## Rational Equation Word Problem

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## 7.5: Solving Rational Equations

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Learning Objectives

• Solve rational equations.
• Solve literal equations, or formulas, involving rational expressions.

## Solving Rational Equations

A rational equation is an equation containing at least one rational expression. Rational expressions typically contain a variable in the denominator. For this reason, we will take care to ensure that the denominator is not 0 by making note of restrictions and checking our solutions.

Solve rational equations by clearing the fractions by multiplying both sides of the equation by the least common denominator (LCD).

Example $$\PageIndex{1}$$

$$\frac{5}{x}-\frac{1}{3} = \frac{1}{x}$$

We first make a note that $$x≠0$$ and then multiply both sides by the LCD, $$3x$$:

Check your answer by substituting 12 for x to see if you obtain a true statement.

\begin{aligned} \frac{5}{x}-\frac{1}{3} &=\frac{1}{x} \\ \color{black}{\frac{5}{\color{OliveGreen}{12}}}-\frac{1}{3} &=\color{black}{\frac{1}{\color{OliveGreen}{12}}} \\ \frac{5}{12}-\frac{4}{12} &=\frac{1}{12} \\ \frac{1}{12} &=\frac{1}{12}\quad\color{Cerulean}{ \checkmark} \end{aligned}

The solution is 12.

After multiplying both sides of the previous example by the LCD, we were left with a linear equation to solve. This is not always the case; sometimes we will be left with a quadratic equation.

Example $$\PageIndex{2}$$

$$2-\frac{1}{x(x+1)}=\frac{3}{x+1}$$

In this example, there are two restrictions, $$x≠0$$ and $$x≠−1$$. Begin by multiplying both sides by the LCD, $$x(x+1)$$.

After distributing and dividing out the common factors, a quadratic equation remains. To solve it, rewrite it in standard form, factor, and then set each factor equal to 0.

$$\begin{array}{rlrl}{2 x+1} & {=0} & {\text { or }} & {x-1=0} \\ {2 x} & {=-1} && {x=1} \\ {x} & {=-\frac{1}{2}}\end{array}$$

Check to see if these values solve the original equation.

$$\begin{array}{c|c}{Check\:x=-\frac{1}{2}}&{Check\:x=1}\\{2-\frac{1}{\color{Cerulean}{(-\frac{1}{2})}\color{black}{(\color{Cerulean}{(-\frac{1}{2})}\color{black}{+1)}}}=\frac{3}{\color{Cerulean}{(-\frac{1}{2})}\color{black}{+1}}}&{2-\frac{1}{\color{Cerulean}{1}\color{black}{(\color{Cerulean}{1}\color{black}{+1)}}=}\frac{3}{\color{Cerulean}{1}\color{black}{+1}}}\\{2-\frac{1}{(-\frac{1}{4})}=\frac{3}{\frac{1}{2}}}&{\frac{4}{2}-\frac{1}{2}=\frac{3}{2}}\\{2+1\cdot\frac{4}{1}=3\cdot\frac{2}{1}}&{\frac{3}{2}=\frac{3}{2}}\quad\color{Cerulean}{\checkmark}\\{2+4=6}\\{6=6}\quad\color{Cerulean}{\checkmark} \end{array}$$

The solutions are $$-\frac{1}{2}$$ and $$1$$.

Up to this point, all of the possible solutions have solved the original equation. However, this may not always be the case. Multiplying both sides of an equation by variable factors may lead to extraneous solutions, which are solutions that do not solve the original equation. A complete list of steps for solving a rational equation is outlined in the following example.

Example $$\PageIndex{3}$$

$$\frac{x}{x+2}+\frac{2}{x^{2}+5 x+6}=\frac{5}{x+3}$$

Step 1 : Factor all denominators and determine the LCD.

\begin{aligned} \frac{x}{x+2}+\frac{2}{x^{2}+5 x+6} &=\frac{5}{x+3} \\ \frac{x}{(x+2)}+\frac{2}{(x+2)(x+3)} &=\frac{5}{(x+3)} \end{aligned}

The LCD is $$(x+2)(x+3)$$.

Step 2 : Identify the restrictions. In this case, they are $$x≠−2$$ and $$x≠−3$$.

Step 3 : Multiply both sides of the equation by the LCD. Distribute carefully and then simplify.

\begin{aligned}\color{Cerulean}{(x+2)(x+3)}\color{black}{\cdot}\left(\frac{x}{(x+2)}+\frac{2}{(x+2)(x+3)} \right)&=\color{Cerulean}{(x+2)(x+3)}\color{black}{\cdot\frac{5}{(x+3)}}\\ \frac{x\cdot\color{Cerulean}{\cancel{(x+2)}(x+3)}}{\color{Cerulean}{\cancel{\color{black}{(x+2)}}}}\color{black}{+}\frac{2\cdot\color{Cerulean}{\cancel{(x+2)}\cancel{(x+3)}}}{\color{Cerulean}{\cancel{\color{black}{(x+2)}}\cancel{\color{black}{(x+3)}}}}&\color{black}{=}\frac{5\cdot\color{Cerulean}{(x+2)\cancel{(x+3)}}}{\color{Cerulean}{\cancel{\color{black}{(x+3)}}}}\\x(x+3)+2&=5(x+2)\\x^{2}+3x+2&=5x+10 \end{aligned}

Step 4 : Solve the resulting equation. Here the result is a quadratic equation. Rewrite it in standard form, factor, and then set each factor equal to $$0$$.

$$\begin{array}{cc}{x+2=0} & {\text { or } \quad x-4=0} \\ {x=-2} & {x=4}\end{array}$$

