Section 2.4: Linear Inequalities
- Page ID
- 186559
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- Evaluate \(2x+5\) when \(x=3\)
- Solve \(2x+5=3x-7\) for \(x\)
- Identify which is larger, \(-7\) or \(9\)
Motivating Problem
Cesar is planning a 4-day trip to visit his friend at a college in another state. It will cost him $198 for airfare, $56 for local transportation, and $45 per day for food. He has $189 in savings and can earn $35 for each lawn he mows. How many lawns must he mow to have more than enough money to pay for the trip?
Fun Fact
Ancient Chinese mathematicians used counting rods to solve inequalities more than 2,000 years ago—long before the Western world formalized the idea. They didn’t have the < or > symbols yet, but they understood the logic just as well!
The Goal
In this section, we will learn how to solve, graph, and interpret linear inequalities. We'll use strategies similar to solving equations, but we must be mindful of the direction of the inequality, especially when multiplying or dividing by a negative number. This skill helps model real-life limits and thresholds in contexts like budgeting, science, and planning.
Graph Inequalities on the Number Line
Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.
What about the solution of an inequality? What number would make the inequality \(x > 3\) true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality \(x > 3\).
We illustrate the solutions to the inequality \(x > 3\) on a number line by shading in all the numbers to the right of 3, indicating that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of \(x > 3\) is shown below. Please note that the following convention is used: light blue arrows indicate the positive direction, and dark blue arrows indicate the negative direction.
The graph of the inequality \(x \geq 3\) is very much like the graph of \(x > 3\), but now we need to show that 3 is a solution, too. We achieve this by placing a bracket at \(x = 3\), as shown below.
Notice that the open parentheses symbol, \((\), shows that the endpoint of the inequality is not included. The open bracket symbol, \([\), shows that the endpoint is included.
Graph on the number line:
a. \(x\leq 1\)
b. \(x<5\)
c. \(x>−1\)
Solution
a. \(x\leq 1\) This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at \(x=1\) to show that it is included.
b. \(x<5\) This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put parentheses at \(x=5\) to show that it is not included.
c. \(x>−1\) This means all numbers greater than −1, but not including −1. We shade in all the numbers on the number line to the right of −1, then put parentheses at \(x=−1\) to show that it is not included.
Graph on the number line:
a. \(x\leq −1\)
b. \(x>2\)
c. \(x<3\)
- Answer
-
a.
b.
c.
We can also represent inequalities using interval notation. As we saw above, the inequality \(x>3\) means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation, we express \(x>3\) as \((3, \infty)\). The symbol \(\infty\) is read as ‘infinity’. It is not an actual number. The figure below shows both the number line and the interval notation.
The inequality \(x\leq 1\) means all numbers less than or equal to 1. There is no lower end to those numbers. We write \(x\leq 1\) in interval notation as \((-\infty, 1]\). The symbol \(-\infty\) is read as ‘negative infinity’. The figure below shows both the number line and interval notation.
Parentheses or brackets in the interval notation match the symbol at the endpoint of the arrow.
Graph on the number line and write in interval notation.
a. \(x \geq -3\)
b. \(x<2.5\)
c. \(x\leq \frac{3}{5}\)
Solution
a.
 |
|
| Shade to the right of −3, and put a bracket at −3. |  |
| Write in interval notation. |  |
 |
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| Shade to the left of 2.5, and put a parenthesis at 2.5. |  |
| Write in interval notation. |  |
 |
|
| Shade to the left of \(-\frac{3}{5}\), and put a bracket at \(-\frac{3}{5}\). |  |
| Write in interval notation. |  |
Graph on the number line and write in interval notation:
a. \(x>2\)
b. \(x\leq −1.5\)
c. \(x\geq \frac{3}{4}\)
- Answer
-
a.
b.
c.
Solve Inequalities using the Subtraction and Addition Properties of Inequality
The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.
\[\begin{array} { l l } { \textbf { Subtraction Property of Equality } } & { \textbf { Addition Property of Equality } } \\ { \text { For any numbers } a , b , \text { and } c , } & { \text { For any numbers } a , b , \text { and } c } \\ { \text { if } \qquad \quad a = b , } & { \text { if } \qquad \quad a = b } \\ { \text { then } a - c = b - c . } & { \text { then } a + c = b + c } \end{array}\nonumber\]
Similar properties hold true for inequalities.
| For example, we know that −4 is less than 2. |  |
| If we subtract 5 from both quantities, is the left side still less than the right side? |  |
| We get −9 on the left and −3 on the right. |  |
| And we know −9 is less than −3. |  |
|
The inequality sign stayed the same. |
Similarly, we can also show that the inequality remains the same for addition.
This leads us to the Subtraction and Addition Properties of Inequality.
\[\begin{array} { l l } { \textbf { Subtraction Property of Inequality } } & { \textbf { Addition Property of Inequality } } \\ { \text { For any numbers } a , b , \text { and } c , } & { \text { For any numbers } a , b , \text { and } c } \\ { \text { if }\qquad \quad a < b } & { \text { if } \qquad \quad a < b } \\ { \text { then } a - c < b - c . } & { \text { then } a + c < b + c } \\\\ { \text { if } \qquad \quad a > b } & { \text { if } \qquad \quad a > b } \\ { \text { then } a - c > b - c . } & { \text { then } a + c > b + c } \end{array}\nonumber\]
We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality \(x+5>9\), the steps would look like this:
\[\begin{array}{rrll} {} &{x + 5} &{ >} &{9} \\ {\text{Subtract 5 from both sides to isolate }x.} &{x + 5 - 5} &{ >} &{9 - 5} \\{} &{x} &{ >} &{4} \\ \end{array}\nonumber\]
Any number greater than 4 is a solution to this inequality.
Solve the inequality \(n-\frac {1}{2} \leq \frac{5}{8}\), graph the solution on the number line, and write the solution in interval notation.
Solution
 |
|
| Add \(\frac{1}{2}\) to both sides of the inequality. |  |
| Simplify. |  |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. | \(\left(-\infty,\frac{9}{8}\right]\) |
Solve each inequality, graphing the solution on the number line, and writing the solution in interval notation.
a. \(p - \frac{3}{4} \geq \frac{1}{6}\)
b. \(r - \frac{1}{3} \leq \frac{7}{12}\)
- Answer
-
a.
b.
Solve Inequalities using the Division and Multiplication Properties of Inequality
The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).
\[\begin{array}{ll} {\textbf{Division Property of Equality}} &{\textbf{Mutiplication Property of Equality}} \\ {\text{For any numbers a, b, c, and c} \neq 0} &{\text{For any numbers a, b, c}} \\ {\text{if } \qquad a = b} &{\text{if} \qquad \quad a = b} \\ {\text{then }\quad \frac{a}{c} = \frac{b}{c}} &{\text{then } \quad ac = bc} \end{array}\nonumber\]
Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?
Consider some numerical examples where we multiply or divide by a positive number.
 |
 |
||
| Divide both sides by 5. |  |
Multiply both sides by 5. |  |
| Simplify. |  |
 |
|
| Fill in the inequality signs. |  |
 |
The inequality signs stayed the same.
Does the inequality stay the same when we divide or multiply by a negative number?
 |
 |
||
| Divide both sides by -5. |  |
Multiply both sides by -5. |  |
| Simplify. |  |
 |
|
| Fill in the inequality signs. |  |
 |
The inequality signs reversed their direction.
When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the direction of the inequality sign reverses.
Here are the Division and Multiplication Properties of Inequality for easy reference.
For any real numbers \(a,b,c\)
\[\begin{array}{ll} {\text{if } a < b \text{ and } c > 0, \text{ then}} &{\frac{a}{c} < \frac{b}{c} \text{ and } ac < bc} \\ {\text{if } a > b \text{ and } c > 0, \text{ then}} &{\frac{a}{c} > \frac{b}{c} \text{ and } ac > bc} \\ {\text{if } a < b \text{ and } c < 0, \text{ then}} &{\frac{a}{c} > \frac{b}{c} \text{ and } ac > bc} \\ {\text{if } a > b \text{ and } c < 0, \text{ then}} &{\frac{a}{c} < \frac{b}{c} \text{ and } ac < bc} \end{array}\nonumber\]
When we divide or multiply an inequality by a:
- If the number is positive, the inequality stays the same.
- If the number is negative, the inequality reverses.
Solve the inequality \(7y<42\), graph the solution on the number line, and write the solution in interval notation.
Solution
 |
|
| Divide both sides of the inequality by 7. Since \(7>0\), the inequality stays the same. |
 |
| Simplify. |  |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. |  |
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
a. \(9c>72\)
b. \(12d\leq 60\)
- Answer
-
a. \(c>8\)
\((8, \infty)\)
b. \(d\leq 5\)
\((-\infty, 5]\)
Solve the inequality \(−10a\geq 50\), graph the solution on the number line, and write the solution in interval notation.
Solution
 |
|
| Divide both sides of the inequality by −10. Since \(−10<0\), the inequality reverses. |
 |
| Simplify. |  |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. |  |
Solve each inequality, graph the solution on the number line, and write the solution in interval notation.
a. \(−8q<32\)
b. \(−7r\leq −70\)
- Answer
-
a. \(q>−4\)
b. \(r\geq10\)
Sometimes, when solving an inequality, the variable ends up on the right side. We can rewrite the inequality in reverse to get the variable to the left.
\[\begin{array}{l} x > a\text{ has the same meaning as } a < x \end{array}\nonumber\]
Think about it as “If Xavier is taller than Alex, then Alex is shorter than Xavier.”
Solve the inequality \(-20 < \frac{4}{5}u\), graph the solution on the number line, and write the solution in interval notation.
Solution
 |
|
| Multiply both sides of the inequality by \(\frac{5}{4}\). Since \(\frac{5}{4} > 0\), the inequality stays the same. |
 |
| Simplify. |  |
| Rewrite the variable on the left. |  |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. |  |
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
a. \(24 \leq \frac{3}{8}m\)
b. \(-24 < \frac{4}{3}n\)
- Answer
-
a.
b.
Solve the inequality \(\frac{t}{-2} \geq 8\), graph the solution on the number line, and write the solution in interval notation.
Solution
 |
|
| Multiply both sides of the inequality by −2. Since \(−2<0\), the inequality reverses. |
 |
| Simplify. |  |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. |  |
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
a. \(\frac{k}{-12}\leq 15\)
b. \(\frac{u}{-4}\geq -16\)
- Answer
-
a.
b.
Solve Inequalities That Require Simplification
Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.
Solve the inequality \(4m\leq 9m+17\), graph the solution on the number line, and write the solution in interval notation.
Solution
 |
|
| Subtract 9m from both sides to collect the variables on the left. |  |
| Simplify. |  |
| Divide both sides of the inequality by −5, and reverse the inequality. |  |
| Simplify. |  |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. |  |
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
a. \(3q\geq 7q−23\)
b. \(6x<10x+19\)
- Answer
-
a.
b.
Solve the inequality \(8p+3(p−12)>7p−28\), graph the solution on the number line, and write the solution in interval notation.
