91Ó°ÊÓ

The given system of inequalities are:

$\left\{\begin{array}{l}y≤-\frac{1}{4}x+2\\ x+4y≤4\end{array}\right.$

We will graph the inequality by graphing $y=-\frac{1}{4}x+2$and the boundary will be a solid line because it contains less than equal to sign.

We will graph the inequality by graphing $x+4y=4$on the same cartesian plane and the boundary will be a solid line because it contains greater than equal to sign.

The graph is given as:

The solution of the given system of inequalities is the common region between inequalities represented by the dark region (dark blue) in the above graph.

Substitute the ordered pair $\left(-5,0\right)$in both equalities as it is in the solution region.

For inequality$y≤\frac{-1}{4}x+2$

$$\begin{gathered} y≤-\frac{1}{4}x+2 \\ 0≤-\frac{1}{4}×(-5)+2 \\ 0≤\frac{13}{5} \end{gathered}$$

This is true

For inequality $x+4y≤4$

$$\begin{gathered} x+4y≤4 \\ -4+4×0≤4 \\ -4≤4 \end{gathered}$$

This is true

Thus, the solution set is correct.