Solve Applications with Rational Equations

By the end of this section, you will be able to: solve proportions, solve similar-figure applications, solve uniform-motion applications, solve work applications, solve direct-variation problems, and solve inverse-variation problems.

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

\tfrac{a}{b}=\tfrac{c}{d}

, where

b\ne0

and

d\ne0

. It is read “

a

is to

b

c

is to

d

.”

The equation $\tfrac{1}{2}=\tfrac{4}{8}$ is a proportion because the two fractions are equal. Since a proportion is an equation with rational expressions, we solve it as we solved rational equations: multiply both sides by the LCD to clear the fractions, then solve the resulting equation.

Example. Solve $\tfrac{n}{n+14}=\tfrac{5}{7}$ .

The restriction is $n\ne-14$ .

\begin{array}{lrcl} \text{Multiply both sides by the LCD.} & 7(n+14)\left(\tfrac{n}{n+14}\right) &=& 7(n+14)\left(\tfrac{5}{7}\right) \\[10pt] \text{Remove common factors.} & 7n &=& 5(n+14) \\[4pt] \text{Distribute.} & 7n &=& 5n+70 \\[4pt] \text{Solve.} & 2n &=& 70 \\[4pt] && n=& 35 \end{array}

Checking gives $\tfrac{35}{35+14}=\tfrac{35}{49}=\tfrac{5}{7}$ .

Solve the proportion $\tfrac{y}{y+55}=\tfrac{3}{8}$ .

Solve the proportion $\tfrac{z}{z-84}=-\tfrac{1}{5}$ .

Clearing the fractions in $\tfrac{a}{b}=\tfrac{c}{d}$ gives $ad=bc$ , the same result as cross-multiplying.

When solving applications with proportions, we use our usual application strategy. The units in the numerators must match one another, and the units in the denominators must match one another.

Example. Pediatricians prescribe 5 milliliters (ml) of acetaminophen for every 25 pounds of a child’s weight. If Zoe weighs 80 pounds, how many milliliters will her doctor prescribe?

Let $a$ be the milliliters of acetaminophen. Keep milliliters in the numerators and pounds in the denominators:

\begin{array}{lrcl} \text{Translate into a proportion.} & \tfrac{5}{25} &=& \tfrac{a}{80} \\[10pt] \text{Multiply by the LCD, }400. & 400\left(\tfrac{5}{25}\right) &=& 400\left(\tfrac{a}{80}\right) \\[10pt] \text{Simplify.} & 16\cdot5 &=& 5a \\[4pt] \text{Solve.} & 16 &=& a \end{array}

Since 80 is about three times 25, about three times as much medicine is reasonable. Substitution gives $\tfrac{5}{25}=\tfrac{16}{80}=\tfrac{1}{5}$ . The pediatrician would prescribe 16 ml of acetaminophen to Zoe.

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

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

Solve similar-figure applications

When you shrink or enlarge a photo, find a distance on a map, or use a pattern to build or sew something, you work with similar figures. Similar figures have exactly the same shape but different sizes; one is a scale model of the other.

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, a triangle with side lengths 12, 16, and 20 is similar to a triangle with corresponding side lengths 3, 4, and 5. Each side of the larger triangle is four times the corresponding side of the smaller triangle:

\frac{16}{4}=\frac{20}{5}=\frac{12}{3}.

Property of Similar Triangles. If $\triangle ABC$ is similar to $\triangle XYZ$ , corresponding angles have equal measures and corresponding sides have the same ratio. If the corresponding sides are $a,x$ ; $b,y$ ; and $c,z$ , then

\frac{a}{x}=\frac{b}{y}=\frac{c}{z}.

Example. On a map, San Francisco, Las Vegas, and Los Angeles form a triangle. The map distances are 2.1 inches from San Francisco to Las Vegas, 1.3 inches from San Francisco to Los Angeles, and 1 inch from Los Angeles to Las Vegas. The actual distance from Los Angeles to Las Vegas is 270 miles. Find the actual distance from Los Angeles to San Francisco.

Let $x$ be the distance from Los Angeles to San Francisco. Keep miles in the numerators and inches in the denominators:

\begin{array}{lrcl} \text{Write a proportion of corresponding sides.} & \tfrac{x}{1.3} &=& \tfrac{270}{1} \\[10pt] \text{Multiply by }1.3. & x &=& 351 \end{array}

On the map, Los Angeles to San Francisco is longer than Los Angeles to Las Vegas; 351 miles is correspondingly more than 270 miles. The actual distance is 351 miles.

For the next two checks, a map shows Seattle, Portland, and Boise forming a triangle. The map distances are 1.5 inches from Seattle to Portland, 3.5 inches from Portland to Boise, and 4 inches from Seattle to Boise. The actual distance from Seattle to Boise is 400 miles.

Using the Seattle–Portland–Boise map measurements above, find the actual distance in miles from Seattle to Portland.

Using the Seattle–Portland–Boise map measurements above, find the actual distance in miles from Portland to Boise.

Similar figures can also find heights we cannot directly measure.

Example. Tyler is 6 feet tall, and his shadow is 8 feet long. At the same time, a tree’s shadow is 24 feet long. Find the tree’s height.

The person and tree form similar right triangles with their shadows. Let $h$ be the height of the tree.

\begin{array}{lrcl} \text{Write a proportion of height to shadow.} & \tfrac{h}{24} &=& \tfrac{6}{8} \\[10pt] \text{Multiply by }24. & h &=& 18 \end{array}

Tyler’s height is less than his shadow, so it makes sense that the tree’s height is less than its shadow. The tree is 18 feet tall.

A telephone pole casts a 50-foot shadow while an 8-foot traffic sign casts a 10-foot shadow. How tall is the telephone pole, in feet?

A pine tree casts an 80-foot shadow while a 30-foot building casts a 40-foot shadow. How tall is the pine tree, in feet?

Solve uniform-motion applications

Uniform motion uses $D=rt$ . If distance $D$ and rate $r$ are known and time $t$ is needed, solve for time: $t=\tfrac{D}{r}$ . A table organizes rate, time, and distance.

Example. An airplane flies 200 miles into a 30-mph headwind in the same time it flies 300 miles with a 30-mph tailwind. Find the airplane’s speed.

Let $r$ be the airplane’s speed. The headwind rate is $r-30$ and the tailwind rate is $r+30$ .

Trip	Rate	Time	Distance
Headwind	$r-30$	$\tfrac{200}{r-30}$	200
Tailwind	$r+30$	$\tfrac{300}{r+30}$	300

The times are equal:

\begin{array}{lrcl} \text{Write the time equation.} & \tfrac{200}{r-30} &=& \tfrac{300}{r+30} \\[10pt] \text{Clear the denominators.} & 200(r+30) &=& 300(r-30) \\[4pt] \text{Distribute.} & 200r+6{,}000 &=& 300r-9{,}000 \\[4pt] \text{Solve.} & 15{,}000 &=& 100r \\[4pt] && r=&150 \end{array}

With the tailwind the rate is 180 mph and time is $\tfrac{300}{180}=\tfrac53$ hours. Against the wind the rate is 120 mph and time is $\tfrac{200}{120}=\tfrac53$ hours. The plane’s speed is 150 mph.

Link rides 20 miles into a 3-mph headwind in the same time he rides 30 miles with a 3-mph tailwind. What is his biking speed in mph?

A river flows at 7 mph. Danica motors 5 miles upstream in the same time she motors 12 miles downstream. What is her boat’s speed in still water, in mph?

Example. Jazmine trained for 3 hours. She ran 8 miles and then biked 24 miles. Her biking speed was 4 mph faster than her running speed. Find her running speed.

Let $r$ be her running speed, so $r+4$ is her biking speed.

Activity	Rate	Time	Distance
Run	$r$	$\tfrac{8}{r}$	8
Bike	$r+4$	$\tfrac{24}{r+4}$	24

Her running time plus biking time is 3 hours:

\begin{array}{lrcl} \text{Write the equation.} & \tfrac{8}{r}+\tfrac{24}{r+4} &=& 3 \\[10pt] \text{Multiply both sides by the LCD.} & r(r+4)\left(\tfrac{8}{r}+\tfrac{24}{r+4}\right) &=& 3r(r+4) \\[10pt] \text{Distribute and remove common factors.} & 8(r+4)+24r &=& 3r(r+4) \\[4pt] \text{Distribute.} & 8r+32+24r &=& 3r^2+12r \\[4pt] \text{Collect like terms.} & 32+32r &=& 3r^2+12r \\[4pt] \text{Write in standard form.} & 0 &=& 3r^2-20r-32 \\[4pt] \text{Factor.} & 0 &=& (3r+4)(r-8) \\[4pt] \text{Use the Zero Product Property.} & 3r+4=0 &\text{or}& r-8=0 \\[4pt] \text{Solve.} & r=-\tfrac43 &\text{or}& r=8 \end{array}

A negative speed does not make sense, so $r=8$ . Running 8 miles at 8 mph takes 1 hour; biking 24 miles at 12 mph takes 2 hours. Jazmine’s running speed is 8 mph.

Dennis skied 20 miles uphill and 20 miles downhill in 6 hours. His uphill speed was 5 mph slower than his downhill speed. Enter his uphill and downhill speeds, separated by commas.

Joon drove for 4 hours: 208 miles on the interstate and 40 miles on country roads. His interstate speed was 15 mph faster. What was his country-road speed in mph?

Example. Hamilton biked downhill 12 miles to the ocean and uphill 12 miles home. His uphill speed was 8 mph slower, and the return took 2 hours longer. Find his downhill speed.

Let $h$ be the downhill speed, so $h-8$ is the uphill speed.

Trip	Rate	Time	Distance
Downhill	$h$	$\tfrac{12}{h}$	12
Uphill	$h-8$	$\tfrac{12}{h-8}$	12

The uphill time is 2 hours more than the downhill time:

\begin{array}{lrcl} \text{Write the equation.} & \tfrac{12}{h-8} &=& \tfrac{12}{h}+2 \\[10pt] \text{Multiply both sides by the LCD.} & h(h-8)\left(\tfrac{12}{h-8}\right) &=& h(h-8)\left(\tfrac{12}{h}+2\right) \\[10pt] \text{Distribute and remove common factors.} & 12h &=& 12(h-8)+2h(h-8) \\[4pt] \text{Distribute.} & 12h &=& 12h-96+2h^2-16h \\[4pt] \text{Write in standard form.} & 0 &=& 2h^2-16h-96 \\[4pt] \text{Factor out the GCF.} & 0 &=& 2(h^2-8h-48) \\[4pt] \text{Factor the trinomial.} & 0 &=& 2(h-12)(h+4) \\[4pt] \text{Use the Zero Product Property.} & h-12=0 &\text{or}& h+4=0 \\[4pt] \text{Solve.} & h=12 &\text{or}& h=-4 \end{array}

At 12 mph downhill, the trip takes 1 hour; at 4 mph uphill, it takes 3 hours. Hamilton’s downhill speed is 12 mph.

Kayla biked 75 miles home and took a bus back. The bus trip took 2 hours less and the bus was 10 mph faster. Find Kayla’s biking speed in mph.

Victoria jogs 12 miles on a flat trail and returns on a 20-mile hilly trail. She is 1 mph slower on the hilly trail, and the return takes 2 hours longer. Find her flat-trail speed in mph.

Solve work applications

Suppose Press 1 completes a job in 6 hours and Press 2 completes it in 12 hours. In one hour they complete $\tfrac16$ and $\tfrac1{12}$ of the job. If $t$ is the time together, they complete $\tfrac1t$ of the job per hour.

Worker	Hours to complete job	Part completed per hour
Press 1	6	$\tfrac16$
Press 2	12	$\tfrac1{12}$
Together	$t$	$\tfrac1t$

The part completed by Press 1 plus the part completed by Press 2 equals the part completed together:

\begin{array}{lrcl} \text{Write the equation.} & \tfrac16+\tfrac1{12} &=& \tfrac1t \\[10pt] \text{Multiply by the LCD, }12t. & 2t+t &=& 12 \\[4pt] \text{Solve.} & t &=& 4 \end{array}

Both presses take 4 hours, less than either press working alone.

Example. Pete can paint a room in 10 hours and Alicia can paint it in 8 hours. How long will they take together?

Worker	Hours to complete job	Part completed per hour
Pete	10	$\tfrac1{10}$
Alicia	8	$\tfrac18$
Together	$t$	$\tfrac1t$

\begin{array}{lrcl} \text{Add their hourly work rates.} & \tfrac1{10}+\tfrac18 &=& \tfrac1t \\[10pt] \text{Multiply by }40t. & 4t+5t &=& 40 \\[4pt] \text{Solve.} & t &=& \tfrac{40}{9}=4\tfrac49 \end{array}

Since $\tfrac49(60)\approx27$ , it would take about 4 hours and 27 minutes.

One gardener mows a golf course in 4 hours and another in 6 hours. How many hours will they take together?

Daria weeds a garden in 7 hours and her mother in 3 hours. How many hours will they take together?

Example. Ra’shon can clean a house in 7 hours. With his sister, the job takes 3 hours. How long does his sister take alone?

Worker	Hours to complete job	Part completed per hour
Ra’shon	7	$\tfrac17$
Sister	$s$	$\tfrac1s$
Together	3	$\tfrac13$

\begin{array}{lrcl} \text{Write the work equation.} & \tfrac17+\tfrac1s &=& \tfrac13 \\[10pt] \text{Multiply by }21s. & 3s+21 &=& 7s \\[4pt] \text{Solve.} & s &=& \tfrac{21}{4}=5\tfrac14 \end{array}

One quarter of an hour is 15 minutes, so Ra’shon’s sister takes 5 hours and 15 minutes alone.

