Elementary Algebra Equations with Fractions

Section 8.4 Equations with Fractions

Subsection "Clearing" Fractions

How can we solve an equation that involves algebraic fractions? Recall that to solve the equation

\begin{equation*} \dfrac{3x}{4} = 9 \end{equation*}

we multiply both sides by 4 (the denominator of the fraction) to "clear" the fraction.

\begin{align*} \cancel{\blert{4}}(\dfrac{3x}{\cancel{4}}) \amp = (9)\blert{4}\\ 3x \amp = 36 \end{align*}

We can now finish the solution to get $x=12\text{.}$

We use the same idea to solve an equation with algebraic fractions.

Example 8.27.

Solve $~~\dfrac{5000+50x}{x} = 75$

Solution.

We multiply both sides of the equation by $x\text{,}$ the denominator of the fraction, to get

\begin{align*} \cancel{\blert{x}}(\dfrac{5000+50x}{\cancel{x}}) \amp = (75)\blert{x}\\ 5000+50x \amp = 75x \end{align*}

Then we proceed as usual to finish the solution. We subtract $50x$ from both sides to find

\begin{align*} 5000 \amp = 25x\\ 200 \amp = x \end{align*}

You can check that $x=200$ does satisfy the equation.

Subsection Using an LCD to Clear Fractions

If an equation contains more than one fraction, we multiply both sides of the equation by the LCD of all the fractions. This will clear all the denominators at once.

Example 8.28.

Solve $~~\dfrac{x}{3} -2 = \dfrac{4}{5} + \dfrac{x}{5}$

Solution.

The LCD of $~\dfrac{x}{3},~ \dfrac{4}{5}$ and $~\dfrac{x}{5}$ is 15. We multiply both sides of the equation by 15.

\begin{align*} \blert{15}\left(\dfrac{x}{3} -2\right) \amp = \left(\dfrac{4}{5} + \dfrac{x}{5}\right)\blert{15} \amp \amp \blert{\text{Apply the distributive law.}}\\ \blert{15}\left(\dfrac{x}{3}\right)- \blert{15}(2) \amp = \blert{15}\left(\dfrac{4}{5}\right) + \blert{15}\left(\dfrac{x}{5}\right) \amp \amp \blert{\text{Simplify each product.}}\\ 5x-30 \amp = 12+3x \end{align*}

Now we can proceed as usual to complete the solution.

\begin{align*} 5x-30 \amp = 12+3x \amp \amp \blert{\text{Subtract}~3x~\text{from both sides;}}\\ \amp \amp\amp \blert{\text{add 30 to both sides.}}\\ 2x \amp = 42 \amp\amp \blert{\text{Divide both sides by 2.}}\\ x \amp = 21 \end{align*}

Caution 8.29.

In Example 8.28, note that we multiplied each term by the LCD, 15, including terms that are not fractions, namely 2 in this example. Be sure to multiply each term of the equation by the LCD.

Reading Questions Reading Questions

1.

When clearing fractions, which terms of the equation should we multiply by the LCD?

Answer.

Every term

Subsection Variables in the Denominator

Equations that involve algebraic fractions can also be solved using an LCD.

Example 8.30.

Solve $~~\dfrac{3}{4} = 8- \dfrac{2x+11}{x-5}$

Solution.

The LCD for the two fractions in the equation is $\blert{4(x-5)}\text{.}$ We multiply both sides of the equation by the LCD.

\begin{align*} \blert{4(x-5)}\left(\dfrac{3}{4}\right) \amp = \left(8- \dfrac{2x+11}{x-5}\right) \cdot \blert{4(x-5)} ~~~~~~~~ \blert{\text{Apply the distributive law.}}\\ \blert{\cancel{4}(x-5)}\left(\dfrac{3}{\cancel{4}}\right) \amp = \blert{4(x-5)}(8)- \blert{4\cancel{(x-5)}}\left(\dfrac{2x+11}{x-5}\right)\\ 3(x-5) \amp = 32(x-5)-4(2x+11) \end{align*}

We proceed as usual to complete the solution. First we use the distributive law to remove parentheses.

\begin{align*} 3x-15 \amp = 32x-160-8x-44 \amp \amp \blert{\text{Combine like terms.}}\\ 3x-15 \amp = 24x-204\\ -21x \amp = -189\\ x \amp = 9 \end{align*}

The solution is $x=9\text{.}$ You can check that $x=9$ satisfies the original equation.

