Different Ways of Solving an Inequality

2.10.3 Different Ways of Solving an Inequality

2.10.3 • Different Ways of Solving an Inequality

Activity

Use the following scenario to answer questions 1 – 2.

Andre and Priya used different strategies to solve the following inequality, but they reached the same solution.

$2 (2 x + 1.5) \leq 18 - x$

Make sense of each strategy until you can explain what each student has done.

Andre	Priya
$\begin{matrix} 2 (2 x + 1.5) & = & 18 - x \\ 4 x + 3 & = & 18 - x \\ 4 x - 15 & = & - x \\ - 15 & = & - 5 x \\ 3 & = & x \end{matrix}$ Testing to see if $x = 4$ is a solution: $\begin{matrix} 2 (2 \cdot 4 + 1.5) & < & 18 - 4 \\ 2 (9.5) & < & 14 \\ 19 & < & 14 \end{matrix}$ The inequality is false, so 4 is not a solution. If a number greater than 3 is not a solution, the solution must be less than 3, or $x < 3$ .	$\begin{matrix} 2 (2 x + 1.5) & = & 18 - x \\ 4 x + 3 & = & 18 - x \\ 5 x + 3 & = & 18 \\ 5 x & = & 15 \\ x & = & 3 \end{matrix}$ In $4 x + 3 = 18 - x$ , there is $4 x$ on the left and $- x$ on the right. If $x$ is a negative number, $4 x + 3$ could be positive or negative, but $18 - x$ will always be positive. For $4 𝑥 + 3 < 18 - 𝑥$ to be true, 𝑥 must include negative numbers or 𝑥 must be less than 3.

Andre

Priya

$\begin{matrix} 2 (2 x + 1.5) & = & 18 - x \\ 4 x + 3 & = & 18 - x \\ 4 x - 15 & = & - x \\ - 15 & = & - 5 x \\ 3 & = & x \end{matrix}$

Testing to see if $x = 4$ is a solution:

$\begin{matrix} 2 (2 \cdot 4 + 1.5) & < & 18 - 4 \\ 2 (9.5) & < & 14 \\ 19 & < & 14 \end{matrix}$

The inequality is false, so 4 is not a solution. If a number greater than 3 is not a solution, the solution must be less than 3, or $x < 3$ .

$\begin{matrix} 2 (2 x + 1.5) & = & 18 - x \\ 4 x + 3 & = & 18 - x \\ 5 x + 3 & = & 18 \\ 5 x & = & 15 \\ x & = & 3 \end{matrix}$

In $4 x + 3 = 18 - x$ , there is $4 x$ on the left and $- x$ on the right.

If $x$ is a negative number, $4 x + 3$ could be positive or negative, but $18 - x$ will always be positive.

For $4 𝑥 + 3 < 18 - 𝑥$ to be true, 𝑥 must include negative numbers or 𝑥 must be less than 3.

Summarize the strategy Andre used to solve the inequality.

Compare your answer: Your answer may vary, but here is a sample. Andre solved the companion equation and found $3 = x$ . Then, he tested a number greater than 3. That created a false mathematical statement so he concluded that any number greater than 3 would not be a solution.

Summarize the strategy Priya used to solve the inequality.

Compare your answer: Your answer may vary, but here is a sample. Priya solved the companion equation and found $x = 3$ . Then, she reasoned that the left side of the equation, $4 x + 3$ , could be either negative or positive while the right side of the equation, $18 - x$ , would always be positive - if negative $x$ values for are substituted into the expressions. So, the possible values for $x$ need to include both positive and negative numbers. That happens when the $x$ values are less than 3.

Examine four inequalities:

Inequality A: $\frac{1}{5} p > - 10$

Inequality B: $4 (x + 7) \leq 4 (2 x + 8)$

Inequality C: $- 9 n < 36$

Inequality D: $\frac{c}{3} < - 2 (c - 7)$

Work with a partner to decide on at least two inequalities to solve. Solve one inequality using Andre's strategy (by testing values on either side of the given solution), while your partner uses Priya's strategy (by reasoning about the parts of the inequality). Switch strategies for the other inequality.

Compare your answer: Your answer may vary, but here are some samples.

Inequality A: $\frac{1}{5} p > - 10$

p>−50p>−50. Sample reasoning (after solving 15p=−1015p=−10 to get p=−50):p=−50):
- Using Andre's strategy: Test a value more than or less than -50, and check to see if that value gives a true statement when substituting back to the original inequality.
- Using Priya's strategy: If $p$ is a negative number that is far away from 0, say -100, then $\frac{1}{5} p$ will be less than, not greater than, -10. This means the solutions must include positive numbers, so the solutions must be greater than -50.

