Two-step Inequalities and Their Applications

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: