## Algebra 2 Common Core

Published by Prentice Hall

# Chapter 3 - Linear Systems - 3-2 Solving Systems Algebraically - Practice and Problem-Solving Exercises: 37

#### Answer

$x = \frac{20}{17}$, $y = \frac{19}{17}$ or $(\frac{20}{17}, \frac{19}{17})$

#### Work Step by Step

Given: $2x - 3y = -1$ $3x + 4y = 8$ Multiply both sides of the first equation, $2x - 3y = -1$, by $3$: $6x - 9y = -3$ Multiply both sides of the second equation, $3x + 4y = 8$, by $2$: $6x + 8y = 16$ Subtract the first equation, $6x - 9y = -3$, from the second equation, $6x + 8y = 16$: $6x + 8y - (6x - 9y) = 16 - (-3)$ $6x + 8y - (6x - 9y) = 19$ Subtract each term of the binomial: $6x - 6x + 8y - (-9y) = 19$ $6x - 6x + 8y + 9y = 19$ $17y = 19$ Divide both sides by $17$: $y = \frac{19}{17}$ Substitute $y = \frac{19}{17}$ into the first equation, $2x - 3y = -1$: $2x - 3y = -1$ $2x - 3(\frac{19}{17}) = -1$ $2x - \frac{57}{17} = -1$ Add $\frac{57}{17}$ to both sides: $2x = \frac{40}{17}$ Divide both sides by $2$: $x = \frac{20}{17}$ $x = \frac{20}{17}$, $y = \frac{19}{17}$

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