Solution
| Simplify each side as much as possible. | \(8p+3(p−12)>7p−28\) |
| Distribute. | \(8p+3p−36>7p−28\) |
| Combine like terms. | \(11p−36>7p−28\) |
| Subtract \(7p\) from both sides to collect the variables on the left. | \(11p−36−7p>7p−28−7p\) |
| Simplify. | \(4p−36>−28\) |
| Add 36 to both sides to collect the constants on the right. | \(4p−36+36>−28+36\) |
| Simplify. | \(4p>8\) |
| Divide both sides of the inequality by 4; the inequality stays the same. | \(\frac{4p}{4}>84\) |
| Simplify. | \(p>2\) |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. | \((2, \infty)\) |
Solve the inequality, graph the solution on the number line, and write the solution in interval notation.
a. \(9y+2(y+6)>5y−24\)
b. \(6u+8(u−1)>10u+32\)
- Answer
-
a.
b.
Just as some equations are identities and some are contradictions, inequalities can also be identities or contradictions. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.
Solve the inequality \(8x−2(5−x)<4(x+9)+6x\), graph the solution on the number line, and write the solution in interval notation.
Solution
| Simplify each side as much as possible. | \(8x−2(5−x)<4(x+9)+6x\) |
| Distribute. | \(8x−10+2x<4x+36+6x\) |
| Combine like terms. | \(10x−10<10x+36\) |
| Subtract 10x from both sides to collect the variables on the left. | \(10x−10−10x<10x+36−10x\) |
| Simplify. | \(−10<36\) |
| The \(x\)’s are gone, and we have a true statement. | The inequality is an identity. The solution is all real numbers. |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. | \((-\infty, \infty)\) |
Solve the inequality \(4b−3(3−b)>5(b−6)+2b\), graph the solution on the number line, and write the solution in interval notation.
- Answer
-
Solve the inequality \(\frac{1}{3}a - \frac{1}{8}a > \frac{5}{24}a + \frac{3}{4}\), graph the solution on the number line, and write the solution in interval notation.
Solution
 |
|
| Multiply both sides by the LCD, 24, to clear the fractions. |  |
| Simplify. |  |
| Combine like terms. |  |
| Subtract 5a from both sides to collect the variables on the left. |  |
| Simplify. |  |
| The statement is false! | The inequality is a contradiction. |
| There is no solution. | |
| Graph the solution on the number line. |  |
| Write the solution in interval notation. | There is no solution. |
Solve the inequality \(\frac{1}{4}x - \frac{1}{12}x > \frac{1}{6}x + \frac{7}{8}\), graph the solution on the number line, and write the solution in interval notation.
- Answer
-
Translate to an Inequality and Solve
To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.
Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase 'at least’ is the same as ‘greater than or equal to’.
The following table shows some common phrases that indicate inequalities.
| > | \(\geq\) | < | \(\leq\) |
|---|---|---|---|
| " class="lt-math-186559">is greater than | is greater than or equal to | is less than | is less than or equal to |
| " class="lt-math-186559">is more than | is at least | is smaller than | is at most |
| " class="lt-math-186559">is larger than | is no less than | has fewer than | is no more than |
| " class="lt-math-186559">exceeds | is the minimum | is lower than | is the maximum |
Translate and solve. Then write the solution in interval notation.
Twelve times \(c\) is no more than 96.
Solution
| Translate. |  |
| Solve—divide both sides by 12. |  |
| Simplify. |  |
| Write in interval notation. | \((-\infty,8]\) |
Translate and solve. Then write the solution in interval notation and graph on the number line.
Nine times \(z\) is no less than 135.
- Answer
-
Translate and solve. Then write the solution in interval notation and graph on the number line.
Thirty less than \(x\) is at least 45.
Solution
| Translate. |  |
| Solve—add 30 to both sides. |  |
| Simplify. |  |
| Write in interval notation. |  |
| Graph on the number line. |  |
Translate and solve. Then write the solution in interval notation and graph on the number line.
Four more than \(a\) is at most 15.
- Answer
-
Solve Applications with Linear Inequalities
Many real-life situations require us to solve inequalities. In fact, inequality applications are so common that we often do not even realize we are doing algebra. For example, how many gallons of gas can be put in the car for $20? Is the rent on an apartment affordable? Is there enough time before class to go get lunch, eat it, and return to class? How much money should each family member’s holiday gift cost without going over budget?
