Chapter 3, Problem 11QC

To determine

To graph: The linear inequality $x+2y\le 5$ .

Explanation of Solution

Given information:

The inequality $x+2y\le 5$ .

Graph:

Consider the inequality $x+2y\le 5$ .

The boundary of the inequality is the graph of the equation $x+2y=5$ .

Recall that the x- intercepts are the points on x -axis where the graph of the equation intersects the x -axis.

Substitute $y=0$ in the equation $x+2y=5$ ,

  $\begin{array}{c}x+2\left(0\right)=5\\ x=5\end{array}$

Therefore, x- intercept is $5$ .

The graph passes through the point $\left(5,0\right)$ .

Recall that the y- intercepts are the points on x -axis where the graph of the equation intersects the y -axis.

Substitute $x=0$ in the equation $x+2y=5$ ,

  $\begin{array}{c}0+2y=5\\ 2y=5\\ y=\frac{5}{2}\end{array}$

Therefore, y- intercept is $\frac{5}{2}$ .

The graph passes through the point $\left(0,\frac{5}{2}\right)$ .

Plot the ordered pairs $\left(5,0\right)$ and $\left(0,\frac{5}{2}\right)$ . Join the line joining the points.

The inequality contains the sign of $\le$ , so the boundary will be solid.

Now, choose a point that does not lie on boundary say $\left(0,0\right)$ .

Substitute x as 0 and y as 0 in the equality $x+2y\le 5$ .

  $\begin{array}{c}0+2\left(0\right)\le 5\\ 0\le 5\end{array}$

The above inequality is true so the shaded region will be towards the point $\left(0,0\right)$ .

The graph obtained is provided below,

  

Interpretation:

The graph of the inequality $x+2y\le 5$ contain all the points whose x -coordinate is less than or equal to 5 and y -coordinate is less than or equal to $\frac{5}{2}$ .

To verify the region select any point from it say $\left(-5,0\right)$ . Substitute the value in the inequality,

  $\begin{array}{c}-5+2\left(0\right)\le 5\\ -5\le 5\end{array}$

The above inequality is true. So the region shaded represents the inequality $x+2y\le 5$ .