Inequality B: $4 (x + 7) \leq 4 (2 x + 8)$

Sample reasoning using Priya's strategy (after solving 4(x+7)=4(2x+8)4(x+7)=4(2x+8) and getting (x=−1)(x=−1):
- After dividing each side by 4, the left side of the inequality has $x$ , while the right side has $2 x$ . This means the expression on the left side will be less than that on the right when $x$ includes positive numbers, so $x$ must be greater than or equal to -1. (In other words, for most negative values of $x$ , $2 x$ will be less than $x$ , which would make the inequality untrue.)

Inequality C: $- 9 n < 36$

Sample reasoning using Priya's strategy (after solving −9n=36−9n=36 to get n=−4n=−4):
- If $n$ includes large positive numbers, then $- 9 n$ will be smaller than 36, so $n$ must include positive numbers which are greater than -4.
- If $n$ is less than -4, then it would include negative numbers like -10, which means the value of $- 9 n$ would be a positive number that is greater than 36, which makes the inequality untrue. So the solution must be greater than -4.

Inequality D: $\frac{c}{3} < - 2 (c - 7)$

Sample reasoning using Priya's strategy (after solving $\frac{c}{3} = - 2 (c - 7)$ and getting $c = 6)$ :
The right side of the inequality has $- 2 c$ while the left side has $\frac{1}{3} c$ . If $c$ is a large positive number, $- 2 c$ will be less than $\frac{1}{3} c$ . If $c$ is a negative number, $- 2 c$ will be positive and greater than $\frac{1}{3} c$ , so the solution must include negative numbers which are less than 6.

Are you ready for more?

Extending Your Thinking

Using positive integers between 1 and 9 and each positive integer at most once, fill in values to get two constraints so that $x = 7$ is the only integer that will satisfy both constraints at the same time. ▢ $x$ + ▢ < ▢ $x$ + ▢

Compare your answer:

Your answers may vary, but here is a sample.

$4 x + 1 < 3 x + 9$

Compare your answer:

Your answers may vary, but here is a sample.

$6 x + 2 > 5 x + 8$

Self Check

Solve the following inequality by solving the related equation and reasoning what the answer may be given the situation.

$\frac13p<4$

$p < \frac43$ ; The related equation is $p = \frac43$ . Since the solution set must contain negative numbers, $p$ must be less than $\frac43$ .
$p < \frac34$ ; The related equation is $p = frac34$ . Since the solution set must contain negative numbers, $p$ must be less than $\frac34$ .
$p < 12$ ; The solution to the related equation is $p = 12$ . Since $\frac13p$ must be less than 4, the solution cannot include large positive numbers, so numbers larger than 12 cannot be included in the solution.
$p > 12$ ; The solution to the related equation is $p = 12$ . For $\frac13p$ to be less than 4, the solution must contain large positive numbers.

Additional Resources

Reasoning about Solution Sets to Linear Inequalities

You can derive a lot of information about the solution set to an inequality by studying the inequality itself.

Let’s look at the inequality shown.

$12 x > - 3 (x - 1)$

The left side of the inequality contains $12 x$ . The right side of the inequality contains $- 3 x$ .

We can think to ourselves: “What values of $x$ would make $12 x$ greater than $- 3 x$ ?”

If $x$ is a large positive number, then $12 x$ will still be positive. If $x$ is a large positive number, then $- 3 x$ will be a negative number. This makes the inequality true, so the solution set must include large positive numbers.

Let’s solve the related equation.

$\begin{matrix} 12 x & > & - 3 (x - 1) \\ 12 x & = & - 3 (x - 1) \\ 12 x & = & - 3 x + 3 \\ 15 x & = & 3 \\ x & = & \frac{3}{15} \end{matrix}$

Since we know that the solution set must include large positive numbers, then $x > \frac{3}{15}$ must be the solution to the inequality.

Try it

Reasoning about Solution Sets to Linear Inequalities

For questions 1 – 2, solve the related equation for each inequality. Use what you know about the inequality to determine the solution set.

$- 5 y \geq 30$

Compare your answer:

The solution $y = - 6$ . Since any positive number substituted for $y$ makes $- 5 y$ a negative value, then the solution set cannot include positive numbers. This means the solution is $y \leq - 6$ .

$4 p < - 4$

Compare your answer:

The solution $p = - 1$ . Since any positive number substituted for $p$ makes $4 p$ a positive value, the solution set cannot include positive numbers. This means the solution is $p < - 1$ .