# Chapter 4 - Section 4.3 - Solving Systems of Linear Equations by Addition - Exercise Set - Page 308: 27

$x=\dfrac{3}{2}$ $y=3$

#### Work Step by Step

$\begin{cases}\dfrac{x}{3}+\dfrac{y}{6}=1\\\dfrac{x}{2}-\dfrac{y}{4}=0\end{cases}$ Multiply the first equation by $6$ and the second equation by $4$ to remove the fractions: $\begin{cases}2x+y=6\\2x-y=0\end{cases}$ Add the two equations to remove the $y$: $4x=6$ Solve for $x$: $x=\dfrac{6}{4}=\dfrac{3}{2}$ Substitute $x=\dfrac{3}{2}$ in the first equation: $2\Big(\dfrac{3}{2}\Big)+y=6$ Solve for $y$: $3+y=6$ $y=6-3=3$ Our two solutions are: $x=\dfrac{3}{2}$ $y=3$

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