 ) gives:
5= 0.2 I1 + 2 I1 − 2 I 2
i.e. 2.2 I1 − 2 I 2 =
5
Rearranging 12 =3I 2 + 0.4 I 2 − 2 ( I1 − I 2 ) gives: 12 = 3I 2 + 0.4 I 2 − 2 I1 + 2 I 2
i.e.
−2 I1 + 5.4 I 2 =
12
Thus,
and
2 × equation (1) gives:
2.2 × equation (2) gives:
(3) + (4) gives:
2.2 I1 − 2 I 2 =
5
(1)
−2 I1 + 5.4 I 2 =
12
(2)
4.4 I1 − 4 I 2 =
10
(3)
−4.4 I1 + 11.88 I 2 =
26.4
7.88 I 2 = 36.4
197
(4)
from which, I 2 =
36.4
= 4.62
7.88
© 2014, John Bird
Substituting in (1) gives:
2.2 I1 − 9.24 =
5 i.e. 2.2 I1 = 14.24
and I1 =
14.24
= 6.47
2.2
Thus, I1 = 6.47 and I 2 = 4.62 and may be checked by substituting into both of the original
equations
3. Velocity v is given by the formula v = u + at. If v = 20 when t = 2 and v = 40 when t = 7, find the
values of u and a. Hence find the velocity when t = 3.5
v = u + at, hence if v = 20 when t = 2, then:
20 = u + 2a
(1)
and
40 = u + 7a
(2)
if v = 40 when t = 7, then:
Equation (2) – equation (1) gives:
20 = 5a
from which,
a=
Substituting in (1) gives:
20
=4
5
20 = u + 8
from which, u = 12
Thus, a = 4 and u = 12 and may be checked by substituting into both of the original equations
When t = 3.5, velocity, v = u + at = 12 + (4)(3.5) = 26
4. Three new cars and four new vans supplied to a dealer together cost £97 700 and five new cars and
two new vans of the same models cost £103 100. Find the cost of a car and a van.
Let a car = C and a van = V then
2 × equation (2) gives:
(3) – (1) gives:
Substituting in (1) gives:
from which,
3C + 4V = 97 700
(1)
5C + 2V = 103 100
(2)
10C + 4V = 206 200
(3)
7C = 108 500 from which, C =
46 500 + 4V = 97 700 i.e. 4V =
V=
108 500
= 15 500
7
108 500
97 700 – 46 500 = 51 200
7
51 200
= 12 800
4
Hence, a car costs £15 500 and a van costs £12 800
198
© 2014, John Bird
5. y = mx + c is the equation of a straight line of slope m and y-axis intercept c. If the line passes
through the point where x = 2 and y = 2, and also through the point where x = 5 and y = 0.5, find
the slope and y-axis intercept of the straight line.
When x = 2 and y = 2, then
When x = 5 and y = 0.5, then
2 = 2m + c
(1)
0.5 = 5m + c
(2)
i.e. m =
1.5
= –0.5
−3
(1) – (2) gives:
1.5 = –3m
In equation (1):
2 = –1 + c i.e. c = 2 + 1 = 3
Hence, slope m = –0.5 and the y-axis intercept c = 3
6. The resistance R ohms of copper wire at t°C is given by R = R 0 (1 + αt), where R0 is the resistance
at 0°C and α is the temperature coefficient of resistance. If R = 25.44 Ω at 30°C and R = 32.17 Ω
at 100°C, find α and R 0
R = R0 (1 + α t ) thus when R = 25.44 Ω at t = 30°C, then:
and when R = 32.17 Ω at t = 100°C, then:
Dividing equation (2) by equation (1) gives:
i.e.
32.17 (1 + 100α )
=
25.44 (1 + 30α )
and
i.e.
i.e.
from which,
Substituting in (1) gives:
from which,
25.44 = R0 (1 + 30α )
(1)
32.17 = R0 (1 + 100α )
(2)
32.17 R0 (1 + 100α )
=
25.44 R0 (1 + 30α )
(32.17)(1 + 30α=
) (25.44)(1 + 100α )
32.17 + 965.1α = 25.44 + 2544α
32.17 – 25.44 = 2544α – 965.1α
6.73 = 1578.9α
α=
6.73
= 0.00426
1578.9
25.44 = R0 [1 + (30)(0.00426) ] =
R0 (1.1278 )
199
© 2014, John Bird
from which,
R0 =
25.44
= 22.56
1.1278
Thus, α = 0.00426 and R0 = 22.56 Ω and may be checked by substituting into both of the original
equations.
7. The molar heat capacity of a solid compound is given by the equation c = a + bT. When c = 52,
T = 100 and when c = 172, T = 400. Find the values of a and b.
c = a + bT, hence if c = 52 when T = 100, then:
and
if c = 172 when T = 400, then:
Equation (2) – equation (1) gives:
52 = a + 100b
(1)
172 = a + 400b
(2)
120 = 300b
120
= 0.40
300
from which,
b=
Substituting in (1) gives:
52 = a + 40
from which, a = 12
Thus, a = 12 and b = 0.40 and may be checked by substituting into both of the original equations
8. In an engineering process two variables p and q are related by: q = ap + b/p, where a and b are
constants. Evaluate a and b if q = 13 when p = 2 and q = 22 when p = 5
If q = 13 when p = 2 then
13 = 2a + b/2
or
26 = 4a + b
If q = 22 when p = 5 then
22 = 5a + b/5
or 110 = 25a + b
(2) – (1) gives:
84 = 21a from which, a =
Substituting in (1) gives:
26 = 16 + b
(1)
(2)
84
=4
21
i.e. b = 26 – 16 = 10
9. In a system of forces, the relationship between two forces F1 and F2 is given by:
5F1 + 3F2 + 6 = 0
3F1 + 5F2 + 18 = 0
Solve for F1 and F2
5F1 + 3F2 + 6 = 0
(1)
200
© 2014, John Bird
3F1 + 5F2 + 18 = 0
(2)
5 × (1) gives:
25F1 + 15F2 + 30 = 0
(3)
3 × (1) gives:
9F1 + 15F2 + 54 = 0
(4)
(3) – (4) gives:
16F1 – 24 = 0
i.e. 16F1 = 24
Substituting in (1) gives:
7.5 + 3F2 + 6 = 0
from which,
F2 =
from which, F1 =
24
= 1.5
16
i.e. 3F2 = –7.5 – 6 = –13.5
−13.5
= –4.5
3
10. For a balanced beam, the equilibrium of forces is given by: R1 + R2 =
12.0 kN
As a result of taking moments:
0.2 R1 + 7 × 0.3 + 3 × 0.6 =
0.8 R2
Determine the values of the reaction forces R1 and R2
Rearranging gives:
5 × (2) gives:
(1) – (3) gives:
from which,
Substituting in (1) gives:
Hence,
R1 + R2 =
12.0
(1)
0.2 R1 − 0.8 R2 =
−3.9
(2)
R1 − 4.0 R2 =
−19.5
(3)
5.0 R2 = 31.5
R2 =
31.5
= 6.3 kN
5
R1 + 6.3 =
12.0
R1 = 12.0 – 6.3 = 5.7 kN
201
© 2014, John Bird
EXERCISE 55 Page 113
1.
Solve the simultaneous equations
x + 2 y + 4z =
16
2 x − y + 5z =
18
3x + 2 y + 2 z =
14
(1) – (3) gives:
2 × (2) gives:
(1) + (5) gives:
7 × (4) gives:
(6) – (7) gives:
Substituting in (6) gives:
i.e.
(1)
2 x − y + 5z =
18
(2)
3x + 2 y + 2 z =
14
(3)
–2x + 2z = 2
i.e.
(4)
4x – 2y + 10z = 36
(5)
5x + 14z = 52
(6)
–14x + 14z = 14
(7)
19x = 38
from which, x =
38
=2
19
from which, z =
42
=3
14
10 + 14z = 52
14z = 52 – 10 = 42
Substituting in (1) gives:
2.
x + 2 y + 4z =
16
2 + 2y + 12 =16
2y = 16 – 2 – 12 = 2 and y =
Solve the simultaneous equations
2
=1
2
2x + y − z =
0
3x + 2 y + z =
4
5x + 3 y + 2 z =
8
(1) + (2) gives:
2 × (1) gives:
2x + y − z =
0
(1)
3x + 2 y + z =
4
(2)
5x + 3 y + 2 z =
8
(3)
5x + 3y = 4
(4)
4x + 2y – 2z = 0
(5)
202
© 2014, John Bird
(3) + (5) gives:
9x + 5y = 8
(6)
5 × (4) gives:
25x + 15y = 20
(7)
3 × (6) gives:
27x + 15y = 24
(8)
(8) – (7) gives:
Substituting in (4) gives:
i.e.
i.e.
from which, x =
4
=2
2
from which, y =
−6
= –2
3
10 + 3y = 4
3y = 4 – 10 = –6
Substituting in (1) gives:
3.
