Exercise 28 Page 96

We are asked to find and graph the solution set for all possible values of g in the given inequality. We will start by subtracting 1 on both sides to isolate the absolute value expression. |- 3g-2|+1<6 ⇔ |- 3g-2|<5 Now we will create a compound inequality by removing the absolute value. The less than symbol < creates an and inequality because absolute value represents a distance and we need our distance from the center to be less than 5 away. - 3g-2 > - 5 and - 3g-2< 5

Let's isolate g in both of these cases before graphing the solution set.

First Inequality

Remember that whenever we multiply or divide an inequality by a negative number, we reverse the direction of the inequality symbol.

- 3g-2>- 5

AddIneq

LHS+2>RHS+2

- 3g>- 3

DivNegIneq

Divide by - 3 and flip inequality sign

g<1

This statement tells us that all values less than 1 will satisfy the inequality.

Second Inequality

Again, remember that whenever we multiply or divide an inequality by a negative number, we reverse the direction of the inequality symbol.

- 3g-2<5

AddIneq

LHS+2

- 3g<7

DivNegIneq

Divide by - 3 and flip inequality sign

g> 7/-3

MoveNegDenomToFrac

Put minus sign in front of fraction

g>- 7/3

FracToMixed

Write fraction as a mixed number

g>- 2 13

RearrangeIneq

Rearrange inequality

- 2 13

This statement tells us that all values greater than - 2 13 will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the combination of the solution sets. First Solution Set:& g< 1 Second Solution Set:& - 2 13 < g Combined Solution Set:& - 2 13

Graph

The graph of this inequality includes all values which are less than 1 and greater than - 2 13. As both are strict inequalities, we will use open circles.