Page 194 - Math Course 2 (Book 1)

Solving Systems of Equations by Multiplying

Multiply Both Equations to Eliminate
Now substitute –1 for x in either equation.
Use elimination to solve the system of equations.
4x + 3y = 8 First equation
3x + 2y = 10
4(–1) + 3y = 8 x = –1 2x + 5y = 3
–4 + 3y = 8 Simplify. A. (–4, 1)
B. (–1, 4)
–4 + 3y + 4 = 8 + 4 Add 4 to each side C. (4, –1)
3y = 12 Simplify. D. (–4. –1)
3y = 12 Divide each side by 3.
3 3 Answer
y = 4 Simplify.

The solution is (–1, 4), which
Answer matches the result obtained with
Method 1.

Your Turn!

Multiply One Equation to Eliminate

Use elimination to solve the system of equations.
x + 7y = 12
3x – 5y = 10
A. (1, 5)
B. (5, 1)
C. (5, 5)
D. (1, 1)

Answer

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