Step 5 : Check for extraneous solutions. Always substitute into the original equation, or the factored equivalent. In this case, choose the factored equivalent to check:

$$\frac{x}{(x+2)}+\frac{2}{(x+2)(x+3)}=\frac{5}{(x+3)}$$

$$\begin{array}{c|c}{Check\:x=-2}&{Check\:x=4}\\{\frac{-2}{(\color{Cerulean}{-2}\color{black}{+2)}}+\frac{2}{(\color{Cerulean}{-2}\color{black}{+2)(}\color{Cerulean}{-2}\color{black}{+3)}}=\frac{5}{(\color{Cerulean}{-2}\color{black}{+3)}}}&{\frac{4}{(\color{Cerulean}{4}\color{black}{+2)}}+\frac{2}{(\color{Cerulean}{4}\color{black}{+2)(}\color{Cerulean}{4}\color{black}{+3)}}=\frac{5}{(\color{Cerulean}{4}\color{black}{+3)}}}\\{-\frac{2}{0}+\frac{2}{0(1)}=\frac{5}{1}\quad\color{red}{x}}&{\frac{4}{6}+\frac{2}{6\cdot7}=\frac{5}{7}}\\{Undefined\:terms!}&{\frac{2}{3}+\frac{1}{21}=\frac{5}{7}}\\{}&{\frac{14}{21}+\frac{1}{21}=\frac{5}{7}}\\{}&{\frac{15}{21}=\frac{5}{7}}\\{}&{\frac{5}{7}=\frac{5}{7}\quad\color{Cerulean}{\checkmark}} \end{array}$$

Here $$−2$$ is an extraneous solution and is not included in the solution set. It is important to note that $$−2$$ is a restriction.

The solution is $$4$$.

If this process produces a solution that happens to be a restriction, then disregard it as an extraneous solution.

Exercise $$\PageIndex{1}$$

Sometimes all potential solutions are extraneous, in which case we say that there is no solution to the original equation. In the next two examples, we demonstrate two ways in which a rational equation can have no solutions.

Example $$\PageIndex{4}$$

$$\frac{3 x}{x^{2}-4}-\frac{2}{x+2}=\frac{1}{x+2}$$

To identify the LCD, first factor the denominators.

\begin{aligned} \frac{3 x}{x^{2}-4}-\frac{2}{x+2} &=\frac{1}{x+2} \\ \frac{3 x}{(x+2)(x-2)}-\frac{2}{(x+2)} &=\frac{1}{(x+2)} \end{aligned}

Multiply both sides by the least common denominator (LCD), $$(x+2)(x−2)$$, distributing carefully.

The equation is a contradiction and thus has no solution.

No solution, $$∅$$

Example $$\PageIndex{5}$$

$$\frac{x}{x-4}-\frac{4}{x+5}=\frac{36}{x^{2}+x-20}$$

First, factor the denominators.

$$\frac{x}{(x-4)}-\frac{4}{(x+5)}=\frac{36}{(x-4)(x+5)}$$

Take note that the restrictions are $$x≠4$$ and $$x≠−5$$. To clear the fractions, multiply by the LCD, $$(x−4)(x+5)$$.

$$\begin{array}{cc}{x-4=0}&{ \text { or } }&{ x+5=0} \\ {x=4} && {x=-5}\end{array}$$

Both of these values are restrictions of the original equation; hence both are extraneous.

No solution, $$\emptyset$$

Exercise $$\PageIndex{2}$$

$$\emptyset$$

It is important to point out that this technique for clearing algebraic fractions only works for equations. Do not try to clear algebraic fractions when simplifying expressions. As a reminder, we have

$$\begin{array}{cc} {\color{Cerulean}{Expression}}&{\color{Cerulean}{Equation}}\\{\frac{1}{x}+\frac{x}{2 x+1}}&{\frac{1}{x}+\frac{x}{2 x+1}=0} \end{array}$$

Expressions are to be simplified and equations are to be solved. If we multiply the expression by the LCD, $$x(2x+1)$$, we obtain another expression that is not equivalent.

$$\begin{array}{c|c}{\color{red}{Incorrect}}&{\color{Cerulean}{Correct}}\\{\frac{1}{x}+\frac{x}{2x+1}}&{\frac{1}{x}+\frac{x}{2x+1}=0}\\{\neq\color{red}{x(2x+1)}\color{black}{\cdot\left(\frac{1}{x}+\frac{x}{2x+1} \right) }}&{\color{Cerulean}{x(2x+1)}\color{black}{\cdot\left(\frac{1}{x}+\frac{x}{2x+1} \right)}\color{black}{=}\color{Cerulean}{x(2x+1)}\color{black}{\cdot 0}}\\{=2x+1+x^{2}\quad\color{red}{x}}&{2x+1+x^{2}=0}\\{}&{x^{2}+2x+1=0\quad\color{Cerulean}{\checkmark}}\end{array}$$

## Literal Equations

Literal equations, or formulas, are often rational equations. Hence the techniques described in this section can be used to solve for particular variables. Assume that all variable expressions in the denominator are nonzero.

