Two-step Inequalities and Their Applications

1. Find the solution set of the inequality:

\(8 x+2 > 34\).

2. Which graph represents the solution set of this inequality?

\(5 a+18 < -27\)

Choose 1 answer:

A.

B.

C.

D.

3. Which graph represents the solution set of this inequality?

\(-11-2 d \geq 1\)

Choose 1 answer:

A.

B.

C.

D.

4. Find the solution set of the inequality:

\(-3 x+8 < 15\).

5. Find the solution set of the inequality:

\(5 x+13 \geq-37\).

6. Which graph represents the solution set of this inequality?

\(12 b-15 > 21\)

A.

B.

C.

D.

7. Which graph represents the solution set of this inequality?

\(-3 b-15 > -24\)

Choose 1 answer:

A.

B.

C.

D.

1. \(x > 4\)

Let’s start by subtracting \(2\) from both sides of the inequality:

\(8 x+2 > 34\)

\(8 x+2-2 > 34-2\)

\(8 x > 32\)

Next, let's divide both sides by \(8\):

\(8 x > 32 \)

\(\frac{8 x}{8} > \frac{32}{8}\)

\(x > 4\)

The solution set of the inequality is \(x > 4\).

2. C.

Let’s start by subtracting \(18\) from both sides of the inequality:

\(5 a+18 < -27\)

\(5 a+18-18 < -27-18\)

\(5 a < -45\)

To isolate \(a\), we need to divide both sides by \(5\).

\(5 a < -45\)

\(\frac{5 a}{5} < \frac{-45}{5}\)

\(a < -9\)

To graph the inequality \(a < -9\), we first draw a circle at \(-9\). This circle divides the number line into two sections: one that contains solutions to the inequality and one that does not.

Since the solution uses a less than sign, the solution does not include the point where \(a= - 9\). So the circle at \(-9\) is not filled in.

Because the solution to the inequality says that \(a < -9\), this means that solutions are numbers to the left of \(-9\).

The graph that represents the solution of the inequality \(a < -9\) is shown in Pink:

3. C.

Let’s start by adding \(11\) to both sides of the inequality:

\(-11-2 d \geq 1\)

\( -11+11-2 d \geq 1+11\)

\(-2 d \geq 12\)

To isolate \(d\) we need to divide both sides by \(-2\):

\(-2 d \geq 12\)

\( \frac{-2 d}{-2} \leq \frac{12}{-2}\)

\(d \leq-6\)

To graph the inequality \(d \leq-6\), we first draw a circle at \(-6\). This circle divides the number line into two sections: one that contains solutions to the inequality and one that does not.

Since the solution uses a less than or equal to sign, the solution includes the point where \(d = -6\). So the circle at \(-6\) is filled in.

Because the solution to the inequality says that \(d \leq-6\), this means that solutions are numbers to the left of \(-6\).

The graph that represents the solution of the inequality \(d \leq-6\), is shown Pink:

4. \(x > -\frac{7}{3}\)

Let’s start by subtracting \(8\) from both sides of the inequality:

\(-3 x+8 < 15\)

\(-3 x+8-8 < 15-8\)

\(-3 x < 7\)

Next, let's divide both sides by \(-3\).When you divide an inequality by a negative number, the inequality sign must be reversed.

\(-3 x < 7\)

\(\frac{-3 x}{-3} > \frac{7}{-3}\)

\(x > -\frac{7}{3}\)

The solution set of the inequality is \(x > -\frac{7}{3}\).

5. \(x \geq-10\)

Let’s start by subtracting \(13\) from both sides of the inequality:

\(5 x+13 \geq-37\)

\( 5 x+13-13 \geq-37-13\)

\(5 x \geq-50\)

Next, let's divide both sides by \(5\):

\(5 x \geq-50\)

\(\frac{5 x}{5} \geq \frac{-50}{5}\)

\(x \geq-10\)

The solution set of the inequality is:

\(x \geq-10\).

6. D.

Let’s start by adding \(15\) to both sides of the inequality:

\(12 b-15 > 21\)

\(12 b-15+15 > 21+15\)

\(12 b > 36\)

To isolate \(b\), we need to divide both sides by \(12\):

\(12 b > 36\)

\(\frac{12 b}{12} > \frac{36}{12}\)

\(b > \frac{36}{12}\)

\(b > 3\)

To graph the inequality \(b > 3\), we first draw a circle at \(3\). This circle divides the number line into two sections: one that contains solutions to the inequality and one that does not.

Since the solution uses a greater than sign, the solution does not include the point where \(b=3\), So the circle at \(3\) is not filled in.

Because the solution to the inequality says that \(b > 3\), this means that solutions are numbers to the right of \(3\).

The graph that represents the solution of the inequality \(b > 3\), is shown in Red:

7. A.

Let’s start by adding \(15\) to both sides of the inequality:

\(-3 b-15 > -24\)

\(-3 b-15+15 > -24+15\)

\(-3 b > -9\)

To isolate \(b\), we need to divide both sides by \(-3\). When you divide an inequality by a negative number, the inequality sign must be reversed.

\(-3 b > -9\)

\(\frac{-3 b}{-3} < \frac{-9}{-3}\)

\(b < 3\)

To graph the inequality \(b < 3\), we first draw a circle at \(3\). This circle divides the number line into two sections: one that contains solutions to the inequality and one that does not.

Since the solution uses a less than sign, the solution does not include the point where \(b=3\). So the circle at \(3\) is not filled in.

Because the solution to the inequality says that \(b < 3\), this means that solutions are numbers to the left of \(3\).

The graph that represents the solution of the inequality \(b < 3\) is shown in Blue:

1. Darcie wants to crochet a minimum of \(3\) blankets to donate to a homeless shelter. Darcie crochets at a rate of \(\frac{1}{15}\) of a blanket per day. She has \(60\) days until when she wants to donate the blankets, but she also wants to skip crocheting some days so she can volunteer in other ways.

Write an inequality to determine the number of days, \(s\), Darcie can skip crocheting and still meet her goal.

Graph the solution set to this inequality.

2. Mustafa, Heloise, and Gia have written more than a combined total of \(22\) articles for the school newspaper. Heloise has written \(\frac{1}{4}\) as many articles as Mustafa has. Gia has written \(\frac {3}{2}\) as many articles as Mustafa has.

Write an inequality to determine the number of articles, \(m\), Mustafa could have written for the school newspaper.

What is the solution set of the inequality?

Choose 1 answer:

A. \(m > \frac{1}{2}\)

B. \(m \geq \frac{1}{2}\)

C. \(m > 8\)

D. \(m \geq 8\)

3. Kim's softball team was playing in the championship game. When there were \(4\) innings left, the team was losing by a score of \(17\) to \(6\) runs. In the last \(4\) innings, her team scored the same number of runs per inning, and the other team did not score any more runs. Kim's team won with the most runs.

Write an inequality to determine the number of runs per inning, \(p\), Kim's team could have scored.

Find the minimum whole number of runs per inning Kim's team could have scored.

4. Janie has \($3\). She earns \($1.20\) for each chore she does and can do fractions of chores. She wants to earn enough money to buy a CD for \($13.50\).

Write an inequality to determine the number of chores, \(c\), Janie could do to have enough money to buy the CD.

Graph the solution set to this inequality.

1. \(\frac{1}{15}(60-s) \geq 3\)

Darcie wants to donate a minimum of \(3\) blankets.

If Darcie skips \(s\) days, then she will crochet for \(60 -s \) days. Darcie crochets at a rate of \(\frac {1}{15}\) of a blanket per day, so she can crochet a total of \(\frac{1}{15}(60-s)\) blankets.

This amount must be greater than or equal to \(3\) blankets, so the inequality is:

\(\frac{1}{15}(60-s) \geq 3\)

To solve the inequality, let's start by multiplying both sides by \(15\). Remember that when we multiply an inequality by a negative number, the inequality sign reverses.

\(\begin{aligned}

\frac{1}{15}(60-s) & \geq 3 \\

15 \cdot \frac{1}{15}(60-s) & \geq 15 \cdot 3 \\

60-s & \geq 45 \\

60-s-60 & \geq 45-60 \\

-s & \geq-15 \\

-s \cdot-1 & \leq-15 \cdot-1 \\

s & \leq 15 \\

\end{aligned}\)

Darcie can skip no more than \(15\) days of crocheting to meet her goal.

