8th Grade
Chapter 8 – Solving Systems of Equations
8.2 By Substitution: Page 362
8.2 Solve Equations by Substitution:
Graphing a solution may not always be accurate or
convenient. A second technique for solving Systems of
Equations is using the SUBSTITUTION Method.
This is not a Guess a number or Trial and Error:
Example: x + y = 6 and x = y + 2. Assume only Positive
whole values, you could figure the possible values are:
(0, 6), (1, 5), (2, 4) and (3, 3) 0r (6, 0), (5, 1), (4, 2). This
would be a simple problem with a simple solution but
you need other methods for when the problems
become more complex!
With Equations with 2 variables, you need two equations:
x+y=6
Set a
Set b
Set c
x=y+2
This set of equations only has coefficients of “1” for x and y so substitution is very simple:
Substitute for x:
(y + 2) + y = 6
Combine terms:
2y + 2 = 6
Solve:
2y = 6 - 2
Divide by 2
2y = 4
y=2
Solve for X in the original problem:
x=y+2
x=2+2
x=4
Check solution is other equation:
x+y=6
4+2=6
6 = 6 : Valid
Try This: Page 362
Equations
Substitute one into the other Solve for 2nd
Check in other equation
variable
x+y=5
(y + 1) + y = 5
x=y+1
x+y=5
and
2y + 1 = 5
x=2+1
3+2=5
x=y+1
2y = 4
x=3
5=5
y=2
a–b=4
a – (2 -5a) = 4
b = 2 – 5a
a–b=4
and
a -2 + 5a = 4
b = 2 – 5(1)
1 – (-3) = 4
b = 2 – 5a
a + 5a = 6
b=2–5
1+3=4
6a
=6
b = -3
4=4
a=1
y=x+2
(2x -1) = x + 2
y = 2x – 1
y=x+2
and
2x – 1 = x + 2
y = 2(3) – 1
5=3+2
y = 2x - 1
2x –x = 2 + 1
y=6–1
5=5
x =3
y=5
8th Grade
Chapter 8 – Solving Systems of Equations
8.2 By Substitution: Page 362
Sometimes the equations and numbers are not as simple or easy to work out:
x – 2y = 6
3x + 2y = 4
Basic Method: x – 2y = 6  x = 2y + 6
Substitute in:
3(2y + 6) + 2y = 4
Solve:
6y + 18 + 2y = 4
Combine Terms:
8y = -14
Solve
y=−
14
8
Substitute into Equation 1:
-
=
𝟕
(yes!, this is an improper fraction)
𝟒
x – 2y = 6
7
x – 2(− 4) = 6
x+
14
4
=6
x =6x =
Substitute into Equation 2:
24
4
14
4
-
14
4
=
10
4
=
5
2
=𝟐
𝟏
𝟐
3x + 2y = 4
5
2
15
( 2 ) + (−
15
( ) + (−
2
8
(2)= 4
7
4
14
)=4
4
7
)=4
2
3 ( ) + 2(− ) = 4
𝟒= 𝟒
There was another substitution for this problem in that both equations had a common term of 2y
So: rewrite: x – 2y = 6  x – 6 = 2y
Substitute into 3x + 2y = 4
3x + (x – 6) = 4
3x + x – 6 = 4
4x = 10
X=
10
4
=
Solve for y: 3x + 2y = 4
5
2
=𝟐
𝟏
𝟐
5
3 (2) + 2y = 4
15
( 2 ) + 2𝑦 = 4
15
2
8
15
−(2)
2
7
−( )
2
2y = 4 - ( )
2y =
2y=
Y=−(
7
7
) ÷ 2 = − (2) 𝑥
2
1
𝟕
= −( )
2
𝟒
Yet a 3rd substitution for this problem was to use the Common Term set the two equations equal to each
other: x – 2y = 6  x – 6 = 2y and 3x + 2y = 4  2y = -3x + 4
2y = 2y or
(x – 6) = (-3x + 4)
4x = 10
10
5
𝟏
X = 4 = 2 = 𝟐𝟐
8th Grade
Chapter 8 – Solving Systems of Equations
8.2 By Substitution: Page 362
Try This : page 363
d) x – 2y = 8 and 2x + y = 8
e) 4x – y = 5 and 2x + y = 10
f) y = x + 5 and 2x + y = 8
f) 3x + 4y = 2 and 2x – y = 5
Exercises 8-2 page 365
1) x + y = 4 and y = 2x + 1
2) x + y = 10 and y = x + 8
3) x = y – 1 and y = 4 – 2x
4) x = y + 6 and y = -2 - x
5) y = 2x – 5 and 3y = x - 5
6) y = 2x + 1 and x + y = -2
7) x = -2y and x = 2 – 4y
8) r = -3s and r = 10 – 4s
9) x = 3y – 4 and 2x – y = 7
10) s + t = - 4 and s – t = 2
11) x – y = 6 and x + y = -2
12) y – 2x = -6 and 2y – x = 5
13) x – y = 5 and x + 2y = 7
14) 2x + 3y = -2 and 2x – y = 9
15) x + 2y – 10 and 3x + 4y = 8
16) x – y = -3 and 2x + 3y = -6
17) 3b + 2a =2 and -2b + a = 8
18) r – 2s = 0 and 4r – 3s =15
19) y – 2x =0 and 3x + 7y = 17
20) x – 3y = 7 and -3x + 16y =28
21) 8x + 4y =6 and 4x = 3 – y
Translate into a system of equations and solve:
22) The sum of two numbers is 28 and one number is 3 times more than the other.
23) The Sum of two numbers is 36 and one number is 2 more than the other
24) Find two numbers whose sum is 58 and whose difference is 16.
25) Find two numbers whose sum is 66 and whose difference is 8.
26) The difference between two numbers is 16. Three times the larger is seven times the smaller.
27) The difference between two numbers is 18. The sum is twice the smaller number and three
times the larger number is 74.
Solve and graph
28) 3y + 3x = 14 and y = -x + 4
29) y = x + 5 and -3x + 3y = 15
30) Is (2, -3) a solution for: x + 3y = -7 and –x + y = -5 and 2x – y = 1?