problem solving involving inequalities

Solving Inequality Word Questions

(You might like to read Introduction to Inequalities and Solving Inequalities first.)

In Algebra we have "inequality" questions like:

Sam and Alex play in the same soccer team. Last Saturday Alex scored 3 more goals than Sam, but together they scored less than 9 goals. What are the possible number of goals Alex scored?

How do we solve them?

The trick is to break the solution into two parts:

Turn the English into Algebra.

Then use Algebra to solve.

Turning English into Algebra

To turn the English into Algebra it helps to:

• Read the whole thing first
• Do a sketch if needed
• Assign letters for the values
• Find or work out formulas

We should also write down what is actually being asked for , so we know where we are going and when we have arrived!

The best way to learn this is by example, so let's try our first example:

Assign Letters:

• the number of goals Alex scored: A
• the number of goals Sam scored: S

We know that Alex scored 3 more goals than Sam did, so: A = S + 3

And we know that together they scored less than 9 goals: S + A < 9

We are being asked for how many goals Alex might have scored: A

Sam scored less than 3 goals, which means that Sam could have scored 0, 1 or 2 goals.

Alex scored 3 more goals than Sam did, so Alex could have scored 3, 4, or 5 goals .

• When S = 0, then A = 3 and S + A = 3, and 3 < 9 is correct
• When S = 1, then A = 4 and S + A = 5, and 5 < 9 is correct
• When S = 2, then A = 5 and S + A = 7, and 7 < 9 is correct
• (But when S = 3, then A = 6 and S + A = 9, and 9 < 9 is incorrect)

Lots More Examples!

Example: Of 8 pups, there are more girls than boys. How many girl pups could there be?

• the number of girls: g
• the number of boys: b

We know that there are 8 pups, so: g + b = 8, which can be rearranged to

We also know there are more girls than boys, so:

We are being asked for the number of girl pups: g

So there could be 5, 6, 7 or 8 girl pups.

Could there be 8 girl pups? Then there would be no boys at all, and the question isn't clear on that point (sometimes questions are like that).

• When g = 8, then b = 0 and g > b is correct (but is b = 0 allowed?)
• When g = 7, then b = 1 and g > b is correct
• When g = 6, then b = 2 and g > b is correct
• When g = 5, then b = 3 and g > b is correct
• (But if g = 4, then b = 4 and g > b is incorrect)

A speedy example:

Example: Joe enters a race where he has to cycle and run. He cycles a distance of 25 km, and then runs for 20 km. His average running speed is half of his average cycling speed. Joe completes the race in less than 2½ hours, what can we say about his average speeds?

• Average running speed: s
• So average cycling speed: 2s
• Speed = Distance Time
• Which can be rearranged to: Time = Distance Speed

We are being asked for his average speeds: s and 2s

The race is divided into two parts:

• Distance = 25 km
• Average speed = 2s km/h
• So Time = Distance Average Speed = 25 2s hours
• Distance = 20 km
• Average speed = s km/h
• So Time = Distance Average Speed = 20 s hours

Joe completes the race in less than 2½ hours

• The total time < 2½
• 25 2s + 20 s < 2½

So his average speed running is greater than 13 km/h and his average speed cycling is greater than 26 km/h

In this example we get to use two inequalities at once:

Example: The velocity v m/s of a ball thrown directly up in the air is given by v = 20 − 10t , where t is the time in seconds. At what times will the velocity be between 10 m/s and 15 m/s?

• velocity in m/s: v
• the time in seconds: t
• v = 20 − 10t

We are being asked for the time t when v is between 5 and 15 m/s:

So the velocity is between 10 m/s and 15 m/s between 0.5 and 1 second after.

And a reasonably hard example to finish with:

Example: A rectangular room fits at least 7 tables that each have 1 square meter of surface area. The perimeter of the room is 16 m. What could the width and length of the room be?

Make a sketch: we don't know the size of the tables, only their area, they may fit perfectly or not!

• the length of the room: L
• the width of the room: W

The formula for the perimeter is 2(W + L) , and we know it is 16 m

• 2(W + L) = 16
• L = 8 − W

We also know the area of a rectangle is the width times the length: Area = W × L

And the area must be greater than or equal to 7:

• W × L ≥ 7

We are being asked for the possible values of W and L

Let's solve:

So the width must be between 1 m and 7 m (inclusive) and the length is 8−width .

• Say W = 1, then L = 8−1 = 7, and A = 1 x 7 = 7 m 2 (fits exactly 7 tables)
• Say W = 0.9 (less than 1), then L = 7.1, and A = 0.9 x 7.1 = 6.39 m 2 (7 won't fit)
• Say W = 1.1 (just above 1), then L = 6.9, and A = 1.1 x 6.9 = 7.59 m 2 (7 fit easily)
• Likewise for W around 7 m

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1.5: Solve Inequalities

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Learning Objectives

Represent inequalities on a number line, represent inequalities using interval notation.

• Use the addition and multiplication properties to solve algebraic inequalities and express their solutions graphically and with interval notation
• Solve inequalities that contain absolute value
• Combine properties of inequality to isolate variables, solve algebraic inequalities, and express their solutions graphically
• Simplify and solve algebraic inequalities using the distributive property to clear parentheses and fractions

First, let’s define some important terminology. An inequality is a mathematical statement that compares two expressions using the ideas of greater than or less than. Special symbols are used in these statements. When you read an inequality, read it from left to right—just like reading text on a page. In algebra, inequalities are used to describe large sets of solutions. Sometimes there are an infinite amount of numbers that will satisfy an inequality, so rather than try to list off an infinite amount of numbers, we have developed some ways to describe very large lists in succinct ways.

The first way you are probably familiar with—the basic inequality. For example:

• \({x}\lt{9}\) or try to list all the possible numbers that are less than 9? (hopefully, your answer is no)

Note how placing the variable on the left or right of the inequality sign can change whether you are looking for greater than or less than.

