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Learning Objectives
- Use the properties of inequality together to isolate variables and solve algebraic inequalities, and express their solutions graphically.
- Simplify and solve algebraic inequalities using the distributive property to clear parentheses and fractions.
Introduction
Solving multi-step inequalities is very similar to solving equations—what you do to one side, you need to do to the other side in order to maintain the “balance” of the inequality. The properties of inequality can help you understand how to add, subtract, multiply, or divide within an inequality.
Using Properties Together to Solve Inequalities
A popular strategy for solving equations, isolating the variable, also applies to solving inequalities. By adding, subtracting, multiplying, and/or dividing, you can rewrite the inequality so that the variable is on one side and everything else is on the other. As with one-step inequalities, the solutions to multi-step inequalities can be graphed on a number line.
Example
Solve for p.
\(\ 4 p+5<29\).
Solution
| \(\ \begin{array}{r} | Begin to isolate the variable by subtracting 5 from both sides of the inequality. |
| \(\ p<6\) | Divide both sides of the inequality by 4 to express the variable with a coefficient of 1. |
\(\ p<6\)
To graph this inequality, you draw an open circle at the end point, 6, on the number line. The circle is open because the inequality is less than 6 and not equal to 6. The values where \(\ p\) is less than 6 are found all along the number line to the left of 6. Draw a blue line with an arrow on the number line pointing in that direction.
To check the solution, substitute the end point, 6, into the original inequality written as an equation, which is called the related equation, to see if you get a true statement. Then check another solution, such as 0, to see if the inequality is correct.
Example
Check that \(\ p<6\) is the solution to the inequality \(\ 4 p+5<29\).
Solution
| \(\ \begin{aligned} 4 p+5 &=29 \\ \text { Does } 4(6)+5 &=29 ? \\ 24+5 &=29 \\ 29 &=29 \text { Yes! } \end{aligned}\) | Check the end point, 6, in the related equation. |
| \(\ \begin{aligned} 4 p+5 &<29 \\ \text { Is } 4(0)+5 &<29 ? \\ 0+5 &<29 \\ 5 &<29 \text { Yes! } \end{aligned}\) | Try another value to check the inequality. Let’s use \(\ p=0\). |
\(\ p<6\) is the solution to the inequality \(\ 4 p+5<29\).
Example
Solve for \(\ x\).
\(\ 3 x-7 \geq 41\)
Solution
| \(\ \begin{array}{r} | Begin to isolate the variable by adding 7 to both sides of the inequality. Divide both sides of the inequality by 3 to express the variable with a coefficient of 1. |
| Check \(\ \begin{aligned} \(\ \begin{aligned} | First, check the end point, 16, in the related equation. Then, try another value to check the inequality. Let’s use \(\ x=20\). |
\(\ x \geq 16\)
When solving multi-step equations, pay attention to situations in which you multiply or divide by a negative number. In these cases, you must reverse the inequality sign.
Example
Solve for \(\ p\).
\(\ -6 p+14<-58\)
Solution
| \(\ \begin{array}{r} -6 p+14<&-58 \\ \ -14\ \ \ \ &-14 \\ \hline \frac{-6 p}{-6} \ \ \ \ \ \ \ \ \ \ >& \frac{-72}{-6}\\ p>&12 \end{array}\) | Begin to isolate the variable by subtracting 14 from both sides of the inequality. Divide both sides of the inequality by -6 to express the variable with a coefficient of 1. Dividing by a negative number results in reversing the inequality sign. |
| Check \(\ \text { Does } \begin{aligned} | Check the solution. First, check the end point, 12, in the related equation. |
| \(\ \begin{aligned} -6 p+14 &<-58 \\ \text { Is }-6(100)+14 &<-58 ? \\ -600+14 &<-58 \\ -586 &<-58 \text { Yes! } \end{aligned}\) | Then, try another value to check the inequality. Try 100. |
\(\ p>12\)
The graph of the inequality \(\ p>12\) has an open circle at 12 with an arrow stretching to the right.
Advanced Example
Solve for \(\ x\).
\(\ \frac{1}{3}(x+3)+\frac{1}{2}>-\frac{9}{2}\).
Solution
| \(\ \begin{aligned} \frac{1}{3}(x+3)+\frac{1}{2}-\frac{1}{2} &>-\frac{9}{2}-\frac{1}{2} \\ \frac{1}{3}(x+3) &>-\frac{10}{2} \\ \frac{1}{3}(x+3) &>-5 \\ 3 \cdot \frac{1}{3}(x+3) &>3 \cdot(-5) \\ x+3 &>-15 \\ x+3-3 &>-15-3 \\ x &>-18 \end{aligned}\) | To isolate the variable, subtract \(\ \frac{1}{2}\) from both sides of the inequality. Then multiply by 3 so that the coefficient in front of the parentheses is 1. Then subtract 3 from both sides. |
| Check \(\ \begin{aligned} | Check the solution. First, check the end point, -18, in the related equation. |
| \(\ \begin{array}{r} \frac{1}{3}(0+3)+\frac{1}{2}>-\frac{9}{2} \\ \frac{1}{3}(3)+\frac{1}{2}>-\frac{9}{2} \\ 1+\frac{1}{2}>-\frac{9}{2} \\ \frac{2}{2}+\frac{1}{2}>-\frac{9}{2} \\ \frac{3}{2}>-\frac{9}{2} \end{array}\) | Now check any value for \(\ x\) that is within the region \(\ x>-18\). We will use \(\ x=0\). |
| The statement is true. |
\(\ x>-18\)
Exercise
A student is solving the inequality \(\ \frac{5}{2}+7 x \leq 4 x-\frac{7}{2}\). If she combines like terms, which of the following inequalities could she see?
