## Precalculus (6th Edition)

Published by Pearson

# Chapter 9 - Systems and Matrices - 9.1 Systems of Linear Equations - 9.1 Exercises - Page 856: 14

#### Answer

$\{(-2,3)\}$

#### Work Step by Step

1. No variable has 1 as a coefficient, there will be fractions... Express y from the second equation (solve for y) $9y=31+2x\qquad/\div 9$ $y=\displaystyle \frac{31+2x}{9}$ 2. Substitute y in the other equation and solve. $4x+5\displaystyle \cdot\frac{31+2x}{9}=7\qquad/\times 9$ ... gets rid of the fraction ... $36x+5(31+2x)=63$ $36x+155+10x=63\qquad/-155$ $46x=-92\qquad/\div 46$ $x=-2$ 3. Back-substitute into the expression of y (step 1). $y=\displaystyle \frac{31+2(-2)}{9}=\frac{27}{9}=3$ 4. Write the solution as an ordered pair. Solution set: $\{(-2,3)\}$

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