The method we will use to solve applications involving linear inequalities is very similar to the one we used when solving applications with equations. We will read the problem and make sure all the words are understood. Next, we will identify what we are looking for and assign a variable to represent it. We will restate the problem in a single sentence to facilitate its translation into an inequality. Then, we will solve the inequality.
Emma has landed a new job and will need to relocate. Her monthly income will be $5,265. To qualify for an apartment rental, Emma’s monthly income must be at least three times the rent amount. What is the highest rent Emma will qualify for?
Solution
\[
\begin{array}{ll}
\textbf{Step 1. Read the problem.} & {} \\
\textbf{Step 2. Identify what we are looking for.} & \text{the highest rent Emma will qualify for} \\
\textbf{Step 3. Name what we are looking for.} & {} \\
{} & \text{Let \(r\) = rent} \\
\text{Choose a variable to represent that quantity.} & {} \\[6pt]
\textbf{Step 4. Translate into an inequality.} &
\begin{aligned}
&\text{Emma’s monthly income must be at least}\\
&\text{three times the rent.}
\end{aligned} \\[6pt]
\textbf{Step 5. Solve the inequality.} & 5265 \ge 3r \\
\text{Remember, } \(a > x\) \text{ has the same meaning} & 1755 \ge r \\
\text{as } \(x < a\) & r \le 1755 \\[6pt]
\textbf{Step 6. Check the answer in the problem} & {} \\
\text{and make sure it makes sense.} & {} \\[3pt]
{} &
\begin{aligned}
&\text{A maximum rent of \$1,755 seems}\\
&\text{reasonable for an income of \$5,265.}
\end{aligned} \\[6pt]
\textbf{Step 7. Answer the question in a complete sentence.}
& \text{The maximum rent is \$1,755.}
\end{array}
\]
Alan is loading a pallet with boxes that each weigh 45 pounds. The pallet can safely support no more than 900 pounds. How many boxes can he safely load onto the pallet?
- Answer
-
There can be no more than 20 boxes.
Sometimes an application requires the solution to be a whole number, but the algebraic solution to the inequality is not a whole number. In that case, we must round the algebraic solution to a whole number. The context of the application will determine whether we round up or down. To check applications like this, we will round our answer to a number that is easy to compute with and ensure that this number makes the inequality true.
Dawn won a mini-grant of $4,000 to buy tablet computers for her classroom. The tablets she would like to buy cost $254.12 each, including tax and delivery. What is the maximum number of tablets Dawn can buy?
Solution
\[
\begin{array}{ll}
\textbf{Step 1. Read the problem.} & {} \\
\textbf{Step 2. Identify what we are looking for.} & \text{the maximum number of tablets Dawn can buy} \\
\textbf{Step 3. Name what we are looking for.} & {} \\
{} & \text{Let \(n\) = the number of tablets.} \\
\text{Choose a variable to represent that quantity.} & {} \\
\textbf{Step 4. Translate. Write a sentence that} & {} \\
\text{gives the information to find it.} & \begin{aligned}
&254.12\text{ times the number of tablets is no}\\
&\text{more than \$4000.}
\end{aligned} \\
\text{Translate into an inequality.} & 254.12n \le 4000 \\
\textbf{Step 5. Solve the inequality.} & n \le 15.74 \\
\text{But \(n\) must be a whole number of tablets,} & {} \\
\text{so round to 15.} & n \le 15 \\[6pt]
\textbf{Step 6. Check the answer in the problem} & {} \\
\text{and make sure it makes sense.} & {} \\[3pt]
{} & \begin{aligned}
&\text{Rounding down the price to \$250,}\\
&\text{15 tablets would cost \$3750, while}\\
&\text{16 tablets would be \$4000.}\\
&\text{So a maximum of 15 tablets at \$254.12}\\
&\text{seems reasonable.}
\end{aligned} \\[6pt]
\textbf{Step 7. Answer the question in a complete sentence.}
& \text{Dawn can buy a maximum of 15 tablets.}
\end{array}
\nonumber\]
Angie has $20 to spend on juice boxes for her son’s preschool picnic. Each pack of juice boxes costs $2.63. What is the maximum number of packs she can buy?
