Title: Lesson 1–1
Author: Teri Willard
Last saved by: MHE

Lesson 3–2

Example 1  Solve by Using Substitution
Use substitution to solve the system of equations.
2x – y = –5
3y – 5x = 14

Solve the first equation for y in terms of x.
2x – y = –5 | First equation
–y = –5 – 2x | Subtract 2x from each side.
y = 5 + 2x | Multiply each side by –1.

Substitute 5 + 2x for y in the second equation and solve for x.
3y – 5x = 14 | Second equation
3(5 + 2x) – 5x = 14 | Substitute 5 + 2x for y.
15 + 6x – 5x = 14 | Distributive Property
15 + x = 14 | Simplify.
x = –1 | Subtract 15 from each side.

Now, substitute the value for x in either original equation and solve for y.
2x – y | = | -5 | First equation
2(–1) – y | = | -5 | Replace x with –1.
–2 – y | = | -5 | Simplify.
–y | = | -3 | Add 2 to each side.
y | = | 3 | Multiply each side by –1.

The solution of the system is (–1, 3).

Standardized Test EXAMPLE
Example 2  Solve by Substitution
GRIDDABLE    Maryn bought chairs and desks for her employees. The price of the chairs were $95 each and the price of the desks were $325 each. She bought a total of 23 desks and chairs and spent $4025. How many chairs did she buy? Show your work.

Read the Test Item
You are asked to find the number of chairs.

Solve the Test Item
Step 1	Define variables and write the system of equations. Let x represent the number of chairs and y represent the number of desks.
x + y = 23 | The total number of desks and chairs was 23.
95x + 325y = 4025 | The total price was $4025.

Step 2	Solve one of the equations for one of the variables in terms of the other. Since the coefficient of y is 1 and you are asked to find the value of x, it makes sense to solve the first equation for y in terms of x.
x + y = 23 | First equation
y = 23 – x | Subtract x from each side.

Step 3	Substitute 23 – x  for y in the second equation.
95x + 325y = 4025 | Second equation
95x + 325(23 – x) = 4025 | Substitute 23 - x for y.
95x + 7475 – 325x = 4025 | Distributive Property
-230x + 7475 = 4025 | Simplify.
-230x = -3450 | Subtract 7475 from each side.
x = 15 | Divide each side by -230.

1

5

Step 4	Maryn bought 15 chairs for her employees. Record your answer in a bubble grid like the one at the right.

Example 3  Solve by Using Elimination
Use the elimination method to solve the system of equations.
3p + 4q = 0
p – 4q = –8

The coefficients of q for the first equation and the second equation are additive inverses. This means that when you add them the result will be 0 and the variable q will be eliminated.

3p + 4q | = | 0 |
(+) p –  4q | = | -8 |
4p | = | -8 | Add the equations.
p | = | -2 | Divide each side by 4.

Now find q by substituting –2 for p in either original equation.
3p + 4q = 0 | First equation
3(–2) + 4q = 0 | Replace p with –2.
–6 + 4q = 0 | Multiply.
4q = 6 | Add 6 to each side.
q = | Divide each side by 4 and simplify.
The solution is .

Example 4  Multiply, Then Use Elimination
Use the elimination method to solve the system of equations.
2x + 3y = 6
5x – 5y = 65

Multiply the first equation by 5 and the second equation by 2. Then subtract the equations to eliminate 
the x variable.
2x + 3y = 6 | Multiply by 5. | 10x + 15y =     30
5x – 5y = 65 | Multiply by 2. | (–) 10x – 10y =   130
| | 25y = –100	Subtract the equations.
| | y = –4	Divide each side by 25.

Replace y with –4 and solve for x.
2x + 3y = 6 | First equation
2x + 3(–4) = 6 | Replace y with –4.
2x – 12 = 6 | Multiply.
2x = 18 | Add 12 to each side.
x = 9 | Divide each side by 2.

The solution is (9, –4).

Example 5  Inconsistent System
Use the elimination method to solve the system of equations.
4x – y = -7
2x - 1/2y = 3
Use multiplication to eliminate x

4x – y = -7	  			4x – y = -7
2x - y = 3	Multiply by 2. 	4x – y = 6
	           0 =  -1	Add the equations.

Since there are no values of x and y that will make the equation  0 = -1 true, there are no solutions for this system of equations.