Exercise 28 Page 150

To solve the given system by graphing, we should first draw each inequality separately. Then we will combine the graphs. The overlapping region will be the solution set. Let's start!

Inequality I

To determine the boundary line of the first inequality, we need to exchange the inequality symbol for an equals sign. Inequality:& y≥ |2x+4|-2 Boundary Line:& y=|2x+4|-2 The graph of this boundary line is the graph of the parent function y=|x| after transformations. Before we can identify them we will rewrite our equation. y=|2x+4|-2 ⇔ y= 2|x+2|-2 Now we can identify a vertical stretch by a factor of 2, translation 2 units to the left and 2 units down. The boundary line will be solid because the inequality is not strict.

Next, we need decide which side of the boundary line we should shade. We can do this by testing a point that does not lie on the boundary line. If the point satisfies the inequality, it lies in the solution set. If not, we will shade the other region. Let's use ( 0, 0).

y≥ |2x+4|-2

SubstituteII

x= 0, y= 0

0? ≥|2( 0)+4|-2

▼

Simplify

ZeroPropMult

Zero Property of Multiplication

0? ≥|0+4|-2

IdPropAdd

Identity Property of Addition

0? ≥|4|-2

AbsPos

|4|=4

0? ≥4-2

SubTerm

Subtract term

0≱ 2

Because (0,0) did not create a true statement, we will shade the region opposite this point.

Inequality II

Now that we've completed the first inequality, let's determine the boundary line of the second inequality. We will follow the same process once more. Inequality:& 3y+x≤ 15 Boundary Line:& 3y+x=15 Let's rewrite this equation in slope-intercept form.

3y+x=15

▼

Write in slope-intercept form

SubEqn

LHS-x=RHS-x

3y=- x+15

DivEqn

.LHS /3.=.RHS /3.

y=-1/3x+5

The equation is in slope-intercept form, below we highlight the slope m and the y-intercept b. y= -1/3x+ 5 Using this information, we will plot the boundary line. Notice that the inequality is not strict, so the boundary line will be solid.

Again, let's use ( 0, 0) as our test point to decide which region we should shade.

3y+x≤ 15

SubstituteII

x= 0, y= 0

3( 0)+ 0? ≤15

▼

Simplify

ZeroPropMult

Zero Property of Multiplication

0+0? ≤15

AddTerms

Add terms

0≤ 15

Because (0,0) created a true statement, we will shade the region containing this point.

Combining the Inequality Graphs

In drawing the inequality graphs on the same coordinate plane, we are able to see the overlapping section. This is the solution set of the system.

Finally, we can view only the solution set by removing the shaded regions that are not overlapping.