Solving Compound Inequalities

a

To solve the compound inequality, we have to solve each of the inequalities separately. Since the word between the individual inequalities is "or," the solution set for the compound inequality consists of both solution sets.

First inequality

Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. $t + 14 \geq 15 \Leftrightarrow t \geq 1$ This inequality tells us that all values greater than or equal to $1$ will satisfy the inequality. Thus, we can graph it on a number line with a closed circle at $1$ and the solution set represented to the right.

Second inequality

Again, we'll solve the inequality by isolating the variable.

t - 9 < - 10

AddIneq

LHS + 9 < RHS + 9

t < - 1

This inequality tells us that all values less than

- 1

will satisfy the inequality. Let's graph this one as well.

Note that the point on $- 1$ is open because it's not included in the solution set.

Compound inequality

The solution to the compound inequality is the combination of the solution sets. $t < - 1 or t \geq 1$ In the same way, the graph is a combination of the to separated graphs.

b

If we solve each inequality separately, we will find two solution sets. The union of those sets is the solution to the compound inequality.

First Inequality

By adding

3

to both sides of the inequality, we can eliminate

3

from the left-hand side. This will help isolate

z .

5 z - 3 > 7

AddIneq

LHS + 3 > RHS + 3

5 z > 10

DivIneq

LHS / 5 > RHS / 5

z > 2

All possible values of

z

that are greater than

2

will satisfy the inequality. Thus, we can graph the inequality with an open circle at

2

and the solution set to the right.

Second Inequality

By adding

6

to both sides of the inequality, we can begin to isolate

z .

4 z - 6 < - 10

AddIneq

LHS + 6 < RHS + 6

4 z < - 4

DivIneq

LHS / 4 < RHS / 4

z < - 1

The second inequality is satisfied for all values of

z

less than

- 1 .

The graph will be a open circle at

- 1

and a shaded region to the left.

Compound inequality

The compound inequality is now represented by the two solution sets. $z < - 1 or z > 2$ The graph will be a combination of the two solution sets as well.