## Precalculus (10th Edition)

$\left\{\left(x,\dfrac{10-2x}{5}\right)|x\text{ is any real number}\right\}$
We are given the system of equations: $\begin{cases} 2x+5y=10\\ 4x+10y=20 \end{cases}$ Use the elimination method. Multiply the first equation by -2 and add it to the second to eliminate $y$ and find $x$: $-2(2x+5y)+4x+10y=-2(10)+20$ $-4x-10y+4x+10y=-20+20$ $0=0$ We got an identity; therefore the system has infinitely many solutions. Determine $y$ in terms of $x$: $2x+5y=10$ $5y=10-2x$ $y=\dfrac{10-2x}{5}$ The solution set is: $\left\{\left(x,\dfrac{10-2x}{5}\right)|x\text{ is any real number}\right\}$