## Precalculus (10th Edition)

$\{(x,y)|x=4-2y,y\text{ is any real number}\}$
We are given the system of equations: $\begin{cases} x+2y=4\\ 2x+4y=8 \end{cases}$ Use the elimination method. Multiply the first equation by $-2$ and add it to the second equation to eliminate $y$ and determine $x$: $\begin{cases} -2(x+2y)=-2(4)\\ 2x+4y=8 \end{cases}$ $\begin{cases} -2x-4y=-8\\ 2x+4y=8 \end{cases}$ $-2x-4y+2x+4y=-8+8$ $0=0$ As we got an identity, the system has infinitely many solutions. We solve the first equation in terms of $y$: $x=4-2y$ The solution set of the system is: $\{(x,y)|x=4-2y,y\text{ is any real number}\}$