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Mathematics LibreTexts

4.4: Solve Rational Inequalities

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    29061
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    Learning Objectives

    • Solve proportions
    • Solve similar figure applications
    • Solve uniform motion applications
    • Solve work applications
    • Solve direct variation problems
    • Solve inverse variation problems

    Be Prepared

    Before you get started, take this readiness quiz.

      Example 2.2.13. Example 2.5.13. Example 2.2.9.

    Solve Proportions

    When two rational expressions are equal, the equation relating them is called a proportion.

    Proportion

    A proportion is an equation of the form \(\dfrac{a}{b}=\dfrac{c}{d}\), where \(b \neq 0, d \neq 0\).

    The proportion is read “\(a\) is to \(b\) as \(c\) is to \(d\).”

    The equation \(\dfrac{1}{2}=\dfrac{4}{8}\) is a proportion because the two fractions are equal. The proportion \(\dfrac{1}{2}=\dfrac{4}{8}\) is read “1 is to 2 as 4 is to 8.”

    Since a proportion is an equation with rational expressions, we will solve proportions the same way we solved rational equations. We’ll multiply both sides of the equation by the LCD to clear the fractions and then solve the resulting equation.

    Example \(\PageIndex{1}\)

    Solve: \(\dfrac{n}{n+14}=\dfrac{5}{7}\).

    Solution

    \[\dfrac{n}{n+14}=\dfrac{5}{7}, \quad n \neq-14 \nonumber \]

    Multiply both sides by LCD.

    \[7(n+14)\left(\dfrac{n}{n+14}\right)=7(n+14)\left(\dfrac{5}{7}\right) \nonumber \]

    Remove common factors on each side.

    \[7 n=5(n+14) \nonumber \]

    Simplify.

    \[7 n=5 n+70 \nonumber \]

    Solve for \(n\).

    \[\begin{aligned} 2n&=70\\ n&=35 \end{aligned} \nonumber \]

    Check.

    \[\dfrac{n}{n+14}=\dfrac{5}{7} \nonumber \]

    Substitute \(n=35\)

    \[\dfrac{35}{35+14} \overset{?}{=} \dfrac{5}{7} \nonumber \]

    Simplify.

    \[\dfrac{35}{49} \overset{?}{=} \dfrac{5}{7} \nonumber \]

    Show common factors.

    \[\dfrac{5 \cdot 7}{7 \cdot 7} \overset{?}{=} \dfrac{5}{7} \nonumber \]

    Simplify.

    \[\dfrac{5}{7}=\dfrac{5}{7}\; \surd \nonumber \]

    Exercise \(\PageIndex{1}\)

    Solve the proportion: \(\dfrac{y}{y+55}=\dfrac{3}{8}\).

    Answer

    \(y=33\)

    Exercise \(\PageIndex{2}\)

    Solve the proportion: \(\dfrac{z}{z-84}=-\dfrac{1}{5}\).

    Answer

    \(z=14\)

    Notice in the last example that when we cleared the fractions by multiplying by the LCD, the result is the same as if we had cross-multiplied.

    \[\begin{aligned} \dfrac{n}{n+14}=\dfrac{5}{7} \quad \quad \quad \dfrac{n}{n+14}=\dfrac{5}{7} \\ 7(n+14)\left(\dfrac{n}{n+14}\right)=7(n+14)\left(\dfrac{5}{7}\right) \quad \quad \quad \dfrac{n}{n+14}=\dfrac{5}{7} \\ 7n=5(n+14) \quad \quad \quad 7n=5(n+14) \end{aligned} \nonumber \]

    For any proportion, \(\dfrac{a}{b}=\dfrac{c}{d}\), we get the same result when we clear the fractions by multiplying by the LCD as when we cross-multiply.

    \[\begin{aligned} \dfrac{a}{b} =\dfrac{c}{d} \quad \quad \quad \dfrac{a}{b}=\dfrac{c}{d} \\ bd\left(\dfrac{a}{b}=\frac{c}{d}\right) bd \quad \quad \quad \frac{a}{b} = \frac{c}{d} \\ ad =bc \quad \quad \quad ad=bc \end{aligned} \nonumber \]

    To solve applications with proportions, we will follow our usual strategy for solving applications But when we set up the proportion, we must make sure to have the units correct—the units in the numerators must match each other and the units in the denominators must also match each other.

    When pediatricians prescribe acetaminophen to children, they prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of the child’s weight. If Zoe weighs 80 pounds, how many milliliters of acetaminophen will her doctor prescribe?

    Solution

    Identify what we are asked to find, and choose a variable to represent it.

    How many ml of acetaminophen will the doctor prescribe?

    Let \(a=ml\) of acetaminophen.

    Write a sentence that gives the information to find it.

    If 5 ml is prescribed for every 25 pounds, how much will be prescribed for 80 pounds?

    Translate into a proportion—be careful of the units.

