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6.3: Multi-Step Inequalities

Difficulty Level: At Grade Created by: CK-12

Learning Objectives

• Solve a two-step inequality.
• Solve a multi-step inequality.
• Identify the number of solutions of an inequality.
• Solve real-world problems using inequalities.

Introduction

In the last two sections, we considered very simple inequalities which required one-step to obtain the solution. However, most inequalities require several steps to arrive at the solution. As with solving equations, we must use the order of operations to find the correct solution. In addition remember that when we multiply or divide the inequality by a negative number the direction of the inequality changes.

The general procedure for solving multi-step inequalities is as follows.

1. Clear parenthesis on both sides of the inequality and collect like terms.
2. Add or subtract terms so the variable is on one side and the constant is on the other side of the inequality sign.
3. Multiply and divide by whatever constants are attached to the variable. Remember to change the direction of the inequality if you multiply or divide by a negative number.

Solve a Two-Step Inequality

Example 1

Solve each of the following inequalities and graph the solution set.

a) \begin{align*}6x-5 <10\end{align*}

b) \begin{align*}-9x<-5x-15\end{align*}

c) \begin{align*}-\frac{9x}{5} \leq 24\end{align*}

Solution

a) \begin{align*}\text{Original problem}: & & 6x - 5 & < 10\\ \text{Add} \ 5 \ \text{to both sides}: & & 6x - 5 + 5 & < 10 + 5\\ \text{Simplify}. & & 6x & < 15\\ \text{Divide both sides by} \ 6. & & \frac{6x}{6} & < \frac{15}{6}\\ \text{Simplify}. & & x & < \frac{5} {2} \ \text{Answer}\end{align*}

b) \begin{align*}\text{Original problem}. & & -9x & \leq -5x-15\\ \text{Add} \ 5x \ \text{to both sides}. & & -9x+5x & \leq -5x + 5x - 15\\ \text{Simplify}. & & -4x & < -15\\ \text{Divide both sides by} \ -4. & & \frac{-4x} {-4} & > \frac{-15} {-4} \ \text{Inequality sign was flipped}\\ \text{Simplify}. & & x & > \frac{15}{4} \ \text{Answer}\end{align*}

c) \begin{align*}\text{Original problem}. & & -9x & \leq 24\\ \text{Multiply both sides by} \ 5. & & \frac{-9x} {5}.5 & \leq 24.5\\ \text{Simplify}. & & -9x & \leq 120\\ \text{Divide both sides by} -9. & & \frac{-9x} {-9}& >\frac{120} {-9}\ \text{Inequality sign was flipped}\\ \text{Simplify}. & & x & \geq -\frac{40} {3} \ \text{Answer}\end{align*}

Solve a Multi-Step Inequality

Example 2

Each of the following inequalities and graph the solution set.

a) \begin{align*} \frac{9x} {5}-7 \geq -3x+12\end{align*}

b) \begin{align*} -25x+12 \leq -10x - 12\end{align*}

Solution

a) \begin{align*}\text{Original problem} & & \frac{9x} {5}-7 & \geq -3x+12\\ \text{Add} \ 3x \ \text{to both sides}. & & \frac{9x} {5}+3x-7 & \geq -3x+3x+12\\ \text{Simplify}. & & \frac{24x} {5}-7 & \geq 12\\ \text{Add} \ 7 \ \text{to both sides}. & & \frac{24x} {5}-7+7 & \geq 12+7\\ \text{Simplify}. & & \frac{24x} {5}-7 & \geq 19\\ \text{Multiply} \ 5 \ \text{to both sides}. & & 5\cdot \frac{24x} {5} & \geq 5.19\\ \text{Simplify}. & & 24x & \geq 95\\ \text{Divide both sides by} \ 24. & & \frac{24x} {24} & \geq \frac{95} {24}\\ \text{Simplify}. & & x & \geq \frac{95} {24} \ \text{Answer}\end{align*}

b) \begin{align*}\text{Original problem} & & -25x+12 & \leq -10x-12\\ \text{Add} \ 5 \ \text{to both sides}. & & -25x+10x+12 & \leq -10x+10x-12\\ \text{Simplify}. & & -15x+12 & \leq -12\\ \text{Subtract} \ 12 \ \text{from both sides}. & & -15x+12-12 & \leq -12-12\\ \text{Simplify}. & & -15x & \leq -24\\ \text{Divide both sides by} \ -15. & & \frac{-15x} {-15} & \geq \frac{-24} {-15} \ \text{Inequality sign was flipped}\\ \text{Simplify}. & & x & \geq \frac{8} {5} \ \text{Answer}\end{align*}