2x = 4
4 + –2 – z = 0
z=2
Solve the simultaneous equations
3x + 5 y + 2 z =
6
x − y + 3z =
0
2 x + 7 y + 3z =
−3
3x + 5 y + 2 z =
6
(1)
x − y + 3z =
0
(2)
2 x + 7 y + 3z =
−3
(3)
x + 8y = –3
(4)
3 × (1) gives:
9x + 15y + 6z = 18
(5)
2 × (2) gives:
2x – 2y + 6z = 0
(6)
(3) – (2) gives:
(5) – (6) gives:
7x + 17y = 18
(7)
7 × (4) gives:
7x + 56y = – 21
(8)
(7) – (8) gives:
Substituting in (7) gives:
i.e.
Substituting in (1) gives:
–39y = 39
from which, y =
39
= –1
− 39
7x – 17 = 18
7x = 18 + 17 = 35
from which, x =
35
=5
7
15 – 5 + 2z = 6
203
© 2014, John Bird
i.e.
4.
2z = 6 – 15 + 5 = –4
Solve the simultaneous equations
from which, z =
−4
= –2
2
2 x + 4 y + 5z =
23
3x − y − 2 z =
6
4 x + 2 y + 5z =
31
2 × (2) gives:
(3) + (4) gives:
2 x + 4 y + 5z =
23
(1)
3x − y − 2 z =
6
(2)
4 x + 2 y + 5z =
31
(3)
6x – 2y – 4z = 12
(4)
10x + z = 43
(5)
8x + 4y + 10z = 62
(6)
(6) – (1) gives:
6x + 5z = 39
(7)
5 × (5) gives:
50x + 5z = 215
(8)
2 × (3) gives:
(8) – (7) gives:
Substituting in (7) gives:
i.e.
44x = 176
5z = 39 – 24 = 15
8 + 4y + 15 = 23
i.e.
4y = 23 – 8 – 15
Solve the simultaneous equations
176
=4
44
24 + 5z = 39
Substituting in (1) gives:
5.
from which, x =
from which, z =
15
=3
5
from which, y = 0
2x + 3y + 4z = 36
3x + 2y + 3z = 29
x + y + z = 11
2 × (2) gives:
2x + 3y + 4z = 36
(1)
3x + 2y + 3z = 29
(2)
x + y + z = 11
(3)
2x + 2y + 2z = 22
(4)
204
© 2014, John Bird
(1) – (4) gives:
y + 2z = 14
3 × (3) gives:
3x + 3y + 3z = 33
(6) – (2) gives:
y=4
Substituting in (5) gives:
4 + 2z = 14
(5)
(6)
i.e.
2z = 14 – 4 = 10
Substituting in (1) gives:
2x + 12 + 20 = 36
i.e.
2x = 36 – 12 – 20 = 4
6.
Solve the simultaneous equations
from which, z =
10
=5
2
from which, x =
4
=2
2
4x + y + 3z = 31
2x – y + 2z = 10
3x + 3y – 2z = 7
(1) + (2) gives:
4x + y + 3z = 31
(1)
2x – y + 2z = 10
(2)
3x + 3y – 2z = 7
(3)
6x + 5z = 41
(4)
6x – 3y + 6z = 30
(5)
9x + 4z = 37
(6)
3 × (4) gives:
18x + 15z = 123
(7)
2 × (6) gives:
18x + 8z = 74
(8)
3 × (2) gives:
(3) + (5) gives:
(7) – (8) gives:
Substituting in (4) gives:
7z = 49
from which, z =
49
=7
7
6x + 35 = 41
i.e.
6x = 41 – 35 = 6
Substituting in (1) gives:
4 + y + 21 = 31
i.e.
y = 31 – 4 – 21 = 6
205
from which, x =
6
=1
6
© 2014, John Bird
7.
Solve the simultaneous equations
5x + 5y – 4z = 37
2x – 2y + 9z = 20
–4x + y + z = – 14
2 × (3) gives:
(2) + (4) gives:
5 × (3) gives:
(1) – (6) gives:
5x + 5y – 4z = 37
(1)
2x – 2y + 9z = 20
(2)
–4x + y + z = –14
(3)
–8x + 2y + 2z = –28
(4)
–6x + 11z = –8
(5)
–20x + 5y + 5z = –70
(6)
25x – 9z = 107
(7)
9 × (5) gives:
–54x + 99z = –72
(8)
11 × (7) gives:
275x – 99z = 1177
(9)
(8) + (9) gives:
221x = 1105
Substituting in (7) gives:
125 – 107 = 9z
Substituting in (1) gives:
25 + 5y – 8 = 37
8.