Reading Questions Reading Questions

2.

How do we "clear" the fractions from an equation?

Answer.

Multiply both sides by the LCD.

Look Ahead.

Remember that we do not obtain an equivalent equation if we multiply both sides of an equation by zero. In Example 8.30 we multiplied by $4(x-5)\text{.}$ Is it possible that $4(x-5)$ equals zero? After solving, we found that $x=9\text{,}$ so $4(x-5) = 16\text{,}$ which is not zero, and the multiplication step was valid. The next example illustrates what can go wrong if we multiply by zero.

Subsection Extraneous Solutions

Example 8.31.

Solve $~~6 +\dfrac{4}{x-3} = \dfrac{x+1}{x-3}$

Solution.

We multiply both sides of the equation by the LCD, $x-3\text{,}$ to clear the fractions.

\begin{align*} \blert{(x-3)}\left(6+\dfrac{4}{x-3}\right) \amp = \left(\dfrac{x+1}{x-3}\right) \cdot \blert{(x-3)} ~~~~~ \blert{\text{Apply the distributive law.}}\\ \blert{(x-3)}(6)+\blert{\cancel{(x-3)}}\left(\dfrac{4}{\cancel{x-3}}\right) \amp = \left(\dfrac{x+1}{x-3}\right)\blert{\cancel{(x-3)}}\\ 6(x-3) + 4 \amp = x+1 \end{align*}

We complete the solution as usual.

\begin{align*} 6x-18 +4 \amp = x+1\\ 6x-14 \amp = x+1\\ 5x \amp = 15\\ x \amp = 3 \end{align*}

The solution appears to be $x=3\text{.}$ But we have a problem, because the LCD, $x-3\text{,}$ equals zero when $x=3\text{.}$ We have multiplied both sides of the equation by zero. When we try to check the solution we find

\begin{align*} 6 + \dfrac{4}{3-3} \amp = \dfrac{3+1}{3-3} ~~~~~~\blert{\text{Simplify each term.}}\\ 6 + \dfrac{4}{0} \amp = \dfrac{3+1}{0} \end{align*}

Because division by zero is undefined, 3 is not a solution after all. The original equation does not have a solution.

In Example 8.31, when we multiplied both sides of the equation by zero we found a false solution for the equation. Such solutions are called extraneous solutions. There is always a danger that an extraneous solution may be introduced when we multiply by an expression that contains the variable.

Reading Questions Reading Questions

3.

When might an extraneous solution be introduced?

Answer.

When we multiply by an expression that contains the variable.

We should always check for extraneous solutions when solving equations that involve algebraic fractions.

To check a solution, we substitute it into the original equation. If a possible solution causes any of the denominators in the equation to equal zero, then that solution is extraneous.

Reading Questions Reading Questions

4.

How do we check for extraneous solutions?

Answer.

See if it causes any of the denominators in the equation to equal zero.

Subsection Formulas

We can also solve formulas that involve algebraic fractions.

Example 8.32.

Solve for $P\text{:}$ $~~\dfrac{1}{T} = \dfrac{PR}{A-P}$

Solution.

The LCD for the two fractions in the equation is $\blert{T(A-P)}\text{.}$ We multiply both sides of the equation by the LCD to obtain

\begin{align*} \blert{\cancel{T}(A-P)} \dfrac{1}{\cancel{T}} \amp = \dfrac{PR}{\cancel{A-P}}\blert{T\cancel{(A-P)}}\\ A-P \amp = PRT \end{align*}

Next, we get all the terms containing the desired variable, $P\text{,}$ on one side of the equation. We add $P$ to both sides to get

\begin{equation*} A=P+PRT \end{equation*}

We now have two unlike terms that contain the desired variable. To proceed, we factor out this variable, and then divide both sides by the remaining factor.

\begin{align*} A \amp = P(1+RT)~~~~~~~~~\blert{\text{Divide both sides by}~1+RT.}\\ \dfrac{A}{\blert{1+RT}} \amp = \dfrac{P\cancel{(1+RT)}}{\blert{\cancel{1+RT}}} \end{align*}

Thus, the new version of the formula is

\begin{equation*} P=\dfrac{A}{1+RT} \end{equation*}

Reading Questions Reading Questions

5.

