Exercise 30 Page 219

The inequality states that the absolute value of 5d+11 is greater than or equal to 3. Since any negative argument inside the absolute value changes signs to be positive, we get a compound inequality when removing the absolute value. Because the distance needs to be greater than or equal to 3, we get an or compound inequality. 5d+11≥ 3 or 5d+11≤- 3 Let's split the compound inequality into two inequalities. First Inequality:& 5d+11≥ 3 Second Inequality:& 5d+11≤- 3

We can solve them one at a time.

First Inequality

We can solve inequalities as if they were equations by using inverse operations to isolate the variable.

5d+11≥ 3

SubIneq

LHS-11≥RHS-11

5d≥- 8

DivIneq

.LHS /5.≥.RHS /5.

d≥- 8/5

MoveNegNumToFrac

Put minus sign in front of fraction

d≥- 8/5

Although d≥- 85 is a perfectly valid expression, we can also rewrite the fraction as a mixed number.

d≥- 8/5

WriteSum

Write as a sum

d≥- 5+3/5

WriteSumFrac

Write as a sum of fractions

d≥-(5/5+3/5)

QuotOne

a/a=1

d≥-(1+3/5)

Rewrite

Rewrite 1+3/5 as 1 35

d≥-1 35

The solution set to this inequality contains all the values greater than or equal to - 1 35. {d | d≥- 1 35}

Second Inequality

We can solve the second inequality in the same way.

5d+11≤- 3

SubIneq

LHS-11≤RHS-11

5d≤- 14

DivIneq

.LHS /5.≤.RHS /5.

d≤-14/5

MoveNegNumToFrac

Put minus sign in front of fraction

d≤-14/5

Although d≤- 145 is a perfectly valid expression, we can also rewrite the fraction as a mixed number.

d≤- 14/5

WriteSum

Write as a sum

d≤- 10+4/5

WriteSumFrac

Write as a sum of fractions

d≤-(10/5+4/5)

CalcQuot

Calculate quotient

d≤-(2+4/5)

Rewrite

Rewrite 2+4/5 as 2 45

d≤-2 45

The solution set to this inequality contains all the values less than or equal to - 2 45. {d | d≤- 2 45}

The Union of Sets

The solution to the compound inequality joined by the word or is the union of the solution sets from the first and second inequalities. {d | d≥- 1 35} ⋃ {d | d≤- 2 45}