Alice paints a room in 6 hours. With Kristina, it takes 4 hours. How many hours would Kristina take alone?

Tracy lays a slab in 3 hours. With Jordan, it takes 2 hours. How many hours would Jordan take alone?

Solve direct-variation problems

When two quantities are related by a proportion, they are proportional. If Lindsay earns $15 per hour, her salary $s$ after $h$ hours is $s=15h$ . Her salary varies directly with the hours worked.

Direct variation. For variables $x$ and $y$ , $y$ varies directly with $x$ if

y=kx,\qquad k\ne0.

The constant $k$ is the constant of variation.

Solve direct-variation problems.

Write the formula for direct variation.
Substitute the given values for the variables.
Solve for the constant of variation.
Write the equation relating $x$ and $y$ using the constant of variation.

Example. The calories $c$ Raoul burns on a treadmill vary directly with the minutes $m$ . He burned 315 calories in 18 minutes. (a) Write the equation relating $c$ and $m$ . (b) How many calories would he burn in 25 minutes?

\begin{array}{lrcl} \text{Use }c\text{ in place of }y\text{ and }m\text{ in place of }x. & c &=& km \\[4pt] \text{Substitute the known pair.} & 315 &=& 18k \\[4pt] \text{Find the constant.} & k &=& 17.5 \\[4pt] \text{Write the variation equation.} & c &=& 17.5m \\[4pt] \text{Evaluate at }m=25. & c &=& 17.5(25)=437.5 \end{array}

Raoul would burn 437.5 calories in 25 minutes.

Calories $c$ vary directly with exercise time $t$ . Arnold burned 312 calories in 65 minutes. Enter the equation relating $c$ and $t$ as an expression for $c$ .

Using Arnold’s equation from the previous check, how many calories would he burn in 90 minutes?

Distance $d$ varies directly with time $t$ . A train travels 100 miles in 2 hours. Enter the equation relating $d$ and $t$ as an expression for $d$ .

Solve inverse-variation problems

Many applications have variables that vary inversely: as one increases, the other decreases.

Inverse variation. For variables $x$ and $y$ , $y$ varies inversely with $x$ if

y=\frac{k}{x},\qquad k\ne0.

The constant $k$ is the constant of variation.

The word inverse refers to the multiplicative inverse; the multiplicative inverse of $x$ is $\tfrac1x$ .

Solve inverse-variation problems.

Write the formula for inverse variation.
Substitute the given values for the variables.
Solve for the constant of variation.
Write the equation relating $x$ and $y$ using the constant of variation.

Example. A guitar string’s frequency varies inversely with its length. A 26-inch string has a frequency of 440 vibrations per second. (a) Write the equation of variation. (b) Find the frequency when its length is 20 inches.

Let $f$ be frequency and $L$ be length.

\begin{array}{lrcl} \text{Write the inverse-variation formula.} & f &=& \tfrac{k}{L} \\[10pt] \text{Substitute the known pair.} & 440 &=& \tfrac{k}{26} \\[10pt] \text{Find the constant.} & k &=& 11{,}440 \\[4pt] \text{Write the variation equation.} & f &=& \tfrac{11{,}440}{L} \\[10pt] \text{Evaluate at }L=20. & f &=& \tfrac{11{,}440}{20}=572 \end{array}

A 20-inch guitar string has a frequency of 572 vibrations per second.

The hours $h$ for ice to melt vary inversely with temperature $t$ . Ice melts in 2 hours at 65°C. Enter the equation relating $h$ and $t$ as an expression for $h$ .

Using the ice-melting equation above, how many hours will the same ice take to melt at 78°C?

Daily demand $x$ varies inversely with price $p$ . At a price of $5, demand is 700 units. Enter the equation relating $x$ and $p$ as an expression for $x$ .

Key terms

proportion — an equation stating that two ratios are equal. similar figures — figures whose corresponding angles are equal and whose corresponding sides have the same ratio. direct variation — a relationship of the form $y=kx$ . constant of variation — the constant $k$ in a direct- or inverse-variation equation. inverse variation — a relationship of the form $y=\tfrac{k}{x}$ .

This section is adapted from Intermediate Algebra 2e, Section 7.5: Solve Applications with Rational Equations by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the source’s application charts as accessible tables and described its map, shadow, and motion diagrams in text while preserving all measurements; omitted the Be Prepared quiz, media link, and end-of-section exercises; and converted the practice problems (“Try Its”) into interactive exercises with instant feedback.

Solve Rational Equations Solve Rational Inequalities