Example 3

Solve the following inequalities.

a) \begin{align*} 4x - 2(3x - 9) \leq -4(2x - 9)\end{align*}

b) \begin{align*} \frac{5x-1} {4} > -2(x+5)\end{align*}

Solution

a) \begin{align*}\text{Original problem} & & 4x-2(3x-9) & \leq -4(2x-9)\\ \text{Simplify parentheses}. & & 4x-6x+18 & \leq -8x+36\\ \text{Collect like terms}. & & --2x+18 & \leq -8x+36\\ \text{Add} \ 8x \ \text{to both sides}. & & -2x+8x+18 & \leq -8x+8x+36\\ \text{Simplify}. & & -6x+18 & \leq 36\\ \text{Subtract} \ 18 \ \text{from both sides}. & & -6x+18-18 & \leq 36-18\\ \text{Simplify}. & & 6x & \leq 18\\ \text{Divide both sides by} \ 6. & & \frac{6x} {6} & \leq \frac{18} {6}\\ \text{Simplify}. & & x & \leq 3 \ \text{Answer}\end{align*}

b) \begin{align*}\text{Original problem} & & \frac{5x-1} {4} & > -2(x+5)\\ \text{Simplify parenthesis}. & & \frac{5x-1} {4} & > -2x-10\\ \text{Multiply both sides by} \ 4. & & 4.\frac{5x-1} {4} & > 4(-2x-10)\\ \text{Simplify}. & & 5x-1 & > -8x-40\\ \text{Add} \ 8x \ \text{to both sides}. & & 5x+8x-1 & > -8x+8x-40\\ \text{Simplify}. & & 13x-1 & > -40\\ \text{Add} \ 1 \ \text{to both sides}. & & 13-1+1 & > - 40+1\\ \text{Simplify}. & & 13x & > -39\\ \text{Divide both sides by} \ 13. & & \frac{13x} {13} & > -\frac{39} {13}\\ \text{Simplify}. & & x & > - 3 \ \text{Answer}\end{align*}

Identify the Number of Solutions of an Inequality

Inequalities can have:

• A set that has an infinite number of solutions.
• No solutions
• A set that has a discrete number of solutions.

Infinite Number of Solutions

The inequalities we have solved so far all have an infinite number of solutions. In the last example, we saw that the inequality

\begin{align*} \frac{5x-1} {4}>-2(x+5)\end{align*} has the solution \begin{align*} x>-3\end{align*}

This solution says that all real numbers greater than -3 make this inequality true. You can see that the solution to this problem is an infinite set of numbers.

No solutions

\begin{align*}\text{Consider the inequality} & & x-5 ( & > x+6 )\\ \text{This simplifies to} & & -5 & >6\end{align*}

This statements is not true for any value of \begin{align*}x\end{align*}. We say that this inequality has no solution.

Discrete solutions

So far we have assumed that the variables in our inequalities are real numbers. However, in many real life situations we are trying to solve for variables that represent integer quantities, such as number of people, number of cars or number of ties.

Example 4

Raul is buying ties and he wants to spend $200 or less on his purchase. The ties he likes the best cost$50. How many ties could he purchase?

Solution

Let \begin{align*}x =\end{align*} the number of ties Raul purchases.

We can write an inequality that describes the purchase amount using the formula.

\begin{align*}\text{(number of ties) }\times \text{(price of a tie)} \leq \200\end{align*} or \begin{align*}50x \leq 200\end{align*}

We simplify our answer. \begin{align*} x \leq 4\end{align*}

This solution says that Raul bought four or less ties. Since ties are discrete objects, the solution set consists of five numbers \begin{align*}\left \{0, 1, 2, 3, 4 \right \}\end{align*}.

Solve Real-World Problems Using Inequalities

Sometimes solving a word problem involves using an inequality.

Example 5

In order to get a bonus this month, Leon must sell at least 120 newspaper subscriptions. He sold 85 subscriptions in the first three weeks of the month. How many subscriptions must Leon sell in the last week of the month?

Solution

Step 1

We know that Leon sold 85 subscriptions and he must sell at least 120 subscriptions.

We want to know the least amount of subscriptions he must sell to get his bonus.

Let \begin{align*}x =\end{align*} the number of subscriptions Leon sells in the last week of the month.

Step 2

The number of subscriptions per month must be greater than 120.

We write

\begin{align*}85 + x \geq 120\end{align*}

Step 3

We solve the inequality by subtracting 85 from both sides \begin{align*}x \geq 35\end{align*}

Answer Leon must sell 35 or more subscriptions in the last week to get his bonus.