Example $$\PageIndex{6}$$

Solve for $$x$$:

$$x=\frac{x-5}{y}$$

The goal is to isolate x . Assuming that y is nonzero, multiply both sides by y and then add $$5$$ to both sides.

$$x=y z+5$$

Example $$\PageIndex{7}$$

Solve for $$c$$:

$$\frac{1}{c}=\frac{1}{a}+\frac{1}{b}$$

In this example, the goal is to isolate $$c$$. We begin by multiplying both sides by the LCD, $$a⋅b⋅c$$, distributing carefully.

\begin{aligned} \frac{1}{c} &=\frac{1}{a}+\frac{1}{b} \\ \color{Cerulean}{a b c}\color{black}{ \cdot \frac{1}{c}} &\color{black}{=}\color{Cerulean}{a b c}\color{black}{ \cdot \frac{1}{a}+}\color{Cerulean}{a b c}\color{black}{ \cdot \frac{1}{b}} \\ a b &=b c+a c \end{aligned}

On the right side of the equation, factor out $$c$$.

$$a b=c(b+a)$$

Next, divide both sides of the equation by the quantity $$(b+a)$$.

$$\begin{array}{c}{\frac{a b}{\color{Cerulean}{(b+a)}}\color{black}{=\frac{c(b+a)}{\color{Cerulean}{(b+a)}}}} \\ {\frac{a b}{b+a}=c}\end{array}$$

$$c=\frac{a b}{b+a}$$

Exercise $$\PageIndex{3}$$

Solve for $$y$$:

$$x=\frac{y+1}{y−1}$$

$$y=\frac{1+x}{x-1}$$

## Key Takeaways

• Begin solving rational equations by multiplying both sides by the LCD. The resulting equivalent equation can be solved using the techniques learned up to this point.
• Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of restrictions. If a solution is a restriction, then it is not part of the domain and is extraneous.
• When multiplying both sides of an equation by an expression, distribute carefully and multiply each term by that expression.
• If all of the resulting solutions are extraneous, then the original equation has no solutions.

Exercise $$\PageIndex{4}$$ Rational Equations

• $$\frac{1}{2}+\frac{1}{x}=\frac{1}{8}$$
• $$\frac{1}{3}−\frac{1}{x}=\frac{2}{9}$$
• $$\frac{1}{3x}−\frac{2}{3}=\frac{1}{x}$$
• $$\frac{2}{5x}−\frac{1}{x}=\frac{3}{10}$$
• $$\frac{1}{2 x+1}=5$$
• $$\frac{3}{3x−1}+4=5$$
• $$\frac{2 x-3}{x+5}=\frac{2}{x+5}$$
• $$\frac{5x}{2x−1}=\frac{x−1}{2x−1}$$
• $$\frac{5}{x−7}=\frac{6}{x−9}$$
• $$\frac{5}{x+5}=\frac{3}{x+1}$$
• $$\frac{x}{6}-\frac{6}{x}=0$$
• $$\frac{5x+x}{5}=−2$$
• $$\frac{x}{x+12}=\frac{2}{x}$$
• $$\frac{2x}{x+5}=\frac{1}{6−x}$$
• $$\frac{1}{x}+\frac{x}{2x+1}=0$$
• $$\frac{9x}{3x−1}−\frac{4}{x}=0$$
• $$1−\frac{2}{x}=\frac{48}{x^{2}}$$
• $$2−\frac{9}{x}=\frac{5}{x^{2}}$$
• $$1+\frac{12}{x}=\frac{12}{x-2}$$
• $$1−\frac{3x−5}{x(3x−4)}=−\frac{1}{x}$$
• $$\frac{x}{2}=\frac{14}{x+3}$$
• $$\frac{3x}{2}=\frac{x+1}{3−x}$$
• $$6=\frac{−3x+3}{x−1}$$
• $$\frac{1}{2x−2}=2+\frac{6(4−x)}{x−2}$$
• $$2+\frac{2x}{x−3}=\frac{3(x−1)}{x−3}$$
• $$\frac{x}{x−1}+\frac{1}{6x−1}=\frac{x(x−1)}{(6x−1)}$$
• $$\frac{12}{x^{2}−81}=\frac{1}{x+9}−\frac{2}{x−9}$$
• $$\frac{14}{x^{2}−49}=\frac{2}{x−7}−\frac{3}{x+7}$$
• $$\frac{6x}{x+3}+\frac{4}{x−3}=\frac{3x}{x^{2}−9}$$
• $$\frac{3x}{x+2}−\frac{17}{x−2}=−\frac{48}{x^{2}−4}$$
• $$x^{-1}+3=0$$
• $$4^{−y}−1=0$$
• $$y^{−2}−4=0$$
• $$9 x^{-2}-1=0$$
• $$3(x−1)^{−1}+5=0$$
• $$5−2(3x+1)^{−1}=0$$
• $$3+2x^{−3}=2x^{−3}$$
• $$\frac{1}{x}=\frac{1}{x+1}$$
• $$\frac{x}{x+1}=\frac{x+1}{x}$$
• $$\frac{3x−1}{3x}=\frac{x}{x+3}$$
• $$\frac{4x−7}{x−5}=\frac{3x−2}{x−5}$$
• $$\frac{x}{x^{2}−9}=\frac{1}{x−3}$$
• $$\frac{3x+4}{x−8}−\frac{2}{8−x}=1$$
• $$\frac{1}{x}=\frac{6}{x(x+3)}$$
• $$\frac{3}{x}=\frac{1}{x+1}+\frac{13}{x(x+1)}$$
• $$\frac{x}{x−1}−\frac{3}{4x−1}=\frac{9x}{(4x-1)(x−1)}$$
• $$\frac{1}{x−4}+\frac{x}{x−2}=2x^{2}−6x+8$$
• $$\frac{x}{x−5}+\frac{x−1}{x^{2}−11x+30}=\frac{5}{x−6}$$
• $$\frac{x}{x+1}−\frac{6}{5x^{2}+4x−1}=−\frac{5}{5x−1}$$
• $$\frac{−8x^{2}−4}{x−12}+\frac{2(x+2)}{x^{2}+4x−60}=\frac{1}{x+2}$$
• $$\frac{x}{x+2}-\frac{20}{x^{2}-6-x}=\frac{-4}{x-3}$$
• $$\frac{x+7}{x−1}+\frac{x−1}{x+1}=\frac{4}{x^{2}−1}$$
• $$\frac{x−1}{x−3}+\frac{x−3}{x−1}=−\frac{x+5}{x−3}$$
• $$\frac{x−2}{x−5}−\frac{x−5}{x−2}=\frac{8−x}{x−5}$$
• $$\frac{x+7}{x−2}−\frac{81}{x^{2}+5x−14}=\frac{9}{x+7}$$
• $$\frac{x}{x−6}+1=\frac{5x+30}{36−x^{2}}$$
• $$\frac{2x}{x+1}−\frac{4}{4x−3}=−\frac{7}{4x^{2}+x−3}$$
• $$\frac{x−5}{x−10}+\frac{5}{x−5}=−\frac{5x}{x^{2}−15x+50}$$
• $$\frac{5}{x^{2}+5 x+4}+\frac{x+1}{x^{2}+3 x-4}=\frac{5}{x^{2}-1}$$
• $$\frac{1}{x^{2}−2x−63}+\frac{x−9x^{2}+10}{x+21}=\frac{1}{x^{2}−6x−27}$$
• $$\frac{4}{x^{2}-4}+\frac{2(x-2)}{x^{2}-4 x-12}=\frac{x+2}{x^{2}-8 x+12}$$
• $$\frac{x+2}{x^{2}−5x+4}+\frac{x+2}{x^{2}+x−2}=\frac{x−1}{x^{2}−2x−8}$$
• $$\frac{6 x}{x-1}-\frac{11 x+1}{2 x^{2}-x-1}=\frac{6 x}{2 x+1}$$
• $$\frac{8x}{2x−3}+\frac{4x^{2}}{x^{2}−7x+6}=\frac{1}{x−2}$$

1. $$−\frac{8}{3}$$

3. $$−1$$

5. $$−\frac{2}{5}$$

7. $$\frac{5}{2}$$

9. $$−3$$

11. $$-6, 6$$

13. $$−4, 6$$

15. $$−1$$

17. $$−6, 8$$

19. $$−4, 6$$

21. $$−7, 4$$

23. $$∅$$

25. $$∅$$

27. $$−39$$

29. $$\frac{4}{3}, \frac{3}{2}$$

31. $$−\frac{1}{3}$$

33. $$−\frac{1}{2}, \frac{1}{2}$$

35. $$\frac{2}{5}$$

37. $$∅$$

39. $$−\frac{1}{2}$$

41. $$∅$$

43. $$−7$$

47. $$−1$$

49. $$∅$$

51. $$−4$$

53. $$\frac{5}{3}$$

55. $$∅$$

57. $$\frac{1}{2}$$

59. $$−6, 4$$

63. $$\frac{1}{3}$$

Exercise $$\PageIndex{5}$$ Literal Equations

Solve for the indicated variable.

• Solve for $$r$$: $$t=\frac{D}{r}$$
• Solve for $$b$$: $$h=\frac{2A}{b}$$
• Solve for $$P$$: $$t=\frac{I}{Pr}$$
• Solve for $$π$$: $$r=\frac{C}{2π}$$
• Solve for $$c$$: $$\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$$
• Solve for $$y$$: $$m=\frac{y−y_{1}}{x−x_{1}}$$
• Solve for $$w$$: $$P=2(l+w)$$
• Solve for $$t$$: $$A=P(1+rt)$$
• Solve for $$m$$: $$s=\frac{1}{n+m}$$
• Solve for $$S$$: $$h=S2πr−r$$
• Solve for $$x$$: $$y=\frac{x}{x+2}$$
• Solve for $$x$$: $$y=2x+15x$$
• Solve for $$R$$: $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$$
• Solve for $$S_{1}$$: $$\frac{1}{f}=\frac{1}{S_{1}}+\frac{1}{S_{2}}$$

1. $$r=\frac{D}{t}$$

3. $$P=\frac{I}{tr}$$

5. $$c=\frac{ab}{b−a}$$

7. $$w=\frac{P}{2}-l$$

9. $$m=\frac{1}{s}-n$$

11. $$x=-\frac{2y}{y-1}$$

13. $$R=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$$

Exercise $$\PageIndex{6}$$ Discussion Board

• Explain why multiplying both sides of an equation by the LCD sometimes produces extraneous solutions.
• Explain the connection between the technique of cross multiplication and multiplying both sides of a rational equation by the LCD.
• Explain how we can tell the difference between a rational expression and a rational equation. How do we treat them differently?

## Word Problems Leading to Rational Equations

Related Topics: Lesson Plans and Worksheets for Algebra II Lesson Plans and Worksheets for all Grades More Lessons for Algebra Common Core For Algebra

Student Outcomes

• Students solve word problems using models that involve rational expressions.

## New York State Common Core Math Algebra II, Module 1, Lesson 27

Worksheets for Algebra 2

• Anne and Maria play tennis almost every weekend. So far, Anne has won 12 out of 20 matches. a. How many matches will Anne have to win in a row to improve her winning percentage to 75%? b. How many matches will Anne have to win in a row to improve her winning percentage to 90%? c. Can Anne reach a winning percentage of 100%? d. After Anne has reached a winning percentage of 90% by winning consecutive matches as in part (b), how many matches can she now lose in a row to have a winning percentage of 50%?

Example Working together, it takes Sam, Jenna, and Francisco two hours to paint one room. When Sam works alone, he can paint one room in 6 hours. When Jenna works alone, she can paint one room in 4 hours. Determine how long it would take Francisco to paint one room on his own.

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## "Work" Word Problems

Painting & Pipes Tubs & Man-Hours Unequal Times Etc.

"Work" problems usually involve situations such as two people working together to paint a house. You are usually told how long each person takes to paint a similarly-sized house, and you are asked how long it will take the two of them to paint the house when they work together.

Many of these problems are not terribly realistic — since when can two laser printers work together on printing one report? — but it's the technique that they want you to learn, not the applicability to "real life".

The method of solution for "work" problems is not obvious, so don't feel bad if you're totally lost at the moment. There is a "trick" to doing work problems: you have to think of the problem in terms of how much each person / machine / whatever does in a given unit of time . For instance:

Content Continues Below

## Suppose one painter can paint the entire house in twelve hours, and the second painter takes eight hours to paint a similarly-sized house. How long would it take the two painters together to paint the house?

To find out how much they can do together per hour , I make the necessary assumption that their labors are additive (in other words, that they never get in each other's way in any manner), and I add together what they can do individually per hour . So, per hour, their labors are:

But the exercise didn't ask me how much they can do per hour; it asked me how long they'll take to finish one whole job, working togets. So now I'll pick the variable " t " to stand for how long they take (that is, the time they take) to do the job together. Then they can do:

This gives me an expression for their combined hourly rate. I already had a numerical expression for their combined hourly rate. So, setting these two expressions equal, I get:

I can solve by flipping the equation; I get:

An hour has sixty minutes, so 0.8 of an hour has forty-eight minutes. Then:

They can complete the job together in 4 hours and 48 minutes.

The important thing to understand about the above example is that the key was in converting how long each person took to complete the task into a rate.

hours to complete job:

first painter: 12

second painter: 8

together: t

Since the unit for completion was "hours", I converted each time to an hourly rate; that is, I restated everything in terms of how much of the entire task could be completed per hour. To do this, I simply inverted each value for "hours to complete job":

completed per hour:

Then, assuming that their per-hour rates were additive, I added the portion that each could do per hour, summed them, and set this equal to the "together" rate:

As you can see in the above example, "work" problems commonly create rational equations . But the equations themselves are usually pretty simple to solve.

## One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours. How long would it take to fill the pool if only the slower pipe is used?

My first step is to list the times taken by each pipe to fill the pool, and how long the two pipes take together. In this case, I know the "together" time, but not the individual times. One of the pipes' times is expressed in terms of the other pipe's time, so I'll pick a variable to stand for one of these times.

Since the faster pipe's time to completion is defined in terms of the second pipe's time, I'll pick a variable for the slower pipe's time, and then use this to create an expression for the faster pipe's time:

slow pipe: s

together: 5

Next, I'll convert all of the completion times to per-hour rates:

Then I make the necessary assumption that the pipes' contributions are additive (which is reasonable, in this case), add the two pipes' contributions, and set this equal to the combined per-hour rate:

multiplying through by 20 s (being the lowest common denominator of all the fractional terms):

20 + 25 = 4 s

45/4 = 11.25 = s

They asked me for the time of the slower pipe, so I don't need to find the time for the faster pipe. My answer is:

The slower pipe takes 11.25 hours.

Note: I could have picked a variable for the faster pipe, and then defined the time for the slower pipe in terms of this variable. If you're not sure how you'd do this, then think about it in terms of nicer numbers: If someone goes twice as fast as you, then you take twice as long as he does; if he goes three times as fast as you, then you take three times as long as him. In this case, if he goes 1.25 times as fast, then you take 1.25 times as long. So the variables could have been " f  " for the number of hours the faster pipe takes, and then the number of hours for the slower pipe would have been " 1.25 f  ".

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## Applications of Rational Expressions and Word Problems - Expii

Applications of rational expressions and word problems, explanations (3).

## Word Problems Involving Rational Equations

Seven divided by the sum of a number and two is equal to half the difference of the number and three. Find all such numbers.

Let's begin by defining our variable. Let x=the number Now we can set up our equation. Seven divided by the sum of a number and two 7x+2 is equal to 7x+2= half the difference of the number and three. 7x+2=x−32 Before we solve, take note that there's a variable in the denominator which means we will have a restricted value. So, x≠−2 because that would make the denominator zero which would make the expression undefined. Now we can cross multiply and solve! 7x+2=x−327⋅2=(x+2)(x−3)14=x2−x−6−14=          −140=x2−x−200=(x−5)(x+4)x=5 or x=−4 None of the solutions are our restricted value so we know both these will work. Finally we state our answer. The numbers are 5 and−4.

## Related Lessons

Distance, rate, and time word problems.

Let's talk about how to solve word problems involving distance, rate, and time. These problems can be tricky, but here is a good guide on how to solve them.

A boat that can travel fifteen miles per hour in still water can travel thirty-six miles downstream in the same amount of time that it can travel twenty-four miles upstream. Find the speed of the current in the river.

The general formula we use for these problems is:

Overall Rate = speed of the vehicle (with no current/wind) ± speed of the water or wind

Since these are distance, rate, and time problems, we will be using the distance formula as well: Distance=Rate×Timed=rt

Here's the skeleton of the table we make to help solve these problems.

Once we make the outline of our table we begin by filling in the information that was given to us. The problem directly tells us that you travel thirty-six miles downstream and twenty-four miles upstream.

Next we're given that the boat can travel fifteen miles per hour in still water so we have a rate of 15 mph. But this is in still water, not going upstream or downstream. So we need to use our formula given above for overall rate.

So we know the speed of the vehicle with no current or wind. That's the 15 mph. So let's write that in under rate.

Now we need to figure out what gets added or subtracted. When going upstream, are you adding speed or losing speed? You're losing speed because you have to fight the current. So it would be a minus. When going downstream you're gaining speed because you're moving with the current. So it would be a plus.

So, we're adding or subtracting the speed of the current. Well, what's the speed of the current? That's what the problem is asking us to find. So, we can define our variable here: Let x=speed of current

Now what about the time? Well, we don't know the time. But we do know that our distance equals our rate times our time: d=r⋅t We also have values for distance and rate. So we can solve this equation for time. d=r⋅tdr=r⋅trt=dr So we can just put in the distance divided by the rate into our slot for time.

Now what does it say about the time in the problem? It can travel thirty-six miles downstream in the same amount of time that it can travel twenty-four miles upstream. The time it takes to travel upstream is equal to the time it takes to travel downstream. 2415−x=3615+x And now we have a rational equation which we can solve with cross multiplication! 2415−x=3615+x24(15+x)=36(15−x)24(15+x)12=36(15−x)122(15+x)=3(15−x)30+2x=45−3x−30         =−302x=15−3x+3x=    +3x5x=155x5=155x=3

Finally, we state our answer. The speed of the current is 3 miles per hour.

## (Video) Word Problems with Rational Equations

by larryschmidt

Here is a really good video by larryschmidt showing you how to work through some word problems involving rational expressions.

One type of problem is when one person can work at a certain rate, and another person can work at a different rate, and you have to calculate how long it would take if these two people worked together.

The first problem he looks at is:

Sandra can paint a kitchen in 6 hours and Roger can paint the same kitchen in 7 hours. How long would it take for both working together to point the kitchen?

We start by setting up some rates per hour. It says that Sandra can paint 1 kitchen in 6 hours, or: 16 kitchen/hour

Roger can paint 1 kitchen in 7 hours, or: 17 kitchen/hour

Finally, let's let t be the time it takes when they work together. So they can paint 1 kitchen in t hours, or: 1t kitchen/hour

If they work together, this is the same as their rates being added, so we get the equation: 16+17=1t Let's solve for t. 16+17=1t77⋅16+17⋅66=1t742+642=1t1342=1t4213=t3.23≈t

So it takes approximately 3.23 hours for them to paint the kitchen together.

Test Yourself: Sam can beat a video game in 4 hours. Sara can beat the game in 3.5 hours. If they team up, how long will it take to beat the game.

2.48 hours

1.87 hours

1.13 hours

0.54 hours

• PROGRAM SUPPORT

## Word Problems Leading to Rational Equations

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• Algebra II Module 1, Topic C, Lesson 27: Student Version
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## WORD PROBLEMS ON RATIONAL EQUATIONS WORKSHEET

Problem 1 :

Decreasing the reciprocal of a number by 7 results ⁻²⁰⁄₃ . What is the number?

Problem 2 :

7 more than three times the reciprocal of a number results  ³⁸⁄₅ . What is the number?

Problem 3 :

Kevin needs 150 ounces of sugar for making 50 pounds cookies. If Kevin currently has 80 ounces of sugar, how many more ounces of sugar does he need to make 30 pounds cookies?

Problem 4 :

John is able to complete a certain work in 10 days, but Peter is abobe to complete the same work in 8 days. How many days will it take them to complete the work, if they work together?

Problem 5 :

The numerator and denominator of a fraction add up to 11. If 5 be added to both numerator and denominator, the fraction becomes ¾ . Find the original fraction.

Problem 6 :

Divide 15 into two parts such that both of them are positive and the difference between their reciprocals is  ¼ .

Let x be the number.

¹⁄ₓ  - 7 = ⁻²⁰⁄₃

Multiply both sides by x to get rid of the denominator x on the left side.

x( ¹⁄ₓ  - 7 ) = x( ⁻²⁰⁄₃ )

x( ¹⁄ₓ )  - x(7) = ⁻²⁰ˣ⁄₃

1  - 7x =  ⁻²⁰ˣ⁄₃

Multiply both sides by 3 to get rid of the denominator 5 on the right side.

3(1  - 7x) = 3( ⁻²⁰ˣ⁄₃ )

3 - 21x = -20x

The number is 3.

3( ¹⁄ₓ )  + 7 =  ³⁸⁄₅

³⁄ₓ  + 7 =  ³⁸⁄₅

x( ³⁄ₓ  + 7 ) = x( ³⁸⁄₅ )

x( ³⁄ₓ )  + x(7) =  ³⁸ˣ⁄₅

3 + 7x =  ³⁸ˣ⁄₅

Multiply both sides by 5 to get rid of the denominator 5 on the right side.

5(3 + 7x) = 5( ³⁸ˣ⁄₅ )

15 + 35x = 38x

Subtract 35x from both sides.

Divide both sides by 3.

The number is 5.

Let x  be the additional ounces ofsugar needed to make 30 pounds of cookies.

150 ounces : 50 pounds = (80 + x) ounces : 30 pounds

Write each ratio in the equation above as a fraction.

¹⁵⁰⁄₅₀   =   ⁽⁸⁰ ⁺ ˣ⁾⁄₃₀

3 =   ⁽⁸⁰ ⁺ ˣ⁾⁄₃₀

Multiply both sides by 30 to get rid of the denominator 30 on the right side.

30(3 ) = 30( ⁽⁸⁰ ⁺ ˣ⁾⁄₃₀ )

90 = 80 + x

Subtract 80 from both sides.

Kevin needs 10 more ounces of sugar to make 30 pounds.

Givren : John is able to complete a certain work in 10 days and Peter is abobe to complete the same work in 8 days.

Part of the work completed by John in 1 hour :

Part of the work completed by Peter in 1 hour :

Let x  be the number of days required to complete the work, if both John and Peter work together.

Part of the work completed by both John and Peter in 1 hour :

=  ¹⁄ₓ

¹⁄ₓ  =  ⅒   +  ⅛

Multiply both sides by x to get rid of the denominator on the left side.

x( ¹⁄ₓ )   = x( ⅒   +  ⅛ )

1 = x( ⅒ )   + x( ⅛ )

1 = ˣ⁄₁₀  +   ˣ⁄₈

The least common multiple of (10, 8) = 40.

