1. 6-2 Solving Systems Using Substitution
Hubarth
Algebra

2. Ex 1 Solve Systems by Substitution
Solve using substitution. y = 2x + 2
y = –3x + 4
Step 1: Write an equation containing only one variable and solve.
y = –3x + 4
y = 2x + 2
–3x + 4 = 2x + 2 Substitute –3x + 4 for y in that equation.
4 = 5x + 2 Add 3x to each side.
2 = 5x Subtract 2 from each side.
0.4 = x Divide each side by 5.
Step 2: Solve for the other variable.
y = 2(0.4) + 2 Substitute 0.4 for x in either equation.
y = Simplify.
y = 2.8
The solution is ( 0.4, 2.8)

3. Ex 2 Using Substitution and the Distributive Property
Solve using substitution.
–2x + y = –1
4x + 2y = 12
Step 1: Solve the first equation for y because it has a coefficient of 1.
–2x + y = –1
y = 2x – Add 2x to each side.
Step 2: Write an equation containing only one variable and solve.
4x + 2y = Start with the other equation.
4x + 2(2x –1) = Substitute 2x –1 for y in that equation.
4x + 4x – 2 = Use the Distributive Property.
8x = Combine like terms and add 2 to each side.
x = Divide each side by 8.
Step 3: Solve for y in the other equation.
–2(1.75) + y = –1 Substitute 1.75 for x.
–3.5 + y = – Simplify.
y = Add 3.5 to each side.
The solution is (1.75, 2.5)

4. Ex 3 Solve System by Using Substitution
y = 4x – 1
5 = 6x – y ( )
( )
5 = 6x – (4x – 1)
y = 4 (2) – 1
5 = 6x – 4x + 1
y = 8 – 1
y = 7
5 = 2x +1
4 = 2x
x = 2
(2, 7) is the solution of the system

5. Practice
y = x – 3
x + y = 5
2. 2x + y = 1
x = -2y + 5
(4, 1)
(-1, 3)
4. x = 4y – 1
3x + 5y = 31
3. x – 5 = y
2x – 3y = 7
(7, 2)
(8, 3)