Lesson 7.5: Solve Applications with Rational Equations

New Concept

This lesson applies the power of rational equations to solve real-world problems. You will learn to model and solve applications involving proportions, similar figures, uniform motion, work rates, and both direct and inverse variation.

What’s next

First, we'll tackle proportions. Get ready for guided examples and practice cards to build your skills before moving on to more complex applications.

Property

A proportion is an equation of the form $\frac{a}{b} = \frac{c}{d}$, where $b \neq 0$, $d \neq 0$. The proportion is read “$a$ is to $b$ as $c$ is to $d$.” Since a proportion is an equation with rational expressions, we solve it by multiplying both sides of the equation by the LCD to clear the fractions. This is equivalent to cross-multiplying, which gives the equation $ad = bc$.

Examples

• To solve $\frac{x}{x+12} = \frac{4}{7}$, we cross-multiply to get $7x = 4(x+12)$. This simplifies to $7x = 4x + 48$, so $3x = 48$, and $x = 16$.
• A recipe for 24 cupcakes requires 3 cups of sugar. To make 40 cupcakes, how many cups of sugar, $s$, are needed? The proportion is $\frac{3}{24} = \frac{s}{40}$. This simplifies to $\frac{1}{8} = \frac{s}{40}$. Cross-multiplying gives $40 = 8s$, so $s = 5$ cups of sugar.
• Solve for $y$: $\frac{y-3}{5} = \frac{y}{10}$. Cross-multiply to get $10(y-3) = 5y$. This gives $10y - 30 = 5y$, which means $5y = 30$, and $y = 6$.

Explanation

A proportion shows that two ratios (or fractions) are equal. Think of it as a balanced scale. To find a missing value, you can use cross-multiplication, which is a quick way to clear the denominators and solve the equation.

Property

Two figures are similar if the measures of their corresponding angles are equal and their corresponding sides have the same ratio. For similar triangles $\triangle ABC$ and $\triangle XYZ$, their corresponding angles are equal ($m\angle A = m\angle X$, $m\angle B = m\angle Y$, $m\angle C = m\angle Z$) and the ratio of their corresponding sides is constant:

Examples

• A man who is 6 feet tall casts a 5-foot shadow. At the same time, a nearby building casts a 40-foot shadow. To find the building's height, $h$, set up the proportion $\frac{h}{40} = \frac{6}{5}$. This gives $5h = 240$, so $h = 48$ feet.
• Two similar rectangular gardens are side-by-side. The smaller garden has a width of 4 meters and a length of 6 meters. If the larger garden has a width of 10 meters, find its length, $L$. The proportion is $\frac{L}{6} = \frac{10}{4}$. This gives $4L = 60$, so $L = 15$ meters.
• On a blueprint, 2 inches represent 15 feet. If a wall is 5 inches long on the blueprint, what is its actual length, $x$? Set up the proportion $\frac{x}{5} = \frac{15}{2}$. This gives $2x = 75$, so $x = 37.5$ feet.

Explanation

Similar figures are scaled versions of each other—same shape, different size. Because their corresponding sides are proportional, you can set up a proportion to find an unknown side length, which is useful for measuring heights or distances indirectly.

Property

The formula for uniform motion, $D = rt$, can be solved for time as $t = \frac{D}{r}$. This rational form is used to solve problems where time is the key connection. Typically, you create a table organizing the distance, rate, and time for different parts of a journey. The equation comes from either setting the times equal ($\frac{D_1}{r_1} = \frac{D_2}{r_2}$) or by relating them through a total ($\frac{D_1}{r_1} + \frac{D_2}{r_2} = \text{Total Time}$).

Examples

• A kayak travels 12 miles upstream against a 2-mph current in the same time it takes to travel 18 miles downstream with the current. Let $r$ be the kayak's speed. The equation is $\frac{12}{r-2} = \frac{18}{r+2}$. This gives $12(r+2) = 18(r-2)$, so $12r + 24 = 18r - 36$, which means $6r = 60$ and $r = 10$ mph.
• Maria spent 4 hours on a hike. She walked 9 miles on a flat trail and then 5 miles on a steep trail. Her speed on the flat trail was 2 mph faster than on the steep trail, $s$. The equation is $\frac{5}{s} + \frac{9}{s+2} = 4$. Multiplying by the LCD $s(s+2)$ gives $5(s+2) + 9s = 4s(s+2)$, which simplifies to $4s^2 - 6s - 10 = 0$, or $2s^2 - 3s - 5 = 0$. Factoring gives $(2s-5)(s+1)=0$. Her steep trail speed was $s=2.5$ mph.
• A train travels 200 miles. A car traveling 25 mph slower takes 2 hours longer to cover the same distance. Let $r$ be the train's speed. The car's time is $\frac{200}{r-25}$ and the train's time is $\frac{200}{r}$. The equation is $\frac{200}{r-25} = \frac{200}{r} + 2$. The solution is $r=50$ mph for the train.

Explanation

These problems involve distance, rate, and time. By expressing time as distance divided by rate, you can create rational equations. This helps solve for an unknown speed when travel times are equal or when you know the total duration of a trip.

Property

If a person can complete a job in $T$ hours, their work rate is $\frac{1}{T}$ of the job per hour. When two people work together, their combined rate is the sum of their individual rates. The formula is:

Examples

• Leo can mow a lawn in 2 hours, and his sister Casey can do it in 3 hours. To find how long it takes them together, $t$, solve $\frac{1}{2} + \frac{1}{3} = \frac{1}{t}$. The LCD is $6t$. The equation becomes $3t + 2t = 6$, so $5t = 6$, and $t = 1.2$ hours.
• A hot water faucet can fill a tub in 10 minutes. When both the hot and cold faucets are on, the tub fills in 4 minutes. How long would it take the cold faucet alone, $c$? The equation is $\frac{1}{10} + \frac{1}{c} = \frac{1}{4}$. The LCD is $20c$. This gives $2c + 20 = 5c$, so $3c = 20$, and $c = \frac{20}{3}$ minutes.
• Two printers are working on a large job. The faster printer can finish it in 6 hours. When the slower printer helps, they finish in 4 hours. How long would the slower printer take alone, $s$? The equation is $\frac{1}{6} + \frac{1}{s} = \frac{1}{4}$. The LCD is $12s$. This gives $2s + 12 = 3s$, so $s = 12$ hours.

Explanation

Work problems are about combining rates. By thinking about how much of a job each person can do in one hour, you can add their efforts together to see how quickly they can finish the job as a team.

Property

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.
To solve direct variation problems:

1. Write the formula for direct variation, $y=kx$.
2. Substitute the given values for the variables.
3. Solve for the constant of variation, $k$.
4. Write the equation that relates $x$ and $y$ using the constant of variation.

Examples

• The weekly pay, $P$, for an employee varies directly with the hours worked, $h$. An employee earns 320 dollars for 20 hours of work. Find the equation. $P=kh \rightarrow 320 = k(20) \rightarrow k=16$. The equation is $P=16h$.
• Using the equation $P=16h$, how much would the employee earn for working 35 hours? $P = 16(35) = 560$. The employee would earn 560 dollars.
• The weight of a certain type of wire, $w$, varies directly with its length, $l$. A 15-foot roll of this wire weighs 6 pounds. What is the weight of a 40-foot roll? First, find $k$: $w=kl \rightarrow 6=k(15) \rightarrow k=0.4$. Then, find the new weight: $w = 0.4(40) = 16$ pounds.

Explanation

Direct variation describes a simple relationship where two quantities increase or decrease together at a constant rate. For example, the more you buy of something, the more it costs. The constant $k$ is that fixed rate.

Property

For any two variables $x$ and $y$, $y$ varies inversely with $x$ if $y = \frac{k}{x}$, where $k \neq 0$. The constant $k$ is called the constant of variation.
To solve inverse variation problems:

1. Write the formula for inverse variation, $y = \frac{k}{x}$.
2. Substitute the given values for the variables.
3. Solve for the constant of variation, $k$.
4. Write the equation that relates $x$ and $y$ using the constant of variation.

Examples

• The time, $t$, it takes to drive between two cities varies inversely with speed, