Exercise 14 Page 396

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 we divide or multiply by a negative number, we must reverse the inequality sign.

5+h > 7

SubIneq

LHS-5>RHS-5

5+h- 5 > 7- 5

CommutativePropAdd

Commutative Property of Addition

h+5-5 > 7-5

DiffZero

5-5=0

h> 7-5

SubTerm

Subtract term

h>2

We found that all values of h greater than 2 will satisfy the inequality. Now let’s graph the inequality on a number line. Since h cannot equal 2, we draw an open circle at this point. Since h is all values greater than 2, we will shade the part of the number line that represents numbers greater than 2. This means that we shade to the right of our point at 2.