Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 5 - Cumulative Review Exercises - Page 417: 14


The solution is $(0 , 5)$.

Work Step by Step

For this system of equations, we can use the addition method to solve for $x$ and $y$. We first need to transform the equations so that the coefficients of one of the variables differs only in sign. We see that if we multiply the first equation by $3$ and the second equation by $2$, we get coefficients of $6$ and $-6$ for the $y$ term: $$3(3x + 2y) = 3(10)$$ $$2(4x - 3y) = 2(-15)$$ Distribute first: $$9x + 6y = 30$$ $$8x - 6y = -30$$ Now we add the two equations together to get the following solution. The $y$ term cancels out: $$x = 0$$ We plug in $0$ for $x$ into the first equation to solve for $y$: $$3(0) + 2y = 10$$ Multiply first: $$2y = 10$$ Divide both sides by $2$ to solve for $y$: $$y = 5$$ The solution is $(0 , 5)$.
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