91Ó°ÊÓ

The boundary line shown in the graph is,

$y=-\frac{1}{2}x+1$

The line $y=-\frac{1}{2}x+1$is the boundary line.

On one side of the line are the points with $y>-\frac{1}{2}x+1$, and on the other side of the line are the points with $y<-\frac{1}{2}x+1$.

Let's test the point $(0,0)$and see which inequality describes its position relative to the boundary line.

At $(0,0)$which inequality is true: $y>-\frac{1}{2}x+1$or $y<-\frac{1}{2}x+1$

Checking for $y>-\frac{1}{2}x+1$, at $x=0,y=0$

$$\begin{gathered} y>-\frac{1}{2}x+1 \\ ⇒0>-\frac{1}{2}.0+1 \\ ⇒0>1 \end{gathered}$$

which is false.

Now, checking $y<-\frac{1}{2}x+1$, at $x=0,y=0$

$$\begin{gathered} y<-\frac{1}{2}x+1 \\ ⇒0<-\frac{1}{2}.0+1 \\ ⇒0<1 \end{gathered}$$

which is true.

Since, $y<-\frac{1}{2}x+1$is true, the side of the line with $(0,0)$is the solution. The shaded region shows the solution of the inequality $y<-\frac{1}{2}x+1$.

Since, the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality$y≤-\frac{1}{2}x+1$