When solving a formula, what should we do if there are two terms that contain the variable?

Answer.

Factor out the variable.

Subsection Applications

Recall the formula $d=rt\text{,}$ which is useful in solving problems about motion. By solving for $r$ or $t\text{,}$ we can write the equation in the forms

\begin{equation*} r = \dfrac{d}{t}~~~~~~\text{or}~~~~~~t = \dfrac{d}{r} \end{equation*}

if either of these is more useful for the problem.

Example 8.33.

A cruise boat travels 18 miles downstream and back in $4\dfrac{1}{2}$ hours. If the speed of the current is 3 miles per hour, what is the speed of the boat in still water?

Solution.

We let $x$ stand for unknown quantity, the speed of the boat in still water. Then we make a table showing the distance, rate, and time for each part of the trip. We begin by filling in the information given in the problem.

$\hphantom{0000}$	Distance	Rate	Time
Downstream trip	$18$	$x+3$	$\hphantom{0000}$
Upstream trip	$18$	$x-3$	$\hphantom{0000}$

We use the formula $t = \dfrac{d}{r}$ to fill in the last column of the table.

$\hphantom{0000}$	Distance	Rate	Time
Downstream trip	$18$	$x+3$	$\blert{\dfrac{18}{x+3}}$
Upstream trip	$18$	$x-3$	$\blert{\dfrac{18}{x-3}}$

Notice that we did not use the $4\dfrac{1}{2}$ hours in the table, because it was not the trip upstream or the trip downstream that took $4\dfrac{1}{2}$ hours, but the total trip. We use the $4\dfrac{1}{2}$ to write an equation. The sum of the times for the upstream and downstream trips was $4\dfrac{1}{2}$ or $\dfrac{9}{2}$ hours.

\begin{equation*} \dfrac{18}{x+3}+\dfrac{18}{x-3} = \dfrac{9}{2} \end{equation*}

To solve the equation, we multiply both sides by the LCD, $2(x-3)(x+3)\text{.}$

\begin{equation*} \blert{2(x-3)(x+3)}(\dfrac{18}{x+3}+\dfrac{18}{x-3}) = (\dfrac{9}{2})\blert{2(x-3)(x+3)} \end{equation*}

Next, we apply the distributive law to multiply each term of the equation by the LCD.

\begin{align*} \blert{2(x-3)\cancel{(x+3)}}(\dfrac{18}{\cancel{x+3}})+\blert{2\cancel{(x-3)}(x+3)}(\dfrac{18}{\cancel{x-3}}) \amp = (\dfrac{9}{\cancel{2}})\blert{\cancel{2}(x-3)(x+3)}\\ 36(x-3)+36(x+3) \amp = 9(x-3)(x+3) \end{align*}

We simplify each side of the equation and write it in standard form.

\begin{align*} 36x-108+36x+108 \amp = 9x^2-81~~~~~~~~\blert{\text{Combine like terms.}}\\ 9x^2-72x-81 \amp =0 \end{align*}

The equation is quadratic, and we can solve it by factoring.

\begin{align*} 9(x^2-8x-9) \amp = 0\\ 9(x-9)(x+1) \amp = 0~~~~~~~~\blert{\text{Set each factor equal to zero.}}\\ x-9=0~~~~x+1 \amp = 0\\ x=9~~~~~~~~~~x \amp =-1 \end{align*}

Finally, we check each solution by substituting into the original equation. Neither solution is extraneous. However, the speed of the boat is not a negative number, so we discard the solution $x=-1\text{.}$ The boat travels at 9 miles per hour in still water.

Problems involving other types of rates can be solved with similar techniques. Suppose it takes you 8 hours to type a term paper for your history class. If you work at a constant rate, in 1 hour you would complete $\dfrac{1}{8}$ of your task. The rate at which you work, or your work rate, is one-eighth job per hour. In 3 hours you would complete

\begin{equation*} 3 \cdot \dfrac{1}{8} = \dfrac{3}{8} \end{equation*}

of the job. In $t$ hours you would complete

\begin{equation*} t \cdot \dfrac{1}{8} = \dfrac{t}{8} \end{equation*}

of the job. In general, the amount of work done, expressed as a fraction of 1 whole job, is given by the following formula.

