## Algebra: A Combined Approach (4th Edition)

Published by Pearson

# Chapter 4 - Section 4.2 - Solving Systems of Linear Equations by Substitution - Exercise Set: 36

#### Answer

x =$\frac{1}{2}$ y = 3

#### Work Step by Step

x = $\frac{5}{6}$y -2 12x - 5y = -9 substitude x to the second equation 12( $\frac{5}{6}$y - 2 ) - 5y = -9 10y - 24 - 5y = -9 5y = 15 y = 3 substitude y = 3 to first equation x = $\frac{5}{6}$(3) -2 x = $\frac{5}{2}$ - $\frac{4}{2}$ x = $\frac{1}{2}$

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