- Answer
-
She can buy at most seven packs.
Sergio and Lizeth have a tight vacation budget. They plan to rent a car from a company that charges $75 a week plus $0.25 a mile. How many miles can they travel and still keep within their $200 budget?
Solution
\[
\begin{array}{ll}
\textbf{Step 1. Read the problem.} & {} \\
\textbf{Step 2. Identify what we are looking for.} & \text{the number of miles Sergio and Lizeth can travel} \\
\textbf{Step 3. Name what we are looking for.} & {} \\
{} & \text{Let \(m\) = the number of miles.} \\
\text{Choose a variable to represent that quantity.} & {} \\[6pt]
\textbf{Step 4. Translate. Write a sentence that gives the information to find it.}
& \begin{aligned}
&\$75 + 0.25m\text{ is}\\
&\text{less than or equal to \$200.}
\end{aligned} \\[6pt]
\textbf{Step 5. Solve the inequality.} & 0.25m \le 125 \\
{} & m \le 500\text{ miles} \\[12pt]
\textbf{Step 6. Check the answer in the problem} & {} \\
\text{and make sure it makes sense.} & {} \\[3pt]
{} &
\begin{aligned}
&\text{Yes, }75 + 0.25(500) = 200.
\end{aligned} \\[6pt]
\textbf{Step 7. Answer the question in a complete sentence.}
& \text{Sergio and Lizeth can travel 500 miles}\\
{} & \text{and still stay on budget.}
\end{array}
\]
Taleisha’s phone plan costs her $28.80 a month plus $0.20 per text message. How many text messages can they use and keep their monthly phone bill no more than $50?
- Answer
-
They can use no more than 106 text messages a month.
A common goal of most businesses is to make a profit. Profit is the money that remains when the expenses have been subtracted from the money earned. In the following example, we will determine the number of jobs a small businessman needs to complete each month to achieve a certain level of profit.
Elliot has a landscape maintenance business. His monthly expenses are $1,100. If he charges $60 per job, how many jobs must he do to earn a profit of at least $4,000 a month?
Solution
\[
\begin{array}{ll}
\textbf{Step 1. Read the problem.} & {} \\
\textbf{Step 2. Identify what we are looking for.} & \text{the number of jobs Elliot needs} \\
\textbf{Step 3. Name what we are looking for.} & {} \\
{} & \text{Let \(j\) = the number of jobs.} \\
\text{Choose a variable to represent it.} & {} \\[6pt]
\textbf{Step 4. Translate. Write a sentence that gives the information to find it.}
& \begin{aligned}
&\$60\text{ times the number of jobs minus \$1,100}\\
&\text{is at least \$4,000.}
\end{aligned} \\[6pt]
\textbf{Step 5. Solve the inequality.} & 60j - 1100 \ge 4000 \\
{} & 60j \ge 5100 \\
{} & j \ge 85\text{ jobs} \\[6pt]
\textbf{Step 6. Check the answer in the problem} & {} \\
\text{and make sure it makes sense.} & {} \\[3pt]
{} &
\begin{aligned}
&\text{If Elliot did 90 jobs, his profit would be}\\
&60(90)-1100 = \$4,300,\text{ which is more than \$4,000.}
\end{aligned} \\[6pt]
\textbf{Step 7. Answer the question in a complete sentence.}
& \text{Elliot must work at least 85 jobs.}
\end{array}
\]
Caleb has a pet sitting business. He charges $32 per hour. His monthly expenses are $2272. How many hours must he work to earn a profit of at least $800 per month?
- Answer
-
He must work at least 96 hours.