Work Formula.

\begin{align*} \blert{\text{work rate} \times \text{time}} \amp \blert{= \text{work completed}}\\ \blert{rt} \amp \blert{= w} \end{align*}

Subsection Skills Warm-Up

Exercises Exercises

Exercise Group.

Write each expression as a single fraction in simplest form.

1.

$\dfrac{1}{x-2} + \dfrac{2}{x}$

2.

$\dfrac{1}{x-2} - \dfrac{2}{x}$

3.

$\dfrac{1}{x-2} \cdot \dfrac{2}{x}$

4.

$\dfrac{1}{x-2} \div \dfrac{2}{x}$

Subsubsection Answers to Skills Warm-Up

$\displaystyle \dfrac{3x-4}{x^2-2x}$
$\displaystyle \dfrac{4-x}{x^2-2x}$
$\displaystyle \dfrac{2}{x^2-2x}$
$\displaystyle \dfrac{x}{2x-4}$

Subsection

Subsubsection Activity 1: Using a Graph

A new health club opened up, and the manager kept track of the number of active members over its first few months of operation. The equation below gives the number $N$ of active members, in hundreds, $t$ months after the club opened.

\begin{equation*} N=\dfrac{10t}{4+t^2} \end{equation*}

The graph of this equation is shown at right.

Use the equation to find out in which months the club had 200 active members.
Verify your answers on the graph.

Subsubsection Activity 2: Using Algebra

Exercises Exercises

1.

Solve $~~\dfrac{x^2}{2} + \dfrac{5x}{4} = 3~~$ by first clearing the fractions.

2.

Solve $~~\dfrac{1}{x-2} + \dfrac{2}{x} = 1~~$

$\blert{\text{Step 1 Find the LCD for all the fractions in the equation.}}$

$\blert{\text{Step 2 Multiply each term of the equation by the LCD, and simplify.}}$

$\blert{\text{Step 3 You now have an equation without fractions. Solve as usual.}}$

3.

Solve $~~\dfrac{15}{x^2-3x} + \dfrac{4}{x} = \dfrac{5}{x-3}~~$

$\blert{\text{Step 1 Factor each denominator and find the LCD for all the fractions in the equation.}}$

$\blert{\text{Step 2 Multiply each term of the equation by the LCD, and simplify.}}$

$\blert{\text{Step 3 You now have an equation without fractions. Solve as usual.}}$

$\blert{\text{Step 4 Check for extraneous solutions.}}$

4.

Solve the formula $~~\dfrac{w}{q} = 1 - \dfrac{T}{H}~~$ for $q\text{.}$

Subsubsection Activity 3: Work Problem

In this application, the work being done is the water flowing into (or out of) the reservoir, so the work rate is the flow rate of water through the pipes.

The city reservoir was completely emptied for repairs, and is now being refilled. Water flows in through the intake pipe at a steady rate that can fill the reservoir in 120 days. However, water is also being drained from the reservoir as it is used by the city, so that it actually takes 150 days to fill. After the reservoir is filled, the water supply is turned off, but the city continues to use water at the same rate. How long will it take to drain the reservoir dry?

Let $d$ stand for the number of days for the outflow pipe to drain the full reservoir.

Fraction of the reservoir that is drained in one day:

This is the rate at which water leaves the reservoir.
Now imagine that the intake pipe is open, but the outflow pipe is closed.

Fraction of the reservoir that is filled in one day:

This is the rate at which water enters the reservoir.
Consider the time period described in the problem when both pipes are open. Fill in the table.

 Flow rate $~~~~$Time$~~~~$ Fraction of reservoir

Water entering   

Water leaving   
Write an equation:

$\blert{\text{Fraction of reservoir filled}} - \blert{\text{Fraction of reservoir drained}} = \blert{\text{One whole reservoir}}$
Solve your equation.

Subsubsection Wrap-Up

Objectives.

In this Lesson we practiced the following skills:

Solving equations with fractions
Identifying extraneous solutions
Solving motion problems
Solving work problems

Questions.

In Activity 1, part (b), explain how you used the graph.
In Activity 2, Problem 2, what was the LCD?
In Activity 2, Problem 4, what was the LCD?

Subsection Homework Preview

Exercises Exercises

1.