Step 4:

To check the answer, we see that \begin{align*}85 + 35 = 120\end{align*}. If he sells 35 or more subscriptions the number of subscriptions sold that month will be 120 or more.

Example 6

Virena's Scout Troup is trying to raise at least $650 this spring. How many boxes of cookies must they sell at$4.50 per box in order to reach their goal?

Solution

Step 1

Virena is trying to raise at least $650 Each box of cookies sells for$4.50

Let \begin{align*}x =\end{align*} number of boxes sold

The inequality describing this problem is:

\begin{align*}450x \geq 650.\end{align*}

Step 3

We solve the inequality by dividing both sides by 4.50

\begin{align*}x \geq 1444.44\end{align*}

Answer We round up the answer to 145 since only whole boxes can be sold.

Step 4

If we multiply 145 by $4.50 we obtain$652.50. If Virena’s Troop sells more than 145 boxes, they raise more that 650. The answer checks out. Example 7 The width of a rectangle is 20 inches. What must the length be if the perimeter is at least 180 inches? Solution Step 1 width = 20 inches Perimeter is at least 180 inches What is the smallest length that gives that perimeter? Let \begin{align*}x =\end{align*} length of the rectangle Step 2 Formula for perimeter is \begin{align*}\text{Perimeter}\ = 2 \times \ \text{length} \ + 2 \times \ \text{width} \end{align*} Since the perimeter must be at least 180 inches, we have the following equation. \begin{align*}2x+ 2(20) \geq 180\end{align*} Step 3 We solve the inequality. Simplify. \begin{align*}2x + 40 \geq 180\end{align*} Subtract 40 from both sides. \begin{align*} 2x\geq 140\end{align*} Divide both sides by 2. \begin{align*}x \geq 70\end{align*} Answer The length must be at least 70 inches. Step 4 If the length is at least 70 inches and the width is 20 inches, then the perimeter can be found by using this equation. \begin{align*}2(70) + 2(20) = 180 \ inches\end{align*} The answer check out. Lesson Summary • The general procedure for solving multi-step inequalities is as follows. 1. Clear parentheses on both sides of the inequality and collect like terms. 2. Add or subtract terms so the variable is on one side and the constant is on the other side of the inequality sign. 3. Multiply and divide by whatever constants are attached to the variable. Remember to change the direction of the inequality if you multiply or divide by a negative number. • Inequalities can have multiple solutions, no solutions, or discrete solutions. Review Questions Solve the following inequalities and give the solution in set notation and show the solution graph. 1. \begin{align*}4x + 3< -1\end{align*} 2. \begin{align*}2x< 7x- 36\end{align*} 3. \begin{align*}5x>8x+27\end{align*} 4. \begin{align*}5-x<9+x\end{align*} 5. \begin{align*}4-6x \leq 2(2x+3) \end{align*} 6. \begin{align*}5(4x+3)\geq 9(x-2)-x\end{align*} 7. \begin{align*}2(2x- 1) + 3 < 5(x + 3) -2x\end{align*} 8. \begin{align*}8x-5(4x+1) \geq -1 +2(4x-3) \end{align*} 9. \begin{align*} 2( 7x- 2) - 3(x + 2) < 4x - (3x + 4) \end{align*} 10. \begin{align*}\frac{2}{3}x-\frac{1}{2}(4x-1) \geq x+2(x-3) \end{align*} 11. At the San Diego Zoo, you can either pay22.75 for the entrance fee or $71 for the yearly pass which entitles you to unlimited admission. At most how many times can you enter the zoo for the$22.75 entrance fee before spending more than the cost of a yearly membership?
12. Proteek’s scores for four tests were 82, 95, 86 and 88. What will he have to score on his last test to average at least 90 for the term?

1. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x<-1\left . \right \}\end{align*}
2. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x>\frac{36}{5}\left . \right \}\end{align*}
3. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x<-9\left . \right \}\end{align*}
4. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x>-2\left . \right \}\end{align*}
5. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x\geq -\frac{1}{5}\left . \right \}\end{align*}
6. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x\geq -\frac{33}{12}\left . \right \}\end{align*}
7. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x<14\left . \right \}\end{align*}
8. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x\leq \frac{1}{10}\left . \right \}\end{align*}
9. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x<\frac{3}{5}\left . \right \}\end{align*}
10. \begin{align*}\left \{ \right .x \mid x\end{align*} is a real number, \begin{align*}x\leq \frac{3}{2}\left . \right \}\end{align*}
11. At most 3 times.
12. At least 99.

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