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: 16

Answer

The solution set is $\left\{(5, 0)\right\}$.

Work Step by Step

By adding $7y$ to both sides of the equation, first equation is equivalent to \begin{align*} 3x-7y+7y&=15+7y\\ 3x&=15+7y\end{align*} Since $3x=15+7y$, substitute $15+7y$ to $3x$ in the second equation to obtain: \begin{align*} 3x+7y&=15\\ 15+7y+7y&=15\\ 15+14y&=15\\ 14y&=0\\ y&=0 \end{align*} Plug $y=0$ into $3x-7y=15$, the solve for $x$ o obtain: \begin{align*}3x-7y&=15\\ 3x-7(0)&=15\\ 3x&=15\\ x&=\frac{15}{3}\\ x&=5\end{align*}
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