Solve $~~~\dfrac{3x}{4}-1=2x+\dfrac{9}{2}$

2.

Solve $~~~\dfrac{3}{x+4}-7=\dfrac{-4}{x+4}$

3.

Solve $~~~\dfrac{1}{a^2+a}=2+\dfrac{1}{a}$

4.

Solve $~~~R=\dfrac{cd}{c+d}~~$ for $c$

Subsubsection Answers to Homework Preview

$\displaystyle \dfrac{-22}{5}$
$\displaystyle -3$
$\displaystyle \dfrac{-3}{2}$
$\displaystyle c=\dfrac{dR}{d-R}$

Exercises Homework 8.4

Exercise Group.

For Problems 1–10, solve the equation.

1.

$\dfrac{5x}{2}-1=x+\dfrac{1}{2}$

Answer.

2.

$\dfrac{t}{6}-\dfrac{7}{3}=\dfrac{2t}{9}-\dfrac{t}{4}$

3.

$\dfrac{2}{3}(x-1)+x=6$

Answer.

4.

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

5.

$2+\dfrac{5}{2x}=\dfrac{3}{x}+\dfrac{3}{2}$

Answer.

6.

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

7.

$\dfrac{4}{x-1}-\dfrac{4}{x+2}=\dfrac{3}{7}$

Answer.

$-6\text{,}$ $5$

8.

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

9.

$\dfrac{15x}{1+x^2}=6$

Answer.

$\dfrac{1}{2} \text{,}$ $2$

10.

$\dfrac{3x+2}{x}=\dfrac{x+9}{x+6}$

Exercise Group.

For Problems 11–18, solve the formula for the specified variable.

11.

$V=\dfrac{hT}{P}~~~~$for $~T$

Answer.

$\dfrac{PV}{h} $

12.

$m=\dfrac{2E}{v^2}~~~~$for $~E$

13.

$m=\dfrac{y-k}{x-h}~~~~$for $~x$

Answer.

$\dfrac{y-k+hm}{m} $

14.

$a=\dfrac{F}{m+M}~~~~$for $~m$

15.

$\dfrac{1}{R}=\dfrac{1}{A}+\dfrac{1}{B}~~~~$for $~A$

Answer.

$\dfrac{BR}{B-R} $

16.

$I=\dfrac{E}{R+\dfrac{r}{n}}~~~~$for $~R$

17.

$r=\dfrac{dc}{1-ec}~~~~$for $~e$

Answer.

$\dfrac{r-dc}{cr} $

18.

$w=0.622\dfrac{e}{P-e}~~~~~~$for $~e$

21.

A small lake in a state park has become polluted by runoff from a factory upstream. The cost for removing $p$ percent of the pollution from the lake is given, in thousands of dollars, by

\begin{equation*} C=\dfrac{25p}{100-p} \end{equation*}

How much of the pollution can be removed for $25,000?

Answer.

50%

22.

During the baseball season so far this year, Pete got hits 44 times out of 164 times at bat.

What is Pete’s batting average so far? (Batting average is the fraction of at-bats that resulted in hits.)
If Pete gets hits on every one of his next $x$ at-bats, write an expression for his new batting average.
How many consecutive hits does Pete need to raise his batting average to $0.350\text{?}$

23.

The rectangle $ABCD$ in the figure is divided into a square and a smaller rectangle, $CDEF\text{.}$ The two rectangles $ABCD$ and $CDEF$ are similar, which means that their corresponding sides are proportional.

$golden rectangle$

A rectangle with this property is called a golden rectangle, and the ratio of its length to its width is called the golden ratio. The golden ratio appears frequently in art and nature, and is considered to give the most pleasing proportions to many figures. We’ll compute the golden ratio as follows.

Let $AB=1$ and $AD=x\text{.}$ What are the lengths of $AE,~ ED,$ and $CD\text{?}$
Write a proportion in terms of $x$ for the similarity of rectangles $ABCD$ and $CDEF\text{.}$ Be careful to match up the corresponding sides.
Solve your proportion for $x\text{.}$ Find the golden ratio, $\dfrac{AD}{AB} = \dfrac{x}{1}.$

Answer.