To graph the inequality \(s \leq 15\), we first draw a circle at \(15\). This circle divides the number line into two sections: one that contains solutions to the inequality and one that does not.

The solution includes the point \(s=15\), so the circle at \(15\) filled in.

Because the solution to the inequality says that \(s \leq 15\), this means that solutions are numbers to the left of \(15\).

The inequality is:

\(\frac{1}{15}(60-s) \geq 3\)

The graph of the solution of the inequality, \(s \leq 15\), looks like this:

2. The inequality is: \(m+\frac{1}{4} m+\frac{3}{2} m > 22\). The solution set is C: \(m > 8\).

Mustafa, Heloise, and Gia have written more than a combined total of \(22\) articles for the school newspaper.

Mustafa has written \(m\) articles. Heloise has written \(\frac{1}{4} m \) articles. Gia has written \(\frac{3}{2} m\) articles.

They have written a combined total of \(m+\frac{1}{4} m+\frac{3}{2} m\) articles.

This must be more than \(22\) articles, so the inequality is:

\(m+\frac{1}{4} m+\frac{3}{2} m > 22\)

Let's start by combining like terms.

\(\begin{aligned}

m+\frac{1}{4} m+\frac{3}{2} m & > 22 \\

\frac{4}{4} m+\frac{1}{4} m+\frac{6}{4} m & > 22 \\

\frac{11}{4} m & > 22 \\

\frac{4}{11} \cdot \frac{11}{4} m & > \frac{4}{11} \cdot 22 \\

m & > 8

\end{aligned}\)

Mustafa has written more than \(8\)articles.

The inequality is:

\(m+\frac{1}{4} m+\frac{3}{2} m > 22\)

The solution set is \(m > 8\).

3. The inequality is: \( 6 +4 p > 17\). Kim's team scored a minimum of \(3\) runs per inning.

Kim's team needed more than \(17\) runs to win.

Kim's team already had \(6\) runs. Since they scored \(p\) runs per inning for \(4\) innings, they scored an additional \(4p\) runs, so Kim's team had a total \(6 + 4p\) runs.

This must be greater than \(17\), so the inequality is:

\(6+4 p > 17\)

We can start by subtracting \(6\) from both sides of the inequality:

\(\begin{aligned}

6+4 p & > 17 \\

6-6+4 p & > 17-6 \\

4 p & > 11 \\

\frac{4 p}{4} & > \frac{11}{4} \\

p & > 2 \frac{3}{4}

\end{aligned}\)

Kim's team must score more than \(2 \frac{3}{4}\) runs per inning to win the game. So Kim's team must have scored a minimum of \(3\) runs per inning.

The inequality is:

\(6+4 p > 17\)

Kim's team scored a minimum of \(3\) runs per inning.

4. The inequality is: \(3+1.2 c \geq 13.50\)

Janey wants to earn enough money to buy a CD for \($13.50\)

If Janie does \(c\) chores, she will earn \($1.2c\) dollars for doing chores, plus \(3\) dollars she already has. The amount she has is:

\(3+1.2 c\)

This amount must be greater than or equal to \($13.50\), so the inequality is:

\(3+1.2 c \geq 13.50\)

To solve the inequality, let's start by subtracting \(3\) from both sides of the inequality:

\(\begin{aligned}

3+1.2 c & \geq 13.5 \\

3+1.2 c-3 & \geq 13.5-3 \\

1.2 c & \geq 10.5 \\

\frac{1.2 c}{1.2} & \geq \frac{10.5}{1.2} \\

c & \geq 8.75

\end{aligned}\)

Janie must do \(8.75\) or more chores to have enough money to purchase the CD.

To graph the inequality \(c \geq 8.75\), we first draw a circle at \(8.75\). This circle divides the number line into two sections: one that contains solutions to the inequality and one that does not.

The solution includes the point \(c=8.75\), so we fill in the circle at \(8.75\).

Because the solution to the inequality says that \(c \geq 8.75\), this means that solutions are numbers to the right of \(8.75\).

The inequality is:

\(3+1.2 c \geq 13.50\)

The graph of the solution of the inequality \(c \geq 8.75\) looks like this: