1. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
Solve using substitution. y = 2x + 2
y = –3x + 4
Step 1: Write an equation containing only one variable and solve.
y = 2x + 2 Start with one equation.
–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
7-2

2. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Since x = 0.4 and y = 2.8, the solution is (0.4, 2.8).
Check: See if (0.4, 2.8) satisfies y = –3x + 4 since y = 2x + 2 was used in Step 2.
–3(0.4) + 4 Substitute (0.4, 2.8) for (x, y) in the equation. –
2.8 = 2.8
7-2

3. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
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 –1 Add 2x to each side.
Step 2: Write an equation containing only one variable and solve.
4x + 2y = 12 Start with the other equation.
4x + 2(2x –1) = 12 Substitute 2x –1 for y in that equation.
4x + 4x –2 = 12 Use the Distributive Property.
8x = 14 Combine like terms and add 2 to each side.
x = 1.75 Divide each side by 8.
7-2

4. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Step 3: Solve for y in the other equation.
–2(1.75) + y = 1 Substitute 1.75 for x.
–3.5 + y = –1 Simplify.
y = 2.5 Add 3.5 to each side.
Since x = 1.75 and y = 2.5, the solution is (1.75, 2.5).
7-2

5. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
A youth group with 26 members is going to the beach. There will also be five chaperones that will each drive a van or a car. Each van seats 7 persons, including the driver. Each car seats 5 persons, including the driver. How many vans and cars will be needed?
Let v = number of vans and c = number of cars.
Drivers v c = 5
Persons v c = 31
Solve using substitution.
Step 1: Write an equation containing only one variable.
v + c = 5 Solve the first equation for c.
c = –v + 5
7-2

6. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Step 2: Write and solve an equation containing the variable v.
7v + 5c = 31
7v + 5(–v + 5) = 31 Substitute –v + 5 for c in the second equation.
7v – 5v + 25 = 31 Solve for v.
2v + 25 = 31
2v = 6
v = 3
Step 3: Solve for c in either equation.
3 + c = 5 Substitute 3 for v in the first equation.
c = 2
7-2

7. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-2
(continued)
Three vans and two cars are needed to transport 31 persons.
Check: Is the answer reasonable? Three vans each transporting 7 persons is 3(7), of 21 persons. Two cars each transporting 5 persons is 2(5), or 10 persons. The total number of persons transported by vans and cars is , or 31. The answer is correct.
7-2

8. Solve each system using substitution.
ALGEBRA 1 LESSON 7-2
Solve each system using substitution.
1. 5x + 4y = x + y = m – 2n = 7
y = 5x 2x – y = 6 3m + n = 4
(0.2, 1)
(2, 2)
(1.25, 0.25)
7-2

9. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-3
Solve each system using substitution.
1. y = 4x – 3 2. y + 5x = y = –2x + 2
y = 2x y = 7x – x – 17 = 2y
7-3

10. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-3
Solutions
1. y = 4x – 3
y = 2x + 13
Substitute 4x – 3 for y in the second equation.
4x – 3 = 2x + 13
4x – 2x – 3 = 2x – 2x + 13
2x – 3 = 13
2x = 16
x = 8
y = 4x – 3 = 4(8) – 3 = 32 – 3 = 29
Since x = 8 and y = 29, the solution is (8, 29).
7-3

11. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-3
Solutions (continued)
2. y + 5x = 4
y = 7x – 20
Substitute 7x – 20 for y in the first equation.
y + 5x = 4
7x – x = 4
12x – 20 = 4
12x = 24
x = 2
y = 7x – 20 = 7(2) – 20 = 14 – 20 = –6
Since x = 2 and y = –6, the solution is (2, –6).
7-3

12. Solving Systems Using Substitution
ALGEBRA 1 LESSON 7-3
Solutions (continued)
3. y = –2x + 2
3x – 17 = 2y
Substitute –2x + 2 for y in the second equation.
3x – 17 = 2(–2x + 2)
3x – 17 = –4x + 4
7x – 17 = 4
7x = 21
x = 3
y = –2x + 2 = –2(3) + 2 = –6 + 2  –4
Since x = 3 and y = –4, the solution is (3, –4).
7-3