Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 4 - Section 4.3 - Solving Systems of Linear Equations by Addition - Exercise Set: 28

Answer

$x=3\\y=12$

Work Step by Step

$\begin{cases} \frac{x}{2}+\frac{y}{8}=3\\ x-\frac{y}{4}=0 \end{cases}$ Multiply the first equation by 8 and the second equation by 4 to get rid of the fractions: $\begin{cases} 4x+y=24\\ 4x-y=0. \end{cases}$ Add the two equations to remove the y: $8x=24$. Solve for x: $x=\frac{24}{8}=3$. Substitute $x=3$ in the second equation: $3-\frac{y}{4}=0$. Solve for y: $\frac{y}{4}=3$ $y=3*4=12$. Our two solutions are $x=3\\y=12$.
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