Multiply both sides by 40 to get rid of the denominators 10 and 8 on the right side.

40(1) = 40( ˣ⁄₁₀  +   ˣ⁄₈ )

40 = 40( ˣ⁄₁₀ )   + 40 ( ˣ⁄₈ )

40 = 4x + 5x

Divide both sides by 9.

⁴⁰⁄₉  = x

If both John and Peter work together, they will be able to complete the work in  4 ⁴⁄₉ days.

Let x be the denominator of the fraction.

Since t he numerator and denominator add up to 11, the numerator is (11 - x).

Fraction =  ⁽¹¹ ⁻ ˣ⁾⁄ₓ

Givren :  If 5 is added to both numerator and denominator, the fraction becomes ¾ .

⁽¹¹ ⁻ ˣ ⁺ ⁵⁾⁄₍ₓ ₊ ₅₎  =  ¾

⁽¹⁶ ⁻ ˣ⁾⁄₍ₓ ₊ ₅₎   = ¾

By cross multiplying,

4(16 - x) = 3(x + 5)

64 - 4x = 3x + 15

64 = 7x + 15

Subtract 15 from both sides.

Divide both sides by 7.

11 - x = 11 - 7 = 4

⁽¹¹ ⁻ ˣ⁾⁄ₓ  =  ⁴⁄₇

The required fraction is ⁴⁄₇ .

Let x  be one of the parts of 15.

Then the other part of 15 is (15 - x).

Given :  Difference between their reciprocals of the parts of 25 is  ¼ .

¹⁄ₓ  -  ¹⁄₍₁₅ ₋ ₓ₎  =  ¼

x[ ¹⁄ₓ  -  ¹⁄₍₁₅ ₋ ₓ₎ ] = x( ¼ )

x( ¹⁄ₓ ) - x[ ¹⁄₍₁₅ ₋ ₓ₎ ] =  ˣ⁄₄

1 -  ˣ ⁄₍₁₅ ₋ ₓ₎  =  ˣ⁄₄

Multiply both sides by (15 - x) to get rid of the denominator (15 - x) on the left side.

(15 - x)[1 -  ˣ ⁄₍₁₅ ₋ ₓ₎ ] = (15 - x)( ˣ⁄₄ )

(15 - x)(1) - (15 - x)[ ˣ⁄₍₁₅ ₋ ₓ₎ ] =  (15 - x)( ˣ⁄₄ )

15 - x - x =  (15 - x)( ˣ⁄₄ )

15 - 2x   =  (15 - x)( ˣ⁄₄ )

Multiply both sides by 4 to get rid of the denominator 4 on the right side.

4(15 - 2x) = 4 (15 - x)( ˣ⁄₄ )

60 - 8x = (15 - x)(x)

60 - 8x = 15x - x 2

Add x 2  to both sides.

x 2  - 8x + 60  = 15x

Subtract 15x from both sides.

x 2  - 23 x + 60 = 0

Factor and solve.

x 2  - 3 x - 20x + 60 = 0

x(x   - 3)  - 20(x - 3) = 0

(x - 3)(x - 20) = 0

x - 3 = 0  or x - 20 = 0

x = 3  or  x = 20

15 - x = 15 - 3

15 - x = 12

15 - x = 15 - 20

15 - x = -5

When x = 20, the other part (15 - x) is -5, which is negative. So, it can be ignored.

Therefore, the two parts of 15 are 3 and 12.

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Word Problem: Rachel has 17 apples. She gives some to Sarah. Sarah now has 8 apples. How many apples did Rachel give her?

Simplified Equation: 17 - x = 8

Word Problem: Rhonda has 12 marbles more than Douglas. Douglas has 6 marbles more than Bertha. Rhonda has twice as many marbles as Bertha has. How many marbles does Douglas have?

Variables: Rhonda's marbles is represented by (r), Douglas' marbles is represented by (d) and Bertha's marbles is represented by (b)

Simplified Equation: {r = d + 12, d = b + 6, r = 2 �� b}

Word Problem: if there are 40 cookies all together and Angela takes 10 and Brett takes 5 how many are left?

Simplified: 40 - 10 - 5

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## A Rational equation  is an equation that contain at least one rational expression. In order to solve a rational expression, one has to do the following steps:

Step 1)  multiply all the terms by the least common denominator to eliminate the denominators., step 2)  simplify the equation obtained in step 1., step 3)  solve the simplified equation for the variable., step 4)  check the solution to make sure that it satisfies the original equation., example:  solve ., step 1)  the least common denominator for the denominators in this example is . therefore, we multiply all the terms by :.

## Since the left-hand-side of the equation is equal to the right-hand-side of the equation, the solution is valid.

Example:  solve, step 1)  we have to multiply everything by the least common denominator. to do so, we have to factor denominators and find the least common denominator for them:.

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#### IMAGES

1. How to Solve a Word Problem Using a Rational Equation

2. Video #3.4b

3. Rational Equations Word Problems Worksheet

4. Rational Equation Word Problems

5. 35 Rational Equations Word Problems Worksheet

6. Solve and Graph Rational Equations (examples, solutions, videos

#### VIDEO

1. Solving Rational Equations

2. Solving Rational Equations

3. Math 30-2

4. Solving Rational Equations

5. Solving Rational Equations

6. SOLVING RATIONAL EQUATIONS || GENERAL MATHEMATICS

1. Rational equations word problem: eliminating solutions

Sal solves a word problem about the combined pool-filling rates of two water hoses, by creating a rational equation that models the situation. The equation has a solution that is eliminated due to the context. Created by Sal Khan and Monterey Institute for Technology and Education.