    Step 1. Write the inequality as one quotient on the left and zero on the right. Our inequality is in this form.

    \[\dfrac{x-1}{x+3} \geq 0 \nonumber \]

    Step 2. Determine the critical points-the points where the rational expression will be zero or undefined.

    The rational expression will be zero when the numerator is zero. Since \(x-1=0\) when \(x=1\), then 1 is a critical point. The rational expression will be undefined when the denominator is zero. Since \(x+3=0\) when \(x=-3\), then -3 is a critical point.

    Step 3. Use the critical points to divide the number line into intervals.

    clipboard_ee2fe708799269f0539168f8443a351d2.png

    Step 4. Above the number line show the sign of each factor of the rational expression in each interval. Below the number line show the sign of the quotient.

    Use values in each interval to determine the value of each factor in the interval. In the interval (-3,1), zero is a good value to test. For example, when \(x=0\) then \(x-1=-1\) and \(x+3=3\) The factor \(x-1\) is marked negative and \(x+3\) marked positive. Since a negative divided by a positive is negative, the quotient is marked negative in that interval.

    Step 5. Determine the intervals where the inequality is correct. Write the solution in interval notation.

    clipboard_edae13eef9fc14a8139b8d2bcd44dd965.png

    We want the quotient to be greater than or equal to zero, so the numbers in the intervals \((-\infty,-3)\) and \((1, \infty)\) are solutions. Since 3 must be excluded since it makes the rational expression 0, we cannot include it in the solution. We can include 1 in our solution.

    \[(-\infty,-3) \cup[1, \infty) \nonumber \]

    Multiply both sides by the LCD, 400. Remove common factors on each side. Simplify, but don’t multiply on the left. Notice what the next step will be.

    \[16 \cdot 5=5 a \nonumber \]

    Solve for \(a\).

    \[\begin{aligned} \dfrac{16 \cdot 5}{5}&=\dfrac{5 a}{5}\\ 16&=a \end{aligned} \nonumber \]

    Check. Is the answer reasonable? Write a complete sentence.

    The pediatrician would prescribe 16 ml of acetaminophen to Zoe.

    Exercise \(\PageIndex{3}\)

    Pediatricians prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of a child’s weight. How many milliliters of acetaminophen will the doctor prescribe for Emilia, who weighs 60 pounds?

    Answer

    The pediatrician will prescribe 12 ml of acetaminophen to Emilia.

    Exercise \(\PageIndex{4}\)

    For every 1 kilogram (kg) of a child’s weight, pediatricians prescribe 15 milligrams (mg) of a fever reducer. If Isabella weighs 12 kg, how many milligrams of the fever reducer will the pediatrician prescribe?

    Answer

    The pediatrician will prescribe 180 mg of fever reducer to Isabella.

    Solve similar figure applications

    When you shrink or enlarge a photo on a phone or tablet, figure out a distance on a map, or use a pattern to build a bookcase or sew a dress, you are working with similar figures. If two figures have exactly the same shape, but different sizes, they are said to be similar. One is a scale model of the other. All their corresponding angles have the same measures and their corresponding sides have the same ratio.

    Similar Figures

    Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides have the same ratio.

    For example, the two triangles in Figure below are similar. Each side of \(\Delta ABC\) is four times the length of the corresponding side of \(\Delta XYZ\).

    clipboard_e856a7de8b3a79dcd08723d9871ccd270.png

    This is summed up in the Property of Similar Triangles.

    Property of Similar Triangles

    If \(\Delta ABC\) is similar to \(\Delta XYZ\), then their corresponding angle measure are equal and their corresponding sides have the same ratio.

    clipboard_e4fdc3b30588979c50d44a48aef0d74a0.png

    To solve applications with similar figures we will follow the Problem-Solving Strategy for Geometry Applications we used earlier.

    Example \(\PageIndex{3}\)

    On a map, San Francisco, Las Vegas, and Los Angeles form a triangle. The distance between the cities is measured in inches. The figure on the left below represents the triangle formed by the cities on the map. If the actual distance from Los Angeles to Las Vegas is 270 miles, find the distance from Los Angeles to San Francisco.

    clipboard_efb0263e1d94381d6b9257775d879ded0.png

    Solution

    Since the triangles are similar, the corresponding sides are proportional.

    Read the problem. Draw the figures and label it with the given information. The figures are shown above.

    Identify what we are looking for: the actual distance from Los Angeles to San Francisco

    Name the variables: Let \(x\) = distance from Los Angeles to San Francisco.

    Translate into an equation. Since the triangles are similar, the corresponding sides are proportional. We’ll make the numerators “miles” and the denominators “inches”.

    \[$\dfrac{x \text { miles }}{1.3 \text { inches }}=\dfrac{270 \text { miles }}{1 \text { inch }}$ \nonumber \]

    Solve the equation.

    \[\begin{aligned} 1.3\left(\dfrac{x}{1.3}\right)&=1.3\left(\dfrac{270}{1}\right) \\ x&=351 \end{aligned} \nonumber \]

    Check. On the map, the distance from Los Angeles to San Francisco is more than the distance from Los Angeles to Las Vegas. Since 351 is more than 270 the answer makes sense.

