Exercise 13 Page 204

We are given a compound inequality. Let's start by splitting it into separate inequalities. Compound Inequality: 3< 4p&-5 ≤ 15 First Inequality: 3< 4p &-5 Second Inequality: 4p &-5≤ 15 Notice that compound inequalities written in this way are equivalent to compound inequalities that involves the word and. 3 < 4p-5 and 4p-5≤ 15Let's solve the inequalities separately.

First Inequality

Inequalities can be solved in the same way as equations, by performing inverse operations on both sides until the variable is isolated. The only difference is that when you divide or multiply by a negative number, you must flip the inequality sign.

3<4p-5

AddIneq

LHS+5

8<4p

DivIneq

.LHS /4.<.RHS /4.

2

RearrangeIneq

Rearrange inequality

p>2

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

Note that the point on 2 is open because it is not included in the solution set.

Second Inequality

Again, we will solve the inequality by isolating the variable.

4p-5≤ 15

AddIneq

LHS+5≤RHS+5

4p≤ 20

DivIneq

.LHS /4.≤.RHS /4.

p≤ 5

This tells us that all values less than or equal to 5 will satisfy the inequality.

Note that the point on 5 is closed because it is included in the solution set.

Compound Inequality

The solution to the compound inequality is the intersection of the solution sets. To help visualize the algebraic expression, we will write p>2 as 2number line.