$AE=1\text{,}$ $ED=x-1\text{,}$ $CD=1$
$\displaystyle \dfrac{x}{1}=\dfrac{1}{x-1} $
$\displaystyle \dfrac{1+\sqrt{5}}{2} $

24.

Compare the procedures for adding fractions and for solving fractional equations in the problems below. Explain how the LCD is used differently for adding and for solving equations.

Add: $~~\dfrac{x-1}{4}+\dfrac{3x}{5}$
Solve: $~~\dfrac{x-1}{4}+\dfrac{3x}{5}=1$

Exercise Group.

For Problems 25–28,

Write an equation to model the problem.
Solve your equation and answer the question posed in the problem.

25.

An express train travels 180 miles in the same time that a freight train travels 120 miles. If the express train travels 20 miles per hour faster than the freight train, find the speed of each.

Answer.

$\displaystyle \dfrac{180}{x+20}=\dfrac{120}{x} $
freight train: 40 mph; express train: 60 mph

26.

Sam Scholarship and Reginald Privilege each travel the 360 miles to Fort Lauderdale on spring break, but Reginald drives his Porsche while Sam hitches a ride on a vegetable truck. Reginald travels 20 miles per hour faster than Sam does and arrives in 3 hours less time. How fast did each travel?

27.

Periwinkle Publishing can print the first run of a volume of poems in 10 hours on their new press; the same job takes 18 hours on their old machine. The new press is finishing another job, so the press manager starts the old machine 4 hours ahead of the new one. How many hours are needed with both presses running to finish the printing?

Answer.

$\displaystyle \dfrac{x}{10}+\dfrac{x+4}{18}=1 $
5 hr

28.

It takes 30 minutes to fill a large water tank. However, the tank has a small leak that would completely drain it in 4 hours. How long will it take to fill the tank if the leak is not plugged?

29.

Find the $\alert{\text{error}}$ in the following "proof" that $1=0\text{:}$ Start by letting $x=1\text{.}$

\begin{align*} x \amp = 1 \amp \amp \blert{\text{Multiply both sides by}~x.}\\ x^2 \amp = x \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ x^2-1 \amp = x-1 \amp \amp \blert{\text{Factor the left side.}}\\ (x-1)(x+1) \amp = x-1 \amp \amp \blert{\text{Divide both sides by}~x-1.}\\ \dfrac{(x-1)(x+1)}{x-1} \amp = \dfrac{x-1}{x-1} \amp \amp \blert{\text{Simplify both sides.}}\\ x+1 \amp = 1 \amp \amp \blert{\text{Subtract 1 from both sides.}}\\ x \amp = 0 \end{align*}

Because $x=1$ and $x=0\text{,}$ we have "proved" that $1=0\text{.}$

Answer.

Because $x=1\text{,}$ dividing by $x-1$ in the fourth step is dividing by $9\text{.}$

Exercise Group.

For Problems 30–31, find the error in the "solution" and correct it.

30.

\begin{align*} \dfrac{a}{2} - \dfrac{a}{3} \amp = 1 \\ 6(\dfrac{a}{2} - \dfrac{a}{3}) \amp = 1 \\ 3a-2a \amp = 1\\ a \amp = 1 \end{align*}

31.

\begin{align*} m^2-6m \amp = 0 \\ m^2 \amp = 6m\\ \dfrac{m^2}{m} \amp = \dfrac{6m}{m}\\ a \amp = 6 \end{align*}

Answer.

In the third equation, dividiing by $m$ is dividing by $0$ because $m=0$ is one of the solutions. $m=6, ~0$

32.

If two hens lay two eggs in 2 days, how long will it take six hens to lay six eggs?

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\(\hphantom{0000}\)	Distance	Rate	Time
Downstream trip	\(18\)	\(x+3\)	\(\hphantom{0000}\)
Upstream trip	\(18\)	\(x-3\)	\(\hphantom{0000}\)

\(\hphantom{0000}\)	Distance	Rate	Time
Downstream trip	\(18\)	\(x+3\)	\(\blert{\dfrac{18}{x+3}}\)
Upstream trip	\(18\)	\(x-3\)	\(\blert{\dfrac{18}{x-3}}\)

\(\)	Flow rate	\(~~~~\)Time\(~~~~\)	Fraction of reservoir
Water entering	\(\)	\(\)	\(\)
Water leaving	\(\)	\(\)	\(\)