For example:

• \(x\lt5\) means all the real numbers that are less than 5, whereas;
• \(x\gt{5}\) note how the inequality is still pointing the same direction relative to x. This statement represents all the real numbers that are greater than 5, which is easier to interpret than 5 is less than x.

The second way is with a graph using the number line:

And the third way is with an interval.

We will explore the second and third ways in depth in this section. Again, those three ways to write solutions to inequalities are:

• an inequality
• an interval

Inequality Signs

The box below shows the symbol, meaning, and an example for each inequality sign. Sometimes it’s easy to get tangled up in inequalities, just remember to read them from left to right.

The inequality \({y}<{x}\). The sides of any inequality can be switched as long as the inequality symbol between them is also reversed.

Graphing an Inequality

Inequalities can also be graphed on a number line. Below are three examples of inequalities and their graphs. Graphs are a very helpful way to visualize information – especially when that information represents an infinite list of numbers!

\(x\leq -4\). This translates to all the real numbers on a number line that are less than or equal to 4.

\({x}\geq{-3}\). This translates to all the real numbers on the number line that are greater than or equal to -3.

Each of these graphs begins with a circle—either an open or closed (shaded) circle. This point is often called the end point of the solution. A closed, or shaded, circle is used to represent the inequalities greater than or equal to \(\displaystyle \left(\leq\right)\). The point is part of the solution. An open circle is used for greater than (>) or less than (<). The point is not part of the solution.

The graph then extends endlessly in one direction. This is shown by a line with an arrow at the end. For example, notice that for the graph of \(−3\), represented with a closed circle since the inequality is greater than or equal to \(−3\). The arrow at the end indicates that the solutions continue infinitely.

Graph the inequality \(x\ge 4\) [reveal-answer q=”797241″]Show Solution[/reveal-answer] [hidden-answer a=”797241″]

We can use a number line as shown. Because the values for x include 4, we place a solid dot on the number line at 4.

[/hidden-answer]

This video shows an example of how to draw the graph of an inequality.

A YouTube element has been excluded from this version of the text. You can view it online here: pb.libretexts.org/ba/?p=52

Write an inequality describing all the real numbers on the number line that are less than 2, then draw the corresponding graph. [reveal-answer q=”867890″]Show Solution[/reveal-answer] [hidden-answer a=”867890″]

We need to start from the left and work right, so we start from negative infinity and end at \(-2\).

Inequality: \(x<2\)

To draw the graph, place an open dot on the number line first, then draw a line extending to the left. Draw an arrow at the leftmost point of the line to indicate that it continues for infinity.

The following video shows how to write an inequality mathematically when it is given in words. We will then graph it.

Another commonly used, and arguably the most concise, method for describing inequalities and solutions to inequalities is called interval notation. With this convention, sets are built with parentheses or brackets, each having a distinct meaning. The solutions to \(\left[4,\infty \right)\). This method is widely used and will be present in other math courses you may take.

The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval , or a set of numbers in which a solution falls, are \(-2\) and \(-2\), but not including \(\left(-1,0\right)\), all real numbers between, but not including \(0\); and \(1\). The table below outlines the possibilities. Remember to read inequalities from left to right, just like text.

The table below describes all the possible inequalities that can occur and how to write them using interval notation, where a and b are real numbers.

Describe the inequality \(x\ge 4\) using interval notation [reveal-answer q=”817362″]Show Solution[/reveal-answer] [hidden-answer a=”817362″]

The solutions to \(\left[4,\infty \right)\).

Note the use of a bracket on the left because 4 is included in the solution set.

In the following video we show another example of using interval notation to describe an inequality.

Use interval notation to indicate all real numbers greater than or equal to \(-2\). [reveal-answer q=”961990″]Show Solution[/reveal-answer] [hidden-answer a=”961990″]

Use a bracket on the left of \(\left[-2,\infty \right)\). The bracket indicates that \(-2\) to infinity.

In the following video we show another example of translating words into an inequality and writing it in interval notation, as well as drawing the graph.

Think About It

In the previous examples you were given an inequality or a description of one with words and asked to draw the corresponding graph and write the interval. In this example you are given an interval and asked to write the inequality and draw the graph.

Given \(\left(-\infty,10\right)\), write the associated inequality and draw the graph.

In the box below, write down whether you think it will be easier to draw the graph first or write the inequality first.

[practice-area rows=”1″][/practice-area] [reveal-answer q=”15120″]Show Solution[/reveal-answer]

[hidden-answer a=”15120″]

We will draw the graph first.

The interval reads “all real numbers less than 10,” so we will start by placing an open dot on 10 and drawing a line to the left with an arrow indicating the solution continues to negative infinity.

To write the inequality, we will use < since the parentheses indicate that 10 is not included. \(x<10\)

In the following video, you will see examples of how to draw a graph given an inequality in interval notation.

And finally, one last video that shows how to write inequalities using a graph, with interval notation and as an inequality.

Solve Single-Step Inequalities

Solve inequalities with addition and subtraction.

You can solve most inequalities using inverse operations as you did for solving equations. This is because when you add or subtract the same value from both sides of an inequality, you have maintained the inequality. These properties are outlined in the box below.

Addition and Subtraction Properties of Inequality

If \(a+c>b+c\).

If \(a−c>b−c\).

Because inequalities have multiple possible solutions, representing the solutions graphically provides a helpful visual of the situation, as we saw in the last section. The example below shows the steps to solve and graph an inequality and express the solution using interval notation.

Solve for x.

\({x}+3\lt{5}\)

[reveal-answer q=”952771″]Show Solution[/reveal-answer] [hidden-answer a=”952771″]

It is helpful to think of this inequality as asking you to find all the values for x , including negative numbers, such that when you add three you will get a number less than 5.

\(\displaystyle \begin{array}{l}x+3<\,\,\,\,5\\\underline{\,\,\,\,\,-3\,\,\,\,-3}\\x\,\,\,\,\,\,\,\,<\,\,\,\,2\,\,\end{array}\)

Isolate the variable by subtracting 3 from both sides of the inequality.