    Check \(x=351\) in the original proportion. Use a calculator.

    \[\begin {aligned} \dfrac{x \text { miles }}{1.3 \text { inches }}&=\dfrac{270 \text { miles }}{1 \text { inch }}\\ \dfrac{351 \text { miles }}{1.3 \text { inches }} &\overset{?}{=} \dfrac{270 \text { miles }}{1 \text { inch }}\\ \dfrac{270 \text { miles }}{1 \text { inch }}&=\dfrac{270 \text { miles }}{1 \text { inch }} \surd \end{aligned} \nonumber \]

    Answer the question: The distance from Los Angeles to San Francisco is 351 miles.

    On the map, Seattle, Portland, and Boise form a triangle. The distance between the cities is measured in inches. The figure on the left below represents the triangle formed by the cities on the map. The actual distance from Seattle to Boise is 400 miles.

    clipboard_e545a49106f741f6a3b36f53e39ed4458.png

    Exercise \(\PageIndex{5}\)

    Find the actual distance from Seattle to Portland.

    Answer

    The distance is 150 miles.

    Exercise \(\PageIndex{6}\)

    Find the actual distance from Portland to Boise.

    Answer

    The distance is 350 miles.

    We can use similar figures to find heights that we cannot directly measure.

    Example \(\PageIndex{4}\)

    Tyler is 6 feet tall. Late one afternoon, his shadow was 8 feet long. At the same time, the shadow of a tree was 24 feet long. Find the height of the tree.

    Solution

    Read the problem and draw a figure. We are looking for \(h\), the height of the tree.

    clipboard_ee900ec650dd61d33cbc8b7d340874f27.png

    We will use similar triangles to write an equation. The small triangle is similar to the large triangle.

    \[\dfrac{h}{24}=\dfrac{6}{8} \nonumber \]

    Solve the proportion.

    \[\begin {aligned} 24\left(\dfrac{6}{8}\right)&=24\left(\dfrac{h}{24}\right)\\ 18&=h \end{aligned} \nonumber \]

    Simplify. Check.

    Tyler's height is less than his shadow's length so it makes sense that the tree's height is less than the length of its shadow. Check \(h=18\) in the original proportion.

    \[\begin{aligned} &\dfrac{6}{8}=\dfrac{h}{24}\\ &\dfrac{6}{8} \overset{?}{=} \dfrac{18}{24}\\ &\dfrac{3}{4}=\dfrac{3}{4} \surd \end{aligned} \nonumber \]

    Exercise \(\PageIndex{7}\)

    A telephone pole casts a shadow that is 50 feet long. Nearby, an 8 foot tall traffic sign casts a shadow that is 10 feet long. How tall is the telephone pole?

    Answer

    The telephone pole is 40 feet tall.

    Exercise \(\PageIndex{8}\)

    A pine tree casts a shadow of 80 feet next to a 30 foot tall building which casts a 40 feet shadow. How tall is the pine tree?

    Answer

    The pine tree is 60 feet tall.

    Solve Uniform Motion Applications

    We have solved uniform motion problems using the formula \(D=r t\) in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

    Rate \(\cdot\) Time = Distance

    The formula \(D=r t\) assumes we know \(r\) and \(t\) and use them to find \(D\). If we know \(D\) and \(r\) and need to find \(t\), we would solve the equation for \(t\) and get the formula \(t=\dfrac{D}{r}\).

    We have also explained how flying with or against the wind affects the speed of a plane. We will revisit that idea in the next example.

    Example \(\PageIndex{5}\)

    An airplane can fly 200 miles into a 30 mph headwind in the same amount of time it takes to fly 300 miles with a 30 mph tailwind. What is the speed of the airplane?

    Solution

    This is a uniform motion situation. A diagram will help us visualize the situation.

    clipboard_e675478d490b6955418a9d40f7346c6b5.png

    We fill in the chart to organize the information.

    We are looking for the speed of the airplane. Let \(r\) = the speed of the airplane.

    When the plane flies with the wind, the wind increases its speed and so the rate is \(r + 30\).

    When the plane flies against the wind, the wind decreases its speed and the rate is \(r − 30\).

    Write in the rates. Write in the distances. Since \(D=r \cdot t\), we solve for \(t\) and get \(t=\dfrac{D}{r}\). We divide the distance by the rate in each row, and place the expression in the time column.

    Rate \(\cdot\) Time = Distance
    Headwind \(r-30\) \(\dfrac{200}{r-30}\) 200
    Tailwind \(r+30\) \(\dfrac{300}{r+30}\) 300

    We know the times are equal and so we write our equation.

    \[\dfrac{200}{r-30}=\dfrac{300}{r+30} \nonumber \]

    We multiply both sides by the LCD.

    \[(r+30)(r-30)\left(\frac{200}{r-30}\right)=(r+30)(r-30)\left(\frac{300}{r+30}\right) \nonumber \]

    Simplify and solve.

    \[\begin{aligned} (r+30)(200)&=(r-30) 300 \\ 200 r+6000&=300 r-9000 \\ 15000&=100 r \end{aligned} \nonumber \]

    Check.

    Is \(150 \mathrm{mph}\) a reasonable speed for an airplane? Yes. If the plane is traveling \(150 \mathrm{mph}\) and the wind is \(30 \mathrm{mph}\),

    \[\text { Tailwind } \quad 150+30=180 \mathrm{mph} \quad \dfrac{300}{180}=\dfrac{5}{3} \text { hours } \nonumber \]

    \[\text { Headwind } 150-30=120 \mathrm{mph} \dfrac{200}{120}=\dfrac{5}{3} \text { hours } \nonumber \]

    The times are equal, so it checks. The plane was traveling \(150 \mathrm{mph}\).

    Exercise \(\PageIndex{9}\)

    Link can ride his bike 20 miles into a 3 mph headwind in the same amount of time he can ride 30 miles with a 3 mph tailwind. What is Link’s biking speed?

    Answer

    Link’s biking speed is 15 mph.

    Exercise \(\PageIndex{10}\)

    Danica can sail her boat 5 miles into a 7 mph headwind in the same amount of time she can sail 12 miles with a 7 mph tailwind. What is the speed of Danica’s boat without a wind?

    Answer

    The speed of Danica’s boat is 17 mph.

    In the next example, we will know the total time resulting from traveling different distances at different speeds.

    Example \(\PageIndex{6}\)

    Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?

    Solution

    This is a uniform motion situation. A diagram will help us visualize the situation.

    clipboard_ea415f63ad2acc69c7ea17326f1ee0337.png

    We fill in the chart to organize the information. We are looking for Jazmine’s running speed. Let \(r\) = Jazmine’s running speed.

    Her biking speed is 4 miles faster than her running speed. \(r + 4\) = her biking speed

    The distances are given, enter them into the chart. Since \(D=r \cdot t\), we solve for \(t\) and get \(t=\dfrac{D}{r}\). We divide the distance by the rate in each row, and place the expression in the time column.

    Rate \(\cdot\) Time = Distance
    Run \(r\) \(\dfrac{8}{r}\) 8
    Bike \(r+4\) \(\dfrac{24}{r+4}\) 24
    3

    Write a word sentence: Her time plus the time biking is 3 hours.

    Translate the sentence to get the equation.

    \[\dfrac{8}{r}+\dfrac{24}{r+4}=3 \nonumber \]

    Solve.

    \[\begin{aligned}
    r(r+4)\left(\dfrac{8}{r}+\dfrac{24}{r+4}\right) &=3 \cdot r(r+4) \\
    8(r+4)+24 r &=3 r(r+4) \\
    8 r+32+24 r &=3 r^{2}+12 r \\
    32+32 r &=3 r^{2}+12 r \\
    0 &=3 r^{2}-20 r-32 \\
    0 &=(3 r+4)(r-8)
    \end{aligned} \nonumber \]

    \[\begin{array}{lc} {(3 r+4)=0} & {(r-8)=0} \\ \cancel{r=\dfrac{4}{3}} \quad & {r=8} \end{array} \nonumber \]

    Check.

    A negative speed does not make sense in this problem, so \(r=8\) is the solution.

    Is 8 mph a reasonable running speed? Yes.

    If Jazmine’s running rate is 4, then her biking rate, \(r+4\), which is \(8+4=12\).

    \[\text { Run } 8 \mathrm{mph} \quad \dfrac{8 \mathrm{miles}}{8 \mathrm{mph}}=1 \text { hour } \nonumber \]

    \[\text { Bike } 12 \text { mph } \quad \dfrac{24 \text { miles }}{12 \mathrm{mph}}=2 \text { hours } \nonumber \]

    Jazmine’s running speed is 8 mph.

    Exercise \(\PageIndex{11}\)

    Dennis went cross-country skiing for 6 hours on Saturday. He skied 20 mile uphill and then 20 miles back downhill, returning to his starting point. His uphill speed was 5 mph slower than his downhill speed. What was Dennis’ speed going uphill and his speed going downhill?

    Answer

    Dennis’s uphill speed was 10 mph and his downhill speed was 5 mph.

    Exercise \(\PageIndex{12}\)

    Joon drove 4 hours to his home, driving 208 miles on the interstate and 40 miles on country roads. If he drove 15 mph faster on the interstate than on the country roads, what was his rate on the country roads?

    Answer

    Joon’s rate on the country roads is 50 mph.

    Once again, we will use the uniform motion formula solved for the variable \(t\).

    Example \(\PageIndex{7}\)

    Hamilton rode his bike downhill 12 miles on the river trail from his house to the ocean and then rode uphill to return home. His uphill speed was 8 miles per hour slower than his downhill speed. It took him 2 hours longer to get home than it took him to get to the ocean. Find Hamilton’s downhill speed.

    Solution

    This is a uniform motion situation. A diagram will help us visualize the situation.

    clipboard_eafa696bc696539ea8097deeeb9c97f3b.png

    We fill in the chart to organize the information.

    We are looking for Hamilton's downhill speed. Let \(h\)= Hamilton's downhill speed.

    His uphill speed is 8 miles per hour slower. \(h-8\)= Hamilton's uphill speed.

    Enter the rates into the chart.

    The distance is the same in both directions: 12 miles.

    Since \(D=r \cdot t\), we solve for \(t\) and get \(t=\dfrac{D}{r}\). We divide the distance by the rate in each row, and place the expression in the time column.

    Rate \(\cdot\) Time = Distance
    Downhill \(h\) \(\dfrac{12}{h}\) 12
    Uphill \(h-8\) \(\dfrac{12}{h-8}\) 12

    Write a word sentence about the line: He took 2 hours longer uphill than downhill. The uphill time is 2 more than the downhill time.

    Translate the sentence to get the equation.

    \[\dfrac{12}{h-8}=\dfrac{12}{h}+2 \nonumber \]

    Solve.

    \[\begin{aligned}
    h(h-8)\left(\dfrac{12}{h-8}\right) &=h(h-8)\left(\dfrac{12}{h}+2\right) \\
    12 h &=12(h-8)+2 h(h-8) \\
    12 h &=12 h-96+2 h^{2}-16 h \\
    0 &=2 h^{2}-16 h-96 \\
    0 &=2\left(h^{2}-8 h-48\right) \\
    0 &=2(h-12)(h+4)
    \end{aligned} \nonumber \]

    \[\begin{array}{lc} h-12=0 & h+4=0 \\ h=12 & \cancel {h=4} \end{array} \nonumber \]

    Check. Is \(12 \mathrm{mph}\) a reasonable speed for biking downhill? Yes.

    \[\text { Downhill } 12 \mathrm{mph} \quad \dfrac{12 \text { miles }}{12 \mathrm{mph}}=1 \text { hour } \nonumber \]

    \[\text { Uphill } 12-8=4 \mathrm{mph} \quad \dfrac{12 \text { miles }}{4 \mathrm{mph}}=3 \text { hours} \nonumber \]

    The uphill time is 2 hours more that the downhill time.

    Hamilton's downhill speed is \(12 \mathrm{mph}\).

    Exercise \(\PageIndex{13}\)

    Kayla rode her bike 75 miles home from college one weekend and then rode the bus back to college. It took her 2 hours less to ride back to college on the bus than it took her to ride home on her bike, and the average speed of the bus was 10 miles per hour faster than Kayla’s biking speed. Find Kayla’s biking speed.

    Answer

    Kayla’s biking speed was 15 mph.

    Exercise \(\PageIndex{14}\)

    Victoria jogs 12 miles to the park along a flat trail and then returns by jogging on an 20 mile hilly trail. She jogs 1 mile per hour slower on the hilly trail than on the flat trail, and her return trip takes her two hours longer. Find her rate of jogging on the flat trail.

    Answer

    Victoria jogged 6 mph on the flat trail.

    Solve Work Applications

    The weekly gossip magazine has a big story about the Princess’ baby and the editor wants the magazine to be printed as soon as possible. She has asked the printer to run an extra printing press to get the printing done more quickly. Press #1 takes 6 hours to do the job and Press #2 takes 12 hours to do the job. How long will it take the printer to get the magazine printed with both presses running together?

    This is a typical ‘work’ application. There are three quantities involved here—the time it would take each of the two presses to do the job alone and the time it would take for them to do the job together.

    If Press #1 can complete the job in 6 hours, in one hour it would complete \(\dfrac{1}{6}\) of the job.

    If Press #2 can complete the job in 12 hours, in one hour it would complete \(\dfrac{1}{12}\) of the job.

    We will let \(t\) be the number of hours it would take the presses to print the magazines with both presses running together. So in 1 hour working together they have completed \(\dfrac{1}{t}\) of the job.

    We can model this with the word equation and then translate to a rational equation. To find the time it would take the presses to complete the job if they worked together, we solve for \(t\).

    Follow the steps to organize the information. We are looking for how many hours it would take to complete the job with both presses running together.

    Step 1: Let \(t\) = the number of hours needed to complete the job together.

    Step 2: Enter the hours per job for Press #1, Press #2, and when they work together.

    If a job on Press #1 takes 6 hours, then in 1 hour \(\dfrac{1}{6}\) of the job is completed.

    Similarly find the part of the job completed/hours for Press #2 and when they both together.

    Number of hours to complete the job. Part of job completed/hour
    Press #1 6 \(\dfrac{1}{6}\)
    Press #2 12 \(\dfrac{1}{12}\)
    Together \(t\) \(\dfrac{1}{t}\)

    Write a word sentence. The part completed by Press #1 plus the part completed by Press #2 equals the amount completed together.

    Step 3: Translate into an equation.

    \[\text {Work completed by}\\ \underbrace{\text {Press } \#1 + \text {Press } \#2 = \text {Together}}\\ \dfrac{1}{6} \qquad+\qquad\dfrac{1}{12}\qquad =\qquad \dfrac{1}{t} \nonumber \]

    Step 4: Solve. Simplify.

    \[\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{1}{t} \nonumber \]

    Multiply by the LCD, \(12t\) and simplify.

    \[\begin{aligned}
    12 t\left(\dfrac{1}{6}+\dfrac{1}{12}\right) &=12 t\left(\dfrac{1}{t}\right) \\
    2 t+t &=12 \\
    3 t &=12 \\
    t &=4
    \end{aligned} \nonumber \]

    When both presses are running it takes 4 hours to do the job.

    Keep in mind, it should take less time for two presses to complete a job working together than for either press to do it alone.

    Example \(\PageIndex{8}\)

    Suppose Pete can paint a room in 10 hours. If he works at a steady pace, in 1 hour he would paint \(\dfrac{1}{10}\) of the room. If Alicia would take 8 hours to paint the same room, then in 1 hour she would paint \(\dfrac{1}{8}\) of the room. How long would it take Pete and Alicia to paint the room if they worked together (and didn’t interfere with each other’s progress)?

    Solution

    This is a ‘work’ application. The steps below will help us organize the information. We are looking for the numbers of hours it will take them to paint the room together.

    In one hour Pete did \(\dfrac{1}{10}\) of the job. Alicia did \(\dfrac{1}{8}\) of the job. And together they did \(\dfrac{1}{t}\) of the job.

    Step 1: Let \(t\) be the number of hours needed to paint the room together.

    Step 2: Enter the hours per job for Pete, Alicia, and when they work together. In 1 hour working together, they have completed \(\dfrac{1}{t}\) of the job. Similarly, find the part of the job completed/hour by Pete and then by Alicia.

    Number of hours to complete the job. Part of job completed/hour
    Pete 10 \(\dfrac{1}{10}\)
    Alicia 8 \(\dfrac{1}{8}\)
    Together \(t\) \(\dfrac{1}{t}\)

    Write a word sentence. The work completed by Pete plus the work completed by Alicia equals the total work completed.

    Step 3: Translate into an equation.

    \[\text {Work completed by}\\ \underbrace{\text {Pete } + \text {Alicia } = \text {Together}}\\ \dfrac{1}{10} \qquad+\qquad\dfrac{1}{8}\qquad =\qquad \dfrac{1}{t} \nonumber \]

    Step 4: Simplify. Solve.

    Multiply by the LCD, \(40t\).

    \[40 t\left(\dfrac{1}{10}+\dfrac{1}{8}\right)=40 t\left(\dfrac{1}{t}\right) \nonumber \]

    Distribute.

    \[40 t \cdot \dfrac{1}{10}+40 t \cdot \dfrac{1}{8}=40 t\left(\dfrac{1}{t}\right) \nonumber \]

    Simplify and solve.

    \[\begin{array}{r}
    {4 t+5 t=40} \\
    {9 t=40} \\
    {t=\dfrac{40}{9}}
    \end{array} \nonumber \]

    We’ll write as a mixed number so that we can convert it to hours and minutes.

    \[t=4 \dfrac{4}{9} \text { hours } \nonumber \]

    Remember, 1 hour = 60 minutes.

    \[t=4 \text { hours }+\dfrac{4}{9}(60 \text { minutes }) \nonumber \]

    Multiply, and then round to the nearest minute.

    \[t=4 \text { hours }+27\text { minutes } \nonumber \]

    It would take Pete and Alica about 4 hours and 27 minutes to paint the room.

    Exercise \(\PageIndex{15}\)

    One gardener can mow a golf course in 4 hours, while another gardener can mow the same golf course in 6 hours. How long would it take if the two gardeners worked together to mow the golf course?

    Answer

    When the two gardeners work together it takes 2 hours and 24 minutes.

    Exercise \(\PageIndex{16}\)

    Daria can weed the garden in 7 hours, while her mother can do it in 3. How long will it take the two of them working together?

    Answer

    When Daria and her mother work together it takes 2 hours and 6 minutes.

    Example \(\PageIndex{9}\)

    Ra’shon can clean the house in 7 hours. When his sister helps him it takes 3 hours. How long does it take his sister when she cleans the house alone?

    Solution

    This is a work problem. The steps below will help us organize the information. We are looking for how many hours it would take Ra’shon’s sister to complete the job by herself.

    Step 1: Let \(s\) be the number of hours Ra’shon’s sister takes to clean the house alone.

    Step 2: Enter the hours per job for Ra’shon, his sister, and when they work together. If Ra’shon takes 7 hours, then in 1 hour \(\dfrac{1}{s}\) of the job is completed. If Ra’shon’s sister takes \(s\) hours, then in 1 hour \(\dfrac{1}{s}\) of the job is completed.

    Number of hours to complete the job. Part of job completed/hour
    Ra’shon 7 \(\dfrac{1}{7}\)
    His sister \(s\) \(\dfrac{1}{s}\)
    Together 3 \(\dfrac{1}{3}\)

    Write a word sentence. The part completed by Ra’shon plus the part by his sister equals the amount completed together.

    Step 3: Translate to an equation.

    \[\text {Work completed by}\\ \underbrace{\text {Ra'shon } + \text {His sister } = \text {Together}}\\ \dfrac{1}{7} \qquad+\qquad\dfrac{1}{s}\qquad =\qquad \dfrac{1}{3} \nonumber \]

    Step 4: Simplify. Solve.

    \[\dfrac{1}{7}+\dfrac{1}{5}=\dfrac{1}{3} \nonumber \]

    Multiply by the LCD, 21s.

    \[\begin{aligned}
    21 s\left(\dfrac{1}{7}+\dfrac{1}{s}\right) &=\left(\dfrac{1}{3}\right) 21 s \\
    3 s+21 &=7 s
    \end{aligned} \nonumber \]

    Simplify.

    \[\begin{aligned}
    -4 s &=-21 \\
    s &=\frac{-21}{-4}=\frac{21}{4}
    \end{aligned} \nonumber \]

    Write as a mixed number to convert it to hours and minutes.

    \[s=5 \dfrac{1}{4} \text { hours } \nonumber \]

    There are 60 minutes in 1 hour.

    \[s=5 \text { hours }+\dfrac{1}{4}(60 \text { minutes })\\ s=5\text { hours }+15\text { minutes } \nonumber \]

    It would take Ra’shon’s sister 5 hours and 15 minutes to clean the house alone.

    Exercise \(\PageIndex{17}\)

    Alice can paint a room in 6 hours. If Kristina helps her it takes them 4 hours to paint the room. How long would it take Kristina to paint the room by herself?

    Answer

    Kristina can paint the room in 12 hours.

    Exercise \(\PageIndex{18}\)

    Tracy can lay a slab of concrete in 3 hours, with Jordan’s help they can do it in 2 hours. If Jordan works alone, how long will it take?

    Answer

    It will take Jordan 6 hours.

    Solve Direct Variation Problems

    When two quantities are related by a proportion, we say they are proportional to each other. Another way to express this relation is to talk about the variation of the two quantities. We will discuss direct variation and inverse variation in this section.

    Lindsay gets paid $15 per hour at her job. If we let \(s\) be her salary and h be the number of hours she has worked, we could model this situation with the equation

    \[s=15 h \nonumber \]

    Lindsay’s salary is the product of a constant, 15, and the number of hours she works. We say that Lindsay’s salary varies directly with the number of hours she works. Two variables vary directly if one is the product of a constant and the other.

    Direct Variation

    For any two variables \(x\) and \(y\), \(y\) varies directly with \(x\) if

    \(y=kx\), where \(k\neq 0\)

    The constant \(k\) is called the constant of variation.

    In applications using direct variation, generally we will know values of one pair of the variables and will be asked to find the equation that relates \(x\) and \(y\). Then we can use that equation to find values of \(y\) for other values of \(x\).

    We’ll list the steps here.

    How to Solve direct variation problems

    Step 1. Write the formula for direct variation.

    Step 2. Substitute the given values for the variables.

    Step 3. Solve for the constant of variation.

    Step 4. Write the equation that relates \(x\) and \(y\) using the constant of variation.

    Now we’ll solve an application of direct variation.

    Example \(\PageIndex{10}\)

    When Raoul runs on the treadmill at the gym, the number of calories, \(c\), he burns varies directly with the number of minutes, \(m\), he uses the treadmill. He burned 315 calories when he used the treadmill for 18 minutes.

    1. Write the equation that relates \(c\) and \(m\).
    2. How many calories would he burn if he ran on the treadmill for 25 minutes?

    Solution

    The number of calories, \(c\), varies directly with the number of minutes, \(m\), on the treadmill, and \(c=315\) when \(m=18\).

    Write the formula for direct variation.

    \[y=kx \nonumber \]

    We will use \(c\) in place of \(y\) and \(m\) in place of \(x\).

    \[c=k m \nonumber \]

    Substitute the given values for the variables.

    \[315=k \cdot 18 \nonumber \]

    Solve for the constant of variation.

    \[\begin{aligned}
    &\dfrac{315}{18}=\dfrac{k \cdot 18}{18}\\
    &17.5=k
    \end{aligned} \nonumber \]

    Write the equation that relates \(c\) and \(m\).

    \[c=k m \nonumber \]

    Substitute in the constant of variation.

    \[c=17.5 m \nonumber \]

    Write the equation that relates \(c\) and \(m\).

    \[c=17.5 m \nonumber \]

    Substitute the given value for \(m\).

    \[c=17.5(25) \nonumber \]

    Simplify.

    \[c=437.5 \nonumber \]

    Raoul would burn 437.5 calories if he used the treadmill for 25 minutes.

    Exercise \(\PageIndex{19}\)

    The number of calories, \(c\), burned varies directly with the amount of time, \(t\), spent exercising. Arnold burned 312 calories in 65 minutes exercising.

    1. Write the equation that relates \(c\) and \(t\).
    2. How many calories would he burn if he exercises for 90 minutes?
    Answer
    1. \(c=4.8 t\)
    2. He would burn 432 calories.

    Exercise \(\PageIndex{20}\)

    The distance a moving body travels, \(d\), varies directly with time, \(t\), it moves. A train travels 100 miles in 2 hours.

    1. Write the equation that relates \(d\) and \(t\).
    2. How many miles would it travel in 5 hours?
    Answer
    1. \(d=50 t\)
    2. It would travel 250 miles.

    Solve Inverse Variation Problems

    Many applications involve two variable that vary inversely. As one variable increases, the other decreases. The equation that relates them is \(y=\dfrac{k}{x}\)

    Inverse Variation

    For any two variables \(x\) and \(y\), \(y\) varies inversely with \(x\) if

    \(y=\dfrac{k}{x}\), where \(k\neq 0\)

    The constant \(k\) is called the constant of variation.

    The word ‘inverse’ in inverse variation refers to the multiplicative inverse. The multiplicative inverse of \(x\) is \(\dfrac{1}{x}\).

    We solve inverse variation problems in the same way we solved direct variation problems. Only the general form of the equation has changed. We will copy the procedure box here and just change ‘direct’ to ‘inverse’.

    how to Solve inverse variation problems

    Step 1. Write the formula for inverse variation.

    Step 2. Substitute the given values for the variables.

    Step 3. Solve for the constant of variation.

    Step 4. Write the equation that relates \(x\) and \(y\) using the constant of variation.

    Example \(\PageIndex{11}\)

    The frequency of a guitar string varies inversely with its length. A 26 in.-long string has a frequency of 440 vibrations per second.

    1. Write the equation of variation.
    2. How many vibrations per second will there be if the string’s length is reduced to 20 inches by putting a finger on a fret?

    Solution

    The frequency varies inversely with the length.

    Name the variables. Let \(f\) = frequency. \(L\) = length

    Write the formula for inverse variation.

    \[y=\dfrac{k}{x} \nonumber \]

    We will use \(f\) in place of \(y\) and \(L\) in place of \(x\).

    \[f=\dfrac{k}{L} \nonumber \]

    \[f=440 \text { when } L=26 \nonumber \]

    Substitute the given values for the variables.

    \[440=\dfrac{k}{26} \nonumber \]

    Solve for the constant of variation.

    \[\begin{aligned}
    &26(440)=26\left(\dfrac{k}{26}\right)\\
    &11,440=k
    \end{aligned} \nonumber \]

    Write the equation that relates \(f\) and \(L\).

    \[f=\dfrac{k}{L} \nonumber \]

    Substitute the constant of variation.

    \[f=\dfrac{11,440}{L} \nonumber \]

    Find \(f\) when \(L=20\).

    Write the equation that relates \(f\) and \(L\).

    \[f=\dfrac{11,440}{L} \nonumber \]

    Substitute the given value forL.

    \[f=\dfrac{11,440}{20} \nonumber \]

    Simplify.

    \[f=572 \nonumber \]

    A 20''-guitar string has frequency 572 vibrations per second.

    Exercise \(\PageIndex{21}\)

    The number of hours it takes for ice to melt varies inversely with the air temperature. Suppose a block of ice melts in 2 hours when the temperature is 65 degrees Celsius.

    1. Write the equation of variation.
    2. How many hours would it take for the same block of ice to melt if the temperature was 78 degrees?
    Answer
    1. \(h=\dfrac{130}{t}\)
    2. \(1 \dfrac{2}{3}\) hours

    Exercise \(\PageIndex{22}\)

    Xander’s new business found that the daily demand for its product was inversely proportional to the price, \(p\). When the price is $5, the demand is 700 units.

    1. Write the equation of variation.
    2. What is the demand if the price is raised to $7?
    Answer
    1. \(x=\dfrac{3500}{p}\)
    2. 500 units

    Media Access Additional Online Resources

    Access this online resource for additional instruction and practice with applications of rational expressions

    • Applications of Rational Expressions