- \(\ -6 \leq 3 x\)
- \(\ 3 x \leq-6\)
- \(\ \frac{19 x}{2} \leq \frac{x}{2}\)
- \(\ x \leq \frac{-6 x}{7}\)
- Answer
-
- Incorrect. It looks like you tried to combine like terms but made sign mistakes while adding and subtracting. To isolate the variable term on the right, you need to subtract \(\ 7 x\) from both sides; \(\ 4 x-7 x=-3 x\). The correct answer is: \(\ 3 x \leq-6\).
- Correct. You correctly combined like terms. \(\ \frac{5}{2}+7 x \leq 4 x-\frac{7}{2}\) becomes \(\ 7 x-4 x \leq-\frac{7}{2}-\frac{5}{2}\). which is the same as \(\ 3 x \leq-6\).
- Incorrect. Remember that you cannot add variable and non-variable terms like you add integers. The correct answer is: \(\ 3 x \leq-6\).
- Incorrect. Try subtracting \(\ \frac{5}{2}\) and \(\ 4x\) from each side before collecting like terms. The correct answer is: \(\ 3 x \leq-6\).
Using the Distributive Property to Clear Parentheses and Fractions
As with equations, the distributive property can be applied to simplify expressions that are part of an inequality. Once the parentheses have been cleared, solving the inequality will be straightforward.
Example
Solve for \(\ x\).
\(\ 2(3 x-5) \leq 4 x+6\)
Solution
| \(\ \begin{array}{rr} 2(3 x-5) & \leq & 4 x+6 \\ 6 x-10 & \leq & 4 x+6 \\ -4 x \ \ \ \ \ \ \ \ \ & & -4 x\ \ \ \ \ \ \ \\ \hline 2 x-10 &\leq & 6 \\ +10 && +10 \\ \hline \frac{2 x}{2}\ \ \ \ \ \ \ \ & \leq & \frac{16}{2} \\ x \ \ \ \ \ \ \ \ \ & \leq & 8 \end{array}\) | Distribute to clear the parentheses. Subtract \(\ 4x\) from both sides to get the variable term on one side only. Add 10 to both sides to isolate the variable. Divide both sides by 2 to express the variable with a coefficient of 1. |
| Check \(\ \begin{aligned} \(\ \begin{aligned}\text { Is } | Check the solution. First, check the end point, 8, in the related equation. Then, choose another solution and evaluate the inequality for that value to make sure it is a true statement. Try 0. |
\(\ x \leq 8\)
Example
Solve for \(\ a\).
\(\ \frac{2 a-4}{6}<2\)
Solution
| \(\ 6 \cdot \frac{2 a-4}{6}<2 \cdot 6\) | Clear the fraction by multiplying both sides of the equation by 6. |
| \(\ \begin{array}{r} 2 a-4<&12 \\ \ +4\ \ \ \ &+4 \\ \hline \frac{2 a}{2}\ \ \ \ \ \ \ <&\frac{16}{2} \end{array}\) | Add 4 to both sides to isolate the variable. |
| \(\ a<8\) | Divide both sides by 2 to express the variable with a coefficient of 1. |
| Check \(\ \begin{aligned} | Check the solution. First, check the end point, 8, in the related equation. |
| \(\ \begin{aligned}\text { Is } \frac{2(5)-4}{6} &<2 ? \\ \frac{10-4}{6} &<2 \\ \frac{6}{6} &<2 \\ 1 &<2 \text { Yes! } \end{aligned}\) | Then, choose another solution and evaluate the inequality for that value to make sure it is a true statement. Try 5. |
\(\ a<8\)
Advanced Example
Solve for \(\ d\).
\(\ \frac{3}{5}(2 d-5) \leq 4\left(7-\frac{1}{5} d\right)\)
Solution
| \(\ \begin{aligned} \frac{3}{5}(2 d-5) & \leq 4\left(7-\frac{1}{5} d\right) \\ \frac{3}{5}(2 d)+\frac{3}{5}(-5) & \leq 4(7)+4\left(-\frac{1}{5} d\right) \\ \frac{6 d}{5}+\left(\frac{-15}{5}\right) & \leq 28+\left(\frac{-4 d}{5}\right) \\ \frac{6 d}{5}-3 & \leq 28-\frac{4 d}{5} \end{aligned}\) | This inequality contains two parentheses. Use the distributive property to expand both sides of the inequality. |
| \(\ \begin{aligned} \frac{6 d}{5}-3+3 & \leq 28-\frac{4 d}{5}+3 \\ \frac{6 d}{5} & \leq 31-\frac{4 d}{5} \\ \frac{6 d}{5}+\frac{4 d}{5} & \leq 31-\frac{4 d}{5}+\frac{4 d}{5} \\ \frac{10 d}{5} & \leq 31 \\ 2 d & \leq 31 \\ \frac{2 d}{2} & \leq \frac{31}{2} \\ d & \leq \frac{31}{2} \end{aligned}\) | Now that both sides have been expanded, combine like terms and find the range of values for \(\ d\). |
| \(\ \begin{aligned} \frac{3}{5}\left(2 \cdot \frac{31}{2}-5\right) &=4\left(7-\frac{1}{5} \cdot \frac{31}{2}\right) \\ \frac{3}{5}\left(\frac{62}{2}-5\right) &=4\left(7-\frac{31}{10}\right) \\ \frac{3}{5}(31-5) &=4\left(\frac{70}{10}-\frac{31}{10}\right) \\ \frac{3}{5}(26) &=4\left(\frac{39}{10}\right) \\ \frac{78}{5} &=4\left(\frac{39}{10}\right) \\ \frac{78}{5} &=\frac{156}{10} \\ \frac{78}{5} &=\frac{78}{5} \cdot \frac{2}{2} \\ \frac{78}{5} &=\frac{78}{5} \end{aligned}\) | Check the solution. First, check the end point, \(\ \frac{31}{2}\), in the related equation. |
| It results in a true statement. | |
| \(\ \begin{aligned} \frac{3}{5}(2 \cdot 0-5) & \leq 4\left(7-\frac{1}{5} \cdot 0\right) \\ \frac{3}{5}(0-5) & \leq 4(7-0) \\ \frac{3}{5}(-5) & \leq 4(7) \\ \frac{-15}{5} & \leq 28 \\ -3 & \leq 28 \end{aligned}\) | Now try any value for \(\ d\) that is within the region \(\ d \leq \frac{31}{2}\). We will try \(\ d=0\). |
| This is also a true statement. |
\(\ d \leq \frac{31}{2}\)
Exercise
Which is the most logical first step for solving for the variable in the inequality: \(\ 8 x+7<3(2 x+1)\)
- Reverse the inequality sign.
- Use the distributive property to clear the parentheses by multiplying each of the terms in the parentheses by 3.
- Subtract \(\ 2x\) from both sides of the inequality.
- Divide both sides of the inequality by 3.
- Answer
-
- Incorrect. The inequality sign is reversed as a result of dividing or multiplying both sides of an inequality by a negative number. This usually occurs as a later step in a multi-step inequality and, in the case of this example, will not happen at all. The correct answer is to use the distributive property to clear the parentheses by multiplying each of the terms in the parentheses by 3.
- Correct. The distributive property clears the parentheses so that you can isolate the variable.
- Incorrect. It is necessary to distribute the 3 before subtracting any terms with \(\ x\) in it. The correct answer is to use the distributive property to clear the parentheses by multiplying each of the terms in the parentheses by 3.
- Incorrect. Dividing both sides by 3, although legal to do, will result in a complicated fraction on the left side of the inequality and is not a logical first step. The correct answer is to use the distributive property to clear the parentheses by multiplying each of the terms in the parentheses by 3.
Exercise
Solve for \(\ x\): \(\ 5[2(3-x)-1] \leq 27\)
- \(\ x \leq-5\)
- \(\ x \leq 27\)
- \(\ x \geq \frac{1}{8}\)
- \(\ x \geq-\frac{1}{5}\)
- Answer
-
- Incorrect. Try substituting any value that is less than or equal to \(\ -5\) for \(\ x\) into the inequality. You will not get a true statement, so \(\ x\) is not less than or equal to -5. The correct answer is: \(\ x \geq-\frac{1}{5}\).
- Incorrect. You need to evaluate the left side of the inequality and then solve for \(\ x\). Remember that \(\ 5[2(3-x)-1]=5[6-2 x-1]\). The correct answer is: \(\ x \geq-\frac{1}{5}\).
- Incorrect. Make sure you are evaluating the left side of the inequality correctly: \(\ 5[2(3-x)-1]=5[6-2 x-1]\). The correct answer is: \(\ x \geq-\frac{1}{5}\).
- Correct. Evaluating the interior parentheses first, you find that \(\ 5[2(3-x)-1]=5[6-2 x-1]=5[5-2 x]=25-10 x\). Solving for \(\ x\), you subtract 25 from both sides and then divide by -10, which requires you to flip the inequality sign from ≤ to ≥.
Summary
Inequalities can have a range of answers. The solutions are often graphed on a number line in order to visualize all of the solutions. Multi-step inequalities are solved using the same processes that work for solving equations with one exception. When you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol. The inequality symbols stay the same whenever you add or subtract either positive or negative numbers to both sides of the inequality.