2. Rational Equations Word Problems Lesson

Motion Word Problems with Rational Expressions Earlier in our course, we learned how to use the distance formula when working with motion word problems. Recall the distance formula relates the distance traveled (d) to the rate of speed (r) times the time traveled (t). d = rt Let's look at an example. Example 2: Solve each word problem.

3. How to Solve a Word Problem Using a Rational Equation

When given a word problem that presents a rational equation, use the following steps to solve it. Step 1: Write the equation indicated by the word problem. Pay attention to words indicating ...

4. Rational Equation Word Problem Lesson

Motion Word Problems with Rational Expressions Another common application of rational expressions involves motion problems. These problem types were studied earlier in algebra and involve our distance formula: d = r • t Where d is the distance, r is the rate of speed, and t is the time. Let's look at an example. Example 3: Solve each word ...

5. PDF Math 101 Review on Rational Equations & Word Problems

Microsoft Word - Math 101 Review on Rational Equations & Word Problems.doc. MATH 101. Review on Rational Equations and Word Problems. Solve the following equations. List any restrictions and check for extraneous solutions. 2 x −. (1) = − 6. 2. x + 3 x + 3.

6. 7.6: Applications of Rational Equations

Set up an algebraic equation. Solve this rational expression by multiplying both sides by the LCD. The LCD is $$30n(n−4)$$. ... In this section, all of the steps outlined for solving general word problems apply. Look for the new key word "reciprocal," which indicates that you should write the quantity in the denominator of a fraction with ...

7. Solving Rational Equations

http://www.greenemath.com/In this course, we will learn how to solve rational equations, rational inequalities, and word problems that involve rational equat...

8. Rational Equation Word Problem

We will cover how to set up and solve common word problems that involve rational expressions. The two most common problems are: motion problems and work/rate problems. When we encounter motion problems, we use the general formula: distance = rate • time. We can use algebraic methods to solve for rate or time. Work rate problems are usually ...

9. 7.5: Solving Rational Equations

Begin solving rational equations by multiplying both sides by the LCD. The resulting equivalent equation can be solved using the techniques learned up to this point. Multiplying both sides of a rational equation by a variable expression introduces the possibility of extraneous solutions. Therefore, we must check the solutions against the set of ...

10. PDF Pre-Calculus 11: Rational Equations Word Problems

Pre-Calculus 11: Rational Equations Word Problems The speed of a plane is seven times as fast as the speed of a car. The car takes 3 hours longer than the plane to travel 315 kilometers. Find the speed of the car and the speed of the plane The speed of the car is 90 k/hr and the speed of the plane is 630 k/hr.

11. PDF Notes, Examples, and practice (with solutions)

Word Problems that use Rational Expressions Example: Underground pipes can fill a swimming pool in 4 hours. A regular garden hose can fill the pool in 15 hours. If both are used at the same thne, how long will it take to fill the pool? Solving Rational Equalities/Equations Step 3: Check Answer! If time is 3.158 hours, the pipes will add

12. Word Problems Leading to Rational Equations

Word Problems Leading to Rational Equations. Student Outcomes. Students solve word problems using models that involve rational expressions. New York State Common Core Math Algebra II, Module 1, Lesson 27. Worksheets for Algebra 2. Classwork. Exercise 1. Anne and Maria play tennis almost every weekend. So far, Anne has won 12 out of 20 matches.

13. "Work" Word Problems

As you can see in the above example, "work" problems commonly create rational equations. But the equations themselves are usually pretty simple to solve. One pipe can fill a pool 1.25 times as fast as a second pipe. When both pipes are opened, they fill the pool in five hours.

14. Applications of Rational Expressions (Work Rate Problems ...

http://www.greenemath.com/http://www.facebook.com/mathematicsbyjgreeneIn this lesson, we review how to solve work rate word problems, otherwise known as rate...

15. Applications of Rational Expressions and Word Problems

Let x=the number Now we can set up our equation. Seven divided by the sum of a number and two 7x+2 is equal to 7x+2= half the difference of the number and three. 7x+2=x−32 Before we solve, take note that there's a variable in the denominator which means we will have a restricted value. So, x≠−2 because that would make the denominator zero ...

16. MATH G11: Word Problems Leading to Rational Equations

Description. In the preceding lessons, students learned to add, subtract, multiply, and divide rational expressions and solve rational equations in order to develop the tools needed for solving application problems involving rational equations in this lesson (A-REI.A.2). Students develop their problem-solving and modeling abilities by carefully ...

17. Word Problems on Rational Equations Worksheet

WORD PROBLEMS ON RATIONAL EQUATIONS WORKSHEET. Problem 1 : Decreasing the reciprocal of a number by 7 results ⁻²⁰⁄₃. What is the number? Problem 2 : 7 more than three times the reciprocal of a number results ³⁸⁄₅. What is the number? Problem 3 : Kevin needs 150 ounces of sugar for making 50 pounds cookies.

18. Math Problem Solver

Math Word Problem Solutions. Math word problems require interpreting what is being asked and simplifying that into a basic math equation. Once you have the equation you can then enter that into the problem solver as a basic math or algebra question to be correctly solved. Below are math word problem examples and their simplified forms. Word ...

19. NWP Learning Commons: Math : Rational Equations and Problem Solving

Rational Equations and Problem Solving. A Rational equation is an equation that contain at least one rational expression.In order to solve a rational expression, one has to do the following steps: Step 1) Multiply all the terms by the least common denominator to eliminate the denominators. Step 2) Simplify the equation obtained in step 1. Step 3) Solve the simplified equation for the variable.

20. Word Problems Calculator

An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age. Show more