Interval: \(\left(-\infty, 2\right)\)

The line represents all the numbers to which you can add 3 and get a number that is less than 5. There’s a lot of numbers that solve this inequality!

Just as you can check the solution to an equation, you can check a solution to an inequality. First, you check the end point by substituting it in the related equation. Then you check to see if the inequality is correct by substituting any other solution to see if it is one of the solutions. Because there are multiple solutions, it is a good practice to check more than one of the possible solutions. This can also help you check that your graph is correct.

The example below shows how you could check that \(x+3<5\) .

Check that \(x+3<5\).

[reveal-answer q=”811564″]Show Solution[/reveal-answer] [hidden-answer a=”811564″]

Substitute the end point 2 into the related equation, \(x+3=5\).

\(\begin{array}{r}x+3=5 \\ 2+3=5 \\ 5=5\end{array}\)

Pick a value less than 2, such as 0, to check into the inequality. (This value will be on the shaded part of the graph.)

\(\displaystyle \begin{array}{r}x+3<5 \\ 0+3<5 \\ 3<5\end{array}\)

\(x+3<5\).[/hidden-answer]

The following examples show inequality problems that include operations with negative numbers. The graph of the solution to the inequality is also shown. Remember to check the solution. This is a good habit to build!

Solve for x : \(x-10\leq-12\) [reveal-answer q=”815894″]Show Solution[/reveal-answer] [hidden-answer a=”815894″]

Isolate the variable by adding 10 to both sides of the inequality.

\(\displaystyle \begin{array}{r}x-10\le -12\\\underline{\,\,\,+10\,\,\,\,\,+10}\\x\,\,\,\,\,\,\,\,\,\,\le \,\,\,-2\end{array}\)

Check the solution to \(x-10\leq -12\) [reveal-answer q=”268062″]Show Solution[/reveal-answer] [hidden-answer a=”268062″]

Substitute the end point \(x-10=−12\)

\(\displaystyle \begin{array}{r}x-10=-12\,\,\,\\\text{Does}\,\,\,-2-10=-12?\\-12=-12\,\,\,\end{array}\)

Pick a value less than \(−5\), to check in the inequality. (This value will be on the shaded part of the graph.)

\(\displaystyle \begin{array}{r}x-10\le -12\,\,\,\\\text{ }\,\text{ Is}\,\,-5-10\le -12?\\-15\le -12\,\,\,\\\text{It}\,\text{checks!}\end{array}\)

\(x-10\leq -12\)

Solve for a . \(a-17>-17\) [reveal-answer q=”343031″]Show Solution[/reveal-answer] [hidden-answer a=”343031″]

Isolate the variable by adding 17 to both sides of the inequality.

\(\displaystyle \begin{array}{r}a-17>-17\\\underline{\,\,\,+17\,\,\,\,\,+17}\\a\,\,\,\,\,\,\,\,\,\,\,>\,\,\,\,\,\,0\end{array}\)

Inequality: \(\displaystyle a\,\,>\,0\)

Interval: \(\left(0,\infty\right)\) Note how we use parentheses on the left to show that the solution does not include 0.

Graph: Note the open circle to show that the solution does not include 0.

Check the solution to \(a-17>-17\) [reveal-answer q=”653357″]Show Solution[/reveal-answer] [hidden-answer a=”653357″]

Is \(a-17>-17\)?

Substitute the end point 0 into the related equation.

\(\displaystyle \begin{array}{r}a-17=-17\,\,\,\\\text{Does}\,\,\,0-17=-17?\\-17=-17\,\,\,\end{array}\)

Pick a value greater than 0, such as 20, to check in the inequality. (This value will be on the shaded part of the graph.)

\(\displaystyle \begin{array}{r}a-17>-17\,\,\,\\\text{Is }\,\,20-17>-17?\\3>-17\,\,\,\\\\\text{It checks!}\,\,\,\,\end{array}\)

\(a-17>-17\)

The previous examples showed you how to solve a one-step inequality with the variable on the left hand side. The following video provides examples of how to solve the same type of inequality.

What would you do if the variable were on the right side of the inequality? In the following example, you will see how to handle this scenario.

Solve for x : \(4\geq{x}+5\) [reveal-answer q=”815893″]Show Solution[/reveal-answer] [hidden-answer a=”815893″]

\(\displaystyle \begin{array}{r}4\geq{x}+5 \\\underline{\,\,\,-5\,\,\,\,\,-5}\\-1\,\,\,\,\,\,\,\,\,\,\ge \,\,\,x\end{array}\)

Rewrite the inequality with the variable on the left – this makes writing the interval and drawing the graph easier.

\(x\le{-1}\)

Note how the the pointy part of the inequality is still directed at the variable, so instead of reading as negative one is greater or equal to x, it now reads as x is less than or equal to negative one.

Check the solution to \(4\geq{x}+5\) [reveal-answer q=”568062″]Show Solution[/reveal-answer] [hidden-answer a=”568062″]

Substitute the end point \(4=x+5\)

\(\displaystyle \begin{array}{r}4=x+5\,\,\,\\\text{Does}\,\,\,4=-1+5?\\-1=-1\,\,\,\end{array}\)

\(\displaystyle \begin{array}{r}4\geq{-5}+5\,\,\,\\\text{ }\,\text{ Is}\,\,4\ge 0?\\\text{It}\,\text{checks!}\end{array}\)

\(4\geq{x}+5\) [/hidden-answer]

The following video show examples of solving inequalities with the variable on the right side.

Solve inequalities with multiplication and division

Solving an inequality with a variable that has a coefficient other than 1 usually involves multiplication or division. The steps are like solving one-step equations involving multiplication or division EXCEPT for the inequality sign. Let’s look at what happens to the inequality when you multiply or divide each side by the same number.

Multiplication and Division Properties of Inequality

Keep in mind that you only change the sign when you are multiplying and dividing by a negative number. If you add or subtract by a negative number, the inequality stays the same.

Solve for x. \(3x>12\)

[reveal-answer q=”691711″]Show Solution[/reveal-answer] [hidden-answer a=”691711″]Divide both sides by 3 to isolate the variable.

\(\displaystyle \begin{array}{r}\underline{3x}>\underline{12}\\3\,\,\,\,\,\,\,\,\,\,\,\,3\\x>4\,\,\,\end{array}\)

Check your solution by first checking the end point 4, and then checking another solution for the inequality.

\(\begin{array}{r}3\cdot4=12\\12=12\\3\cdot10>12\\30>12\\\text{It checks!}\end{array}\)

Inequality: \(\displaystyle x>4\)

Interval: \(\left(4,\infty\right)\)

There was no need to make any changes to the inequality sign because both sides of the inequality were divided by positive 3. In the next example, there is division by a negative number, so there is an additional step in the solution!

Solve for x . \(−2x>6\)

[reveal-answer q=”604033″]Show Solution[/reveal-answer] [hidden-answer a=”604033″]Divide each side of the inequality by \(−2\) to isolate the variable, and change the direction of the inequality sign because of the division by a negative number.

\(\displaystyle \begin{array}{r}\underline{-2x}<\underline{\,6\,}\\-2\,\,\,\,-2\,\\x<-3\end{array}\)

Check your solution by first checking the end point \(−3\), and then checking another solution for the inequality.

\(\begin{array}{r}-2\left(-3\right)=6 \\6=6\\ -2\left(-6\right)>6 \\ 12>6\end{array}\)

Inequality: \(\displaystyle x<-3\)

Interval: \(\left(-\infty, -3\right)\)

Graph: \(−2\), the inequality symbol was switched from > to <. [/hidden-answer]

The following video shows examples of solving one step inequalities using the multiplication property of equality where the variable is on the left hand side.

Before you read the solution to the next example, think about what properties of inequalities you may need to use to solve the inequality. What is different about this example from the previous one? Write your ideas in the box below.

Solve for x . \(-\frac{1}{2}>-12x\)

[practice-area rows=”1″][/practice-area] [reveal-answer q=”811465″]Show Solution[/reveal-answer] [hidden-answer a=”811465″]

This inequality has the variable on the right hand side, which is different from the previous examples. Start the solution process as before, and at the end, you can move the variable to the left to write the final solution.

Divide both sides by \(-12\) to isolate the variable. Since you are dividing by a negative number, you need to change the direction of the inequality sign.

\(\displaystyle\begin{array}{l}-\frac{1}{2}\gt{-12x}\\\\\frac{-\frac{1}{2}}{-12}\gt\frac{-12x}{-12}\\\end{array}\)

Dividing a fraction by an integer requires you to multiply by the reciprocal, and the reciprocal of \(\frac{1}{-12}\)

\(\displaystyle\begin{array}{r}\left(-\frac{1}{12}\right)\left(-\frac{1}{2}\right)\lt\frac{-12x}{-12}\,\,\\\\ \frac{1}{24}\lt\frac{\cancel{-12}x}{\cancel{-12}}\\\\ \frac{1}{24}\lt{x}\,\,\,\,\,\,\,\,\,\,\end{array}\)

Inequality: \(x\gt\frac{1}{24}\). Writing the inequality with the variable on the left requires a little thinking, but helps you write the interval and draw the graph correctly.

Interval: \(\left(\frac{1}{24},\infty\right)\)

The following video gives examples of how to solve an inequality with the multiplication property of equality where the variable is on the right hand side.

Combine properties of inequality to solve algebraic 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.

Solve for p . \(4p+5<29\)

[reveal-answer q=”211828″]Show Solution[/reveal-answer] [hidden-answer a=”211828″]

Begin to isolate the variable by subtracting 5 from both sides of the inequality.

\(\displaystyle \begin{array}{l}4p+5<\,\,\,29\\\underline{\,\,\,\,\,\,\,\,\,-5\,\,\,\,\,-5}\\4p\,\,\,\,\,\,\,\,\,<\,\,24\,\,\end{array}\)

Divide both sides of the inequality by 4 to express the variable with a coefficient of 1.

\(\begin{array}{l}\underline{4p}\,<\,\,\underline{24}\,\,\\\,4\,\,\,\,<\,\,4\\\,\,\,\,\,p<6\end{array}\)

Inequality: \(p<6\)

Interval: \(\left(-\infty,6\right)\)

Graph: Note the open circle at the end point 6 to show that solutions to the inequality do not include 6. The values where p is less than 6 are found all along the number line to the left of 6.

Check the solution. [reveal-answer q=”291597″]Show Solution[/reveal-answer] [hidden-answer a=”291597″]

Check the end point 6 in the related equation.

\(\displaystyle \begin{array}{r}4p+5=29\,\,\,\\\text{Does}\,\,\,4(6)+5=29?\\24+5=29\,\,\,\\29=29\,\,\,\\\text{Yes!}\,\,\,\,\,\,\end{array}\)

Try another value to check the inequality. Let’s use \(p=0\).

\(\displaystyle \begin{array}{r}4p+5<29\,\,\,\\\text{Is}\,\,\,4(0)+5<29?\\0+5<29\,\,\,\\5<29\,\,\,\\\text{Yes!}\,\,\,\,\,\end{array}\)

\(4p+5<29\)

Solve for x : \(3x–7\ge 41\) [reveal-answer q=”238157″]Show Solution[/reveal-answer] [hidden-answer a=”238157″]

Begin to isolate the variable by adding 7 to both sides of the inequality, then divide both sides of the inequality by 3 to express the variable with a coefficient of 1.

\(\displaystyle \begin{array}{l}3x-7\ge 41\\\underline{\,\,\,\,\,\,\,+7\,\,\,\,+7}\\\frac{3x}{3}\,\,\,\,\,\,\,\,\ge \frac{48}{3}\\\,\,\,\,\,\,\,\,\,\,x\ge 16\end{array}\)

Inequality: \(x\ge 16\)

Interval: \(\left[16,\infty\right)\)

Check the solution. [reveal-answer q=”437341″]Show Solution[/reveal-answer]

[hidden-answer a=”437341″]

First, check the end point 16 in the related equation.

\(\displaystyle \begin{array}{r}3x-7=41\,\,\,\\\text{Does}\,\,\,3(16)-7=41?\\48-7=41\,\,\,\\41=41\,\,\,\\\text{Yes!}\,\,\,\,\,\end{array}\)

Then, try another value to check the inequality. Let’s use \(x = 20\).

\(\displaystyle \begin{array}{r}\,\,\,\,3x-7\ge 41\,\,\,\\\text{Is}\,\,\,\,\,3(20)-7\ge 41?\\60-7\ge 41\,\,\,\\53\ge 41\,\,\,\\\text{Yes!}\,\,\,\,\,\end{array}\)

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.

Solve for p . \(−58>14−6p\)

[reveal-answer q=”424351″]Show Solution[/reveal-answer] [hidden-answer a=”424351″]

Note how the variable is on the right hand side of the inequality, the method for solving does not change in this case.

Begin to isolate the variable by subtracting 14 from both sides of the inequality.

\(\displaystyle \begin{array}{l}−58\,\,>14−6p\\\underline{\,\,\,\,\,\,\,\,\,\,\,\,-14\,\,\,\,\,\,\,-14}\\-72\,\,\,\,\,\,\,\,\,\,\,>-6p\end{array}\)

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.

\(\begin{array}{l}\underline{-72}>\underline{-6p}\\-6\,\,\,\,\,\,\,\,\,\,-6\\\,\,\,\,\,\,12\lt{p}\end{array}\)

We can also write this as \(p>12\). Notice how the inequality sign is still opening up toward the variable p.

Inequality: \(\left(12,\infty\right)\) Graph: The graph of the inequality p > 12 has an open circle at 12 with an arrow stretching to the right.

Check the solution. [reveal-answer q=”500309″]Show Solution[/reveal-answer] [hidden-answer a=”500309″]

First, check the end point 12 in the related equation.

\(\begin{array}{r}-58=14-6p\\-58=14-6\left(12\right)\\-58=14-72\\-58=-58\end{array}\)

Then, try another value to check the inequality. Try 100.

\(\begin{array}{r}-58>14-6p\\-58>14-6\left(100\right)\\-58>14-600\\-58>-586\end{array}\)

In the following video, you will see an example of solving a linear inequality with the variable on the left side of the inequality, and an example of switching the direction of the inequality after dividing by a negative number.

In the following video, you will see an example of solving a linear inequality with the variable on the right side of the inequality, and an example of switching the direction of the inequality after dividing by a negative number.

Simplify and solve algebraic inequalities using the distributive property

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.

Solve for x . \(2\left(3x–5\right)\leq 4x+6\)

[reveal-answer q=”587737″]Show Solution[/reveal-answer] [hidden-answer a=”587737″]

Distribute to clear the parentheses.

\(\displaystyle \begin{array}{r}\,2(3x-5)\leq 4x+6\\\,\,\,\,6x-10\leq 4x+6\end{array}\)

Subtract 4 x from both sides to get the variable term on one side only.

\(\begin{array}{r}6x-10\le 4x+6\\\underline{-4x\,\,\,\,\,\,\,\,\,\,\,\,\,\,-4x}\,\,\,\,\,\,\,\,\,\\\,\,\,2x-10\,\,\leq \,\,\,\,\,\,\,\,\,\,\,\,6\end{array}\)

Add 10 to both sides to isolate the variable.

\(\begin{array}{r}\\\,\,\,2x-10\,\,\le \,\,\,\,\,\,\,\,6\,\,\,\\\underline{\,\,\,\,\,\,+10\,\,\,\,\,\,\,\,\,+10}\\\,\,\,2x\,\,\,\,\,\,\,\,\,\,\,\le \,\,\,\,\,16\,\,\,\end{array}\)

Divide both sides by 2 to express the variable with a coefficient of 1.

\(\begin{array}{r}\underline{2x}\le \,\,\,\underline{16}\\\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\le \,\,\,\,\,8\end{array}\)

Inequality: \(\left(-\infty,8\right]\) Graph: The graph of this solution set includes 8 and everything left of 8 on the number line.

Check the solution. [reveal-answer q=”808701″]Show Solution[/reveal-answer] [hidden-answer a=”808701″]

First, check the end point 8 in the related equation.

\(\displaystyle \begin{array}{r}2(3x-5)=4x+6\,\,\,\,\,\,\\2(3\,\cdot \,8-5)=4\,\cdot \,8+6\\\,\,\,\,\,\,\,\,\,\,\,2(24-5)=32+6\,\,\,\,\,\,\\2(19)=38\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\38=38\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}\)

Then, choose another solution and evaluate the inequality for that value to make sure it is a true statement. Try 0.

\(\displaystyle \begin{array}{l}2(3\,\cdot \,0-5)\le 4\,\cdot \,0+6?\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2(-5)\le 6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-10\le 6\,\,\end{array}\)

\(\left(-\infty,8\right]\)

In the following video, you are given an example of how to solve a multi-step inequality that requires using the distributive property.

In the next example, you are given an inequality with a term that looks complicated. If you pause and think about how to use the order of operations to solve the inequality, it will hopefully seem like a straightforward problem. Use the textbox to write down what you think is the best first step to take.

Solve for a. \(\displaystyle\frac ParseError: EOF expected (click for details) Callstack: at (Courses/Lumen_Learning/Beginning_Algebra_(Lumen)/01:_Solving_Equations_and_Inequalities/1.05:_Solve_Inequalities), /content/body/div[7]/div/div/div[3]/div/p[2]/span[1], line 1, column 2 6{<2}\)

[practice-area rows=”1″][/practice-area]

[reveal-answer q=”701072″]Show Solution[/reveal-answer]

[hidden-answer a=”701072″]

Clear the fraction by multiplying both sides of the equation by 6.

\(\displaystyle \begin{array}{r}\frac ParseError: EOF expected (click for details) Callstack: at (Courses/Lumen_Learning/Beginning_Algebra_(Lumen)/01:_Solving_Equations_and_Inequalities/1.05:_Solve_Inequalities), /content/body/div[7]/div/div/div[3]/div/p[7]/span[1], line 1, column 2 6{<2}\,\,\,\,\,\,\,\,\\\\6\,\cdot \,\frac{2a-4}{6}<2\,\cdot \,6\\\\{2a-4}<12\,\,\,\,\,\,\end{array}\)

Add 4 to both sides to isolate the variable.

\(\displaystyle \begin{array}{r}2a-4<12\\\underline{\,\,\,+4\,\,\,\,+4}\\2a<16\end{array}\)

\(\displaystyle \begin{array}{c}\frac{2a}{2}<\,\frac{16}{2}\\\\a<8\end{array}\)

Inequality: \(a<8\)

Interval: \(\left(-\infty,8\right)\)

Check the solution. [reveal-answer q=”905072″]Show Solution[/reveal-answer]

[hidden-answer a=”905072″] First, check the end point 8 in the related equation.

\(\displaystyle \begin{array}{r}\frac{2a-4}{6}=2\,\,\,\,\\\\\text{Does}\,\,\,\frac{2(8)-4}{6}=2?\\\\\frac{16-4}{6}=2\,\,\,\,\\\\\frac{12}{6}=2\,\,\,\,\\\\2=2\,\,\,\,\\\\\text{Yes!}\,\,\,\,\,\end{array}\)

Then choose another solution and evaluate the inequality for that value to make sure it is a true statement. Try 5.

\(\displaystyle \begin{array}{r}\text{Is}\,\,\,\frac{2(5)-4}{6}<2?\\\\\frac{10-4}{6}<2\,\,\,\\\\\,\,\,\,\frac{6}{6}<2\,\,\,\\\\1<2\,\,\,\\\\\text{Yes!}\,\,\,\,\,\end{array}\)

Solving inequalities is very similar to solving equations, except you have to reverse the inequality symbols when you multiply or divide both sides of an inequality by a negative number. There are three ways to represent solutions to inequalities: an interval, a graph, and an inequality. Because there is usually more than one solution to an inequality, when you check your answer you should check the end point and one other value to check the direction of the inequality.

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.

• Revision and Adaptation. Provided by : Lumen Learning. License : CC BY: Attribution
• Graph Linear Inequalities in One Variable (Basic). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/-kiAeGbSe5c . License : CC BY: Attribution
• Given Interval in Words, Graph and Give Inequality. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/E_ZWNVNEvOg . License : CC BY: Attribution
• Given an Inequality, Graph and Give Interval Notation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/BKhDzNKjVBc . License : CC BY: Attribution
• Given Interval in Words, Graph and Give Interval Notation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/OYkQ-McI2qg . License : CC BY: Attribution
• Given Interval Notation, Graph and Give Inequality. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/lkhILNEPbfk . License : CC BY: Attribution
• Screenshot: Cecilia Venn Diagram. Authored by : Lumen Learning. License : CC BY: Attribution
• Screenshot: Internet Privacy. Authored by : Lumen Learning. License : CC BY: Attribution
• Solutions to Basic OR Compound Inequalities. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/nKarzhZOFIk . License : CC BY: Attribution
• Solutions to Basic AND Compound Inequalities. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/LP3fsZNjJkc . License : CC BY: Attribution
• Unit 10: Solving Equations and Inequalities, from Developmental Math: An Open Program. Provided by : Monterey Institute of Technology and Education. Located at : nrocnetwork.org/resources/downloads/nroc-math-open-textbook-units-1-12-pdf-and-word-formats/. License : CC BY: Attribution
• Ex: Solving One Step Inequalities by Adding and Subtracting (Variable Left Side). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/1Z22Xh66VFM . License : CC BY: Attribution
• Ex: Solving One Step Inequalities by Adding and Subtracting (Variable Right Side). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/RBonYKvTCLU . License : CC BY: Attribution
• Ex: Solve One Step Linear Inequality by Dividing (Variable Left). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/IajiD3R7U-0 . License : CC BY: Attribution
• Ex: Solve One Step Linear Inequality by Dividing (Variable Right). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/s9fJOnVTHhs . License : CC BY: Attribution
• Ex: Graph Basic Inequalities and Express Using Interval Notation. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/X0xrHKgbDT0 . License : CC BY: Attribution
• College Algebra. Authored by : Jay Abramson, et al.. Located at : courses.candelalearning.com/collegealgebra1xmaster/. License : CC BY: Attribution
• Ex: Graph Basic Inequalities and Express Using Interval Notation. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/X0xrHKgbDT0 . License : CC BY: Attribution
• Ex: Solve a Two Step Linear Inequality (Variable Left). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/RB9wvIogoEM . License : CC BY: Attribution
• Ex: Solve a Two Step Linear Inequality (Variable Right). Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/9D2g_FaNBkY . License : CC BY: Attribution
• Ex: Solve a Linear Inequality Requiring Multiple Steps (One Var). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/vjZ3rQFVkh8 . License : Public Domain: No Known Copyright
• Ex: Solve a Compound Inequality Involving OR (Union). Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/oRlJ8G7trR8 . License : CC BY: Attribution
• Ex 1: Solve and Graph Basic Absolute Value inequalities. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/0cXxATY2S-k . License : CC BY: Attribution
• Ex 2: Solve and Graph Absolute Value inequalities Mathispower4u . Authored by : James Sousa (Mathispower4u.com) for Lumen Learning. Located at : https://youtu.be/d-hUviSkmqE . License : CC BY: Attribution
• Ex 3: Solve and Graph Absolute Value inequalitie. Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/ttUaRf-GzpM . License : CC BY: Attribution
• Ex 4: Solve and Graph Absolute Value inequalities (Requires Isolating Abs. Value). Authored by : James Sousa (Mathispower4u.com) . Located at : https://youtu.be/5jRUuiMUxWQ . License : CC BY: Attribution
• College Algebra. Authored by : Jay Abramson, et. al. Located at : courses.candelalearning.com/collegealgebra1xmaster/. License : CC BY: Attribution

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Examples of Solving Harder Linear Inequalities

Intro & Formatting Worked Examples Harder Examples & Word Prob's

Once you'd learned how to solve one-variable linear equations, you were then given word problems. To solve these problems, you'd have to figure out a linear equation that modelled the situation, and then you'd have to solve that equation to find the answer to the word problem.

Content Continues Below

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Solving Inequalities

So, now that you know how to solve linear inequalities — you guessed it! — they give you word problems.

• The velocity of an object fired directly upward is given by V = 80 − 32 t , where the time t is measured in seconds. When will the velocity be between 32 and 64 feet per second (inclusive)?

This question is asking when the velocity, V , will be between two given values. So I'll take the expression for the velocity,, and put it between the two values they've given me. They've specified that the interval of velocities is inclusive, which means that the interval endpoints are included. Mathematically, this means that the inequality for this model will be an "or equal to" inequality. Because the solution is a bracket (that is, the solution is within an interval), I'll need to set up a three-part (that is, a compound) inequality.

I will set up the compound inequality, and then solve for the range of times t :

32 ≤ 80 − 32 t ≤ 64

32 − 80 ≤ 80 − 80 − 32 t ≤ 64 − 80

−48 ≤ −32 t ≤ −16

−48 / −32 ≥ −32 t / −32 ≥ −16 / −32

1.5 ≥ t ≥ 0.5

Note that, since I had to divide through by a negative, I had to flip the inequality signs.

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Note also that you might (as I do) find the above answer to be more easily understood if it's written the other way around, with "less than" inequalities.

And, because this is a (sort of) real world problem, my working should show the fractions, but my answer should probably be converted to decimal form, because it's more natural to say "one and a half seconds" than it is to say "three-halves seconds". So I convert the last line above to the following:

0.5 ≤ t ≤ 1.5

Looking back at the original question, it did not ask for the value of the variable " t ", but asked for the times when the velocity was between certain values. So the actual answer is:

The velocity will be between 32 and 64 feet per second between 0.5 seconds after launch and 1.5 seconds after launch.

Okay; my answer above was *extremely* verbose and "complete"; you don't likely need to be so extreme. You can probably safely get away with saying something simpler like, "between 0.5 seconds and 1.5 seconds". Just make sure that you do indeed include the approprioate units (in this case, "seconds").

Always remember when doing word problems, that, once you've found the value for the variable, you need to go back and re-read the problem to make sure that you're answering the actual question. The inequality 0.5 ≤  t  ≤ 1.5 did not answer the actual question regarding time. I had to interpret the inequality and express the values in terms of the original question.

• Solve 5 x + 7 < 3( x  + 1) .

First I'll multiply through on the right-hand side, and then solve as usual:

5 x + 7 < 3( x + 1)

5 x + 7 < 3 x + 3

2 x + 7 < 3

2 x < −4

x < −2

In solving this inequality, I divided through by a positive 2 to get the final answer; as a result (that is, because I did *not* divide through by a minus), I didn't have to flip the inequality sign.

• You want to invest $30,000 . Part of this will be invested in a stable 5% -simple-interest account. The remainder will be "invested" in your father's business, and he says that he'll pay you back with 7% interest. Your father knows that you're making these investments in order to pay your child's college tuition with the interest income. What is the least you can "invest" with your father, and still (assuming he really pays you back) get at least $1900 in interest?

First, I have to set up equations for this. The interest formula for simple interest is I = Prt , where I is the interest, P is the beginning principal, r is the interest rate expressed as a decimal, and t is the time in years.

Since no time-frame is specified for this problem, I'll assume that t  = 1 ; that is, I'll assume (hope) that he's promising to pay me at the end of one year. I'll let x be the amount that I'm going to "invest" with my father. Then the rest of my money, being however much is left after whatever I give to him, will be represented by "the total, less what I've already given him", so 30000 −  x will be left to invest in the safe account.

Then the interest on the business investment, assuming that I get paid back, will be:

I = ( x )(0.07)(1) = 0.07 x

The interest on the safe investment will be:

(30 000 − x )(0.05)(1) = 1500 − 0.05 x

The total interest is the sum of what is earned on each of the two separate investments, so my expression for the total interest is:

0.07 x + (1500 − 0.05 x ) = 0.02 x + 1500

I need to get at least $1900 ; that is, the sum of the two investments' interest must be greater than, or at least equal to, $1,900 . This allows me to create my inequality:

0.02 x + 1500 ≥ 1900

0.02 x ≥ 400

x ≥ 20 000

That is, I will need to "invest" at least $20,000 with my father in order to get $1,900 in interest income. Since I want to give him as little money as possible, I will give him the minimum amount:

I will invest $20,000 at 7% .

• An alloy needs to contain between 46% copper and 50% copper. Find the least and greatest amounts of a 60% copper alloy that should be mixed with a 40% copper alloy in order to end up with thirty pounds of an alloy containing an allowable percentage of copper.

This is similar to a mixture word problem , except that this will involve inequality symbols rather than "equals" signs. I'll set it up the same way, though, starting with picking a variable for the unknown that I'm seeking. I will use x to stand for the pounds of 60% copper alloy that I need to use. Then 30 −  x will be the number of pounds, out of total of thirty pounds needed, that will come from the 40% alloy.

Of course, I'll remember to convert the percentages to decimal form for doing the algebra.

How did I get those values in the bottom right-hand box? I multiplied the total number of pounds in the mixture ( 30 ) by the minimum and maximum percentages ( 46% and 50% , respectively). That is, I multiplied across the bottom row, just as I did in the " 60% alloy" row and the " 40% alloy" row, to get the right-hand column's value.

The total amount of copper in the mixture will be the sum of the copper contributed by each of the two alloys that are being put into the mixture. So I'll add the expressions for the amount of copper from each of the alloys, and place the expression for the total amount of copper in the mixture as being between the minimum and the maximum allowable amounts of copper:

13.8 ≤ 0.6 x + (12 − 0.4 x ) ≤ 15

13.8 ≤ 0.2 x + 12 ≤ 15

1.8 ≤ 0.2 x ≤ 3

9 ≤ x ≤ 15

Checking back to my set-up, I see that I chose my variable to stand for the number of pounds that I need to use of the 60% copper alloy. And they'd only asked me for this amount, so I can ignore the other alloy in my answer.

I will need to use between 9 and 15 pounds of the 60% alloy.

Per yoozh, I'm verbose in my answer. You can answer simply as " between 9 and 15 pounds ".

• Solve 3( x − 2) + 4 ≥ 2(2 x − 3)

First I'll multiply through and simplify; then I'll solve:

3( x − 2) + 4 ≥ 2(2 x − 3)

3 x − 6 + 4 ≥ 4 x − 6

3 x − 2 ≥ 4 x − 6

−2 ≥ x − 6            (*)

Why did I move the 3 x over to the right-hand side (to get to the line marked with a star), instead of moving the 4 x to the left-hand side? Because by moving the smaller term, I was able to avoid having a negative coefficient on the variable, and therefore I was able to avoid having to remember to flip the inequality when I divided through by that negative coefficient. I find it simpler to work this way; I make fewer errors. But it's just a matter of taste; you do what works for you.

Why did I switch the inequality in the last line and put the variable on the left? Because I'm more comfortable with inequalities when the answers are formatted this way. Again, it's only a matter of taste. The form of the answer in the previous line, 4 ≥ x , is perfectly acceptable.

As long as you remember to flip the inequality sign when you multiply or divide through by a negative, you shouldn't have any trouble with solving linear inequalities.

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Solving Word Problems in Algebra Inequality Word Problems

How are you with solving word problems in Algebra? Are you ready to dive into the "real world" of inequalities? I know that solving word problems in Algebra is probably not your favorite, but there's no point in learning the skill if you don't apply it.

I promise to make this as easy as possible. Pay close attention to the key words given below, as this will help you to write the inequality. Once the inequality is written, you can solve the inequality using the skills you learned in our past lessons.

I've tried to provide you with examples that could pertain to your life and come in handy one day. Think about others ways you might use inequalities in real world problems. I'd love to hear about them if you do!

Before we look at the examples let's go over some of the rules and key words for solving word problems in Algebra (or any math class).

Word Problem Solving Strategies

• Read through the entire problem.
• Highlight the important information and key words that you need to solve the problem.
• Identify your variables.
• Write the equation or inequality.
• Write your answer in a complete sentence.
• Check or justify your answer.

I know it always helps too, if you have key words that help you to write the equation or inequality. Here are a few key words that we associate with inequalities! Keep these handy as a reference.

Inequality Key Words

• at least - means greater than or equal to
• no more than - means less than or equal to
• more than - means greater than
• less than - means less than

Ok... let's put it into action and look at our examples.

Example 1: Inequality Word Problems

Keith has $500 in a savings account at the beginning of the summer. He wants to have at least $200 in the account by the end of the summer. He withdraws $25 each week for food, clothes, and movie tickets.

• Write an inequality that represents Keith's situation.
• How many weeks can Keith withdraw money from his account? Justify your answer.

Step 1: Highlight the important information in this problem.

Note:  At least is a key word that notes that this problem must be written as an inequality.

Step 2 : Identify your variable. What don't you know? The question verifies that you don't know how many weeks.

Let w = the number of weeks

Step 3: Write your inequality.

500 - 25w > 200

I know you are saying, "How did you get that inequality?"

I know the "at least" part is tricky. You would probably think that at least means less than.

But... he wants the amount in his account to be at least $200 which means $200 or greater. So, we must use the greater than or equal to symbol.

Step 4 : Solve the inequality.

The number of weeks that Keith can withdraw money from his account is 12 weeks or less.

Step 5: Justify (prove your answer mathematically).

I'm going to prove that the largest number of weeks is 12 by substituting 12 into the inequality for w. You could also substitute any number less than 12.

Since 200 is equal to 200, my answer is correct. Any more than 12 weeks and his account balance would be less than $200.  Any number of weeks less than 12 and his account would stay above $200.

That wasn't too bad, was it? Let's take a look at another example.

Example 2: More Inequality Word Problems

Yellow Cab Taxi charges a $1.75 flat rat e in addition to $0.65 per mile . Katie has no more than $10 to spend on a ride.

• Write an inequality that represents Katie's situation.
• How many miles can Katie travel without exceeding her budget? Justify your answer.

Note:  No more than are key words that note that this problem must be written as an inequality.

Step 2 : Identify your variable. What don't you know? The question verifies that you don't know the number of miles Katie can travel.

Let m = the number of miles

Step 3:  Write the inequality.

0.65m + 1.75 < 10

Are you thinking, "How did you write that inequality?"

The "no more than" can also be tricky. "No more than" means that you can't have more than something, so that means you must have less than!

Step 4: Solve the inequality.

Since this is a real world problem and taxi's usually charge by the mile, we can say that Katie can travel 12 miles or less before reaching her limit of $10.

Are you ready to try some on your own now? Yes... of course you are! Click here to move onto the word problem practice problems.

Take a look at the questions that other students have submitted:

Quite a complicated problem about perimeter and area of a rectangle

This is a toughie.... a compound inequalities word problem

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