College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 5 - Section 5.1 - Systems of Linear Equations - 5.1 Exercises - Page 486: 19

Answer

The solution set is $\left\{(0, 4)\right\}$.

Work Step by Step

To eliminate $y$, we multiply $3$ to both sides of the first euation to obtain: $$3(3x-y)=3(-4) \implies 9x-3y=-12$$ The system of equations becomes \begin{align*} 9x-3y&=-12\\ x+3y&=12\end{align*} Now, we can add the two equations together to eliminate $y$, to obtain: \begin{align*} &9x-3y=-12\\ &\underline{\space \space x+3y=12}\\ &10x \space \space \space \space \space \space =0 \end{align*} Dividing both sides by $10$ gives $x=0$. Substitute $0$ to $x$ in the first equation to obtain: $3(0)-y=-4 \\ \implies -y=-4\\ \implies y=4$
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