Exercise 127 Page 413

Practice makes perfect

We are asked to graph the solution set for all possible values of x in the given inequality. |x+1|≥ 3 To do this, we will create a compound inequality by removing the absolute value. In this case, and since x+1 can be written as x-(- 1), the solution set contains the numbers that make the distance between x and - 1 greater than or equal to 3 in the positive direction or in the negative direction.

x+1 ≥ 3 or x+1≤ - 3 Let's isolate x in both of these cases before graphing the solution set.

Case 1

x+1 ≥ 3

SubIneq

LHS-1≥RHS-1

x≥ 2

This inequality tells us that all values greater than or equal to 2 will satisfy the inequality.

Case 2

x+1 ≤ - 3

SubIneq

LHS-1≤RHS-1

x≤ -4

This inequality tells us that all values less than or equal to - 4 will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the combination of the solution sets. First Solution Set:& x≥ 2 Second Solution Set:& x≤ -4 Combined Solution Set:& x≤ - 4 or x≥ 2

Graph

The graph of this inequality includes all values less than or equal to - 4 or greater than or equal to 2. We show this by keeping the endpoints closed.

b Graphing a single inequality involves two main steps.

Plotting the boundary line.
Shading half of the plane to show the solution set.

For this exercise, we need to do this process for each of the inequalities in the system. y≤ -2x+3 & (I) y≥ x & (II) x≥ -1 & (III)

The system's solution set will be the intersection of the shaded regions in the graphs of (I), (II), and (III).

Boundary Lines

We can tell a lot of information about the boundary lines from the inequalities given in the system.

Exchanging the inequality symbols for equals signs gives us the boundary line equations.
Observing the inequality symbols tells us whether the inequalities are strict.
Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.

Let's find each of these key pieces of information for the inequalities in the system. Note that since Inequality (III) is in the form x≥-1, the slope of its boundary line will be undefined and therefore the line is vertical.

Information	Inequality (I)	Inequality (II)	Inequality (III)
Given Inequality	y ≤-2x+3	y ≥ x	x ≥ -1
Boundary Line Equation	y =-2x+3	y =x	x =-1
Solid or Dashed?	≤ ⇒ Solid	≥ ⇒ Solid	≥ ⇒ Solid
y= mx+ b	y= -2x+ 3	y= 1x+ 0	x=-1

Great! With all of this information, we can plot the boundary lines.

Shading the Solution Sets

Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on any of the boundary lines.

It looks like the point ( 2, 1) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region as the test point. Otherwise, we shade the opposite region.

Information	Inequality (I)	Inequality (II)	Inequality (III)
Given Inequality	y≤ -2x+3	y≥ x	x≥ -1
Substitute ( 2, 1)	1? ≤ -2( 2)+3	1? ≥ 2	2? ≥ -1
Simplify	1≰ -1	1≱2	2 ≥ -1
Shaded Region	opposite	opposite	same

For Inequality (I), we will shade the region opposite our test point, or below the boundary line. For Inequality (II), we will shade the region opposite our test point, or above the boundary line. For Inequality (III), however, we will shade the region containing the test point, or above the boundary line.