5y = 37 – 25 + 8 = 20
Solve the simultaneous equations
1105
=5
221
125 – 9z = 107
i.e.
i.e.
from which, x =
from which, z =
18
=2
9
from which, y =
20
=4
4
6x + 7y + 8z = 13
3x + y – z = –11
2x – 2y – 2z = –18
6x + 7y + 8z = 13
2 × (2) gives:
(1)
3x + y – z = –11
(2)
2x – 2y – 2z = –18
(3)
6x + 2y – 2z = –22
(4)
206
© 2014, John Bird
(3) + (4) gives:
7 × (2) gives:
8x – 4z = –40
(5)
21x + 7y – 7z = –77
(6)
(1) – (6) gives:
–15x + 15z = 90
(7)
15 × (5) gives:
120x – 60z = –600
(8)
4 × (7) gives:
–60x + 60z = 360
(9)
(8) + (9) gives:
Substituting in (5) gives:
i.e.
i.e.
9.
from which, z =
8
=2
4
–24 + 7y + 16 = 13
7y = 13 + 24 – 16 = 21
Solve the simultaneous equations
240
= –4
60
–32 – 4z = –40
40 – 32 = 4z
Substituting in (1) gives:
from which, x = −
60x = –240
from which, y =
21
=3
7
3x + 2y + z = 14
7x + 3y + z = 22.5
4x – 4y – z = – 8.5
3x + 2y + z = 14
(1)
7x + 3y + z = 22.5
(2)
4x – 4y – z = – 8.5
(3)
(2) + (3) gives:
11x – y = 14
(4)
(3) + (1) gives:
7x – 2y = 5.5
(5)
22x – 2y = 28
(6)
2 × (4) gives:
(6) – (5) gives:
Substituting in (4) gives:
15x = 22.5
from which, x =
22.5
= 1.5
15
16.5 – y = 14
i.e.
16.5 – 14 = y
Substituting in (1) gives:
4.5 + 5 + z = 14
i.e.
z = 14 – 4.5 – 5 = 4.5
207
i.e. y = 2.5
© 2014, John Bird
10. Kirchhoff’s laws are used to determine the current equations in an electrical network and result
in the following:
−31
i1 + 8 i2 + 3 i3 =
−5
3 i1 − 2 i2 + i3 =
2 i1 − 3 i2 + 2 i3 =
6
Determine the values of i1 , i2 and i3
i1 + 8 i2 + 3 i3 = –31
(1)
3 i1 – 2 i2 + i3 = –5
(2)
2 i1 – 3 i2 + 2 i3 = 6
(3)
(2) × (2) gives:
6 i1 – 4 i2 + 2 i3 = –10
(4)
(4) – (3) gives:
4 i1 – i2 = –16
(5)
9 i1 – 6 i2 + 3 i3 = –15
(6)
3 × (2) gives:
(6) – (1) gives:
2 × (5) gives:
(7) – (8) gives:
Substituting in (5) gives:
8 i1 – 14 i2 = 16
(7)
8 i1 – 2 i2 = –32
–12 i2 = 48
(8)
from which, i2 = −
48
= –4
12
4 i1 + 4 = –16
i.e.
4 i1 = –16 – 4 = –20
Substituting in (1) gives:
–5 – 32 + 3 i3 = –31
i.e.
3 i3 = –31 + 5 + 32 = 6
208
i.e. i1 = −
i.e. i3 =
20
= –5
4
6
=2
3
© 2014, John Bird
11. The forces in three members of a framework are F1 , F2 and F3 . They are related by the
simultaneous equations shown below.
1.4F1 + 2.8F2 + 2.8F3 = 5.6
4.2F1 – 1.4F2 + 5.6F3 = 35.0
4.2F1 + 2.8F2 – 1.4F3 = –5.6
Find the values of F1, F2 and F3 .
(1) – (2) gives:
2 × (2) gives:
(1) + (5) gives:
1.4F1 + 2.8F2 + 2.8F3 = 5.6
(1)
4.2F1 – 1.4F2 + 5.6F3 = 35.0
(2)
4.2F1 + 2.8F2 – 1.4F3 = –5.6
(3)
–2.8F1 + 4.2F3 = 11.2
(4)
8.4F1 – 2.8F2 + 11.2F3 = 70.0
(5)
9.8F1 + 14F3 = 75.6
(6)
9.8 × (4) gives:
–27.44F1 + 41.16F3 = 109.76
(7)
2.8 × (6) gives:
27.44F1 + 39.20F3 = 211.68
(8)
(7) + (8) gives:
Substituting in (4) gives:
i.e.
Substituting in (1) gives:
i.e.
80.36F3 = 321.44
from which, F3 =
321.44
=4
80.36
–2.8F1 + 16.8 = 11.2
16.8 – 11.2 = 2.8F1
i.e. F1 =
5.6
=2
2.8
2.8 + 2.8F2 + 11.2 = 5.6
2.8F2 = 5.6 – 2.8 – 11.2 = –8.4
209
i.e. F2 =
8.4
= –3
2.8
© 2014, John Bird