## Precalculus (10th Edition)

$\left\{\left(\dfrac{28-y}{4},y\right)|y\text{ is any real number}\right\}$
We are given the system of equations: $\begin{cases} x+\dfrac{1}{4}y=7\\ 8x+2y=56 \end{cases}$ The augmented matrix associated with the system is: $A=\begin{bmatrix}1&\dfrac{1}{4}&|&7\\8&2&|&56\end{bmatrix}$ Multiply $R_1$ by -8 and add it to $R_2$: $\begin{bmatrix}1&\dfrac{1}{4}&|&7\\0&0&|&0\end{bmatrix}$ As the last line contains only zeros, the system has infinitely many solutions. $x+\dfrac{1}{4}y=7$ $x=7-\dfrac{1}{4}y$ $x=\dfrac{28-y}{4}$ The solution set is: $\left\{\left(\dfrac{28-y}{4},y\right)|y\text{ is any real number}\right\}$