Houghton Mifflin Harcourt Algebra 1, 2015
HM
Houghton Mifflin Harcourt Algebra 1, 2015 View details
5. Creating and Solving Compound Inequalities
Continue to next subchapter

Exercise 23 Page 73

Start by decomposing the compound inequalities and simplifying all individual inequalities. Notice that some signs are strict so interpret them carefully.

a. C
b. A
c. D
d. B

Practice makes perfect

We need to match the compound inequalities with the corresponding graphs for their solutions. We will work each case individually.

Inequality A

The first given inequality is -26 < 6x - 8 < 16, which is a compound inequality. Thus, it can be decomposed into two individual inequalities. -26 < 6x - 8 and 6x - 8 < 16 We can simplify each individual inequality by isolating x. We can achieve this by working with it as if it were an equation. We will use inverse operations. Let's first simplify -26 < 6x - 8.
-26 < 6x - 8
â–Ľ
Solve for x
-18 < 6x
-3 < x
x> - 3
Similarly, we will solve 6x - 8 < 16.
6x - 8 < 16
â–Ľ
Solve for x
6x < 24
x < 4
Therefore, the solution set for Inequality A contains all numbers greater than -3 AND less than 4, not including 4 and -3 themselves. x> -3 AND x < 4

The graph showing this solution set is Graph b.

Inequality B

We now have the inequality - 13 ≤ x + 23 ≤ 2. Again, we are working with a compound inequality. Let's then decompose it and simplify each individual inequality. We can write it as an AND inequality. \begin{gathered} \text{-} \dfrac{1}{3} \leq \dfrac {x + 2}{3} \quad \textbf{AND} \quad \dfrac {x + 2}{3} \leq 2 \end{gathered} We will now simplify - 13 ≤ x + 23.
\text{-} \dfrac{1}{3} \leq \dfrac {x + 2}{3}
â–Ľ
Solve for x
\text{-} \dfrac{1}{3}\cdot 3 \leq \dfrac {x + 2}{3}\cdot 3
- 1 ≤ x + 2
-3 ≤ x
x ≥ - 3
Now we will simplify x + 23 ≤ 2.
\dfrac {x + 2}{3} \leq 2
â–Ľ
Solve for x
\dfrac {x + 2}{3}\cdot 3 \leq 6
x + 2 ≤ 6
x ≤ 4
Therefore, the solution set of Inequality B contains all numbers greater than or equal to -3, and less than or equal to 4. x ≥ - 3 AND x ≤ 4 The graph showing this solution set is Graph d.

Inequality C

We have the inequality 4x+1<-11 OR x2 -5 > - 3. Let's simplify each individual inequality. We will start with 4x + 1 < - 11.
4x + 1 < - 11
â–Ľ
Solve for x
4x < - 12
x < -3
Now we will simplify x2 - 5> -3.
x/2 - 5> -3
â–Ľ
Solve for x
x/2 > 2
x/2* 2 > 4
x > 4
Therefore, the solution set for Inequality C contains all numbers which are greater than 4 or less than -3. Note that 4 and -3 are not included in the solution set. x < - 3 OR x > 4 The graph showing this solution set is Graph a.

Inequality D

Finally we have x -63 ≤ -3 OR 2x + 8≥ 16. Let's simplify each individual inequality. We will start with x -63 ≤ -3.
\dfrac {x -6}{3} \leq \text{-}3
â–Ľ
Solve for x
\dfrac {x -6}{3}\cdot 3 \leq \text{-} 9
x -6 ≤ -9
x ≤ -3
Now we will simplify 2x + 8 ≥ 16.
2x + 8 ≥ 16
â–Ľ
Solve for x
2x ≥ 8
x ≥ 4
The solution set for Inequality D contains all numbers which are less than or equal to -3 OR greater than or equal to 4. x ≤ - 3 OR x ≥ 4 The graph showing this solution set is Graph c.