Exercise 32 Page 404

Practice makes perfect

a We are given a system of linear inequalities and are asked to graph them.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x \geq 2 y \geq - 3 x + y \leq 4 (I) (II) (III)

The boundary line of Inequality (I) is going to be a vertical line at

x = 2 .

The inequality is non-strict, so the boundary line will be solid. Since

x -

values greater than or equal to

2

lie to the right of

2

on a number line, the shaded region will be to the right of the boundary line.

The boundary line of Inequality (II), $y \geq - 3,$ is going to be a horizontal line at $y = - 3 .$ The inequality greater than or equal to is non-strict, so we will use a solid line, and the shaded region will be above the line.

For Inequality (III), we will need to isolate the

y -

variable in order to graph the boundary line.

x + y = 4 \Leftrightarrow y = - x + 4

In order to find which region to shade, we will use the test point

(0, 0) .

If substituting the point into the inequality results in a true statement, we will shade the region that contains the point. Otherwise, we will shade the region that does not contain the point.

x + y \leq 4

SubstituteII

$x = 0$ , $y = 0$

0 + 0 \leq ? 4

AddTerms

Add terms

0 \leq 4 ✓

The inequality is non-strict, so the boundary line will be solid. Since substituting the test point made the inequality true, we will shade the region that contains it.

The solution set of the system is the intersection of all the shaded regions.

b Let's take a look at just the solution set of the system we have graphed in Part A.

We can see that the solution set is a polygon with three sides. Therefore, the solution region can be described as a triangle.

c The vertices of a polygon are where the sides meet, at the corners. Let's take a closer look at those points.

The vertices are located at $(2, 2),$ $(2, - 3),$ and $(7, - 3) .$

d Let's recall the formula for the area of a triangle.

A = \frac{1}{2} b h .

In this formula,

A

is the area,

b

is the length of the base, and

h

is the height of the triangle. To calculate the area, we will need to find the length of the base and the height.

Notice that the two sides are perpendicular. This means we can use one of them as the base and the other as the height of our triangle. Now that we have our base and height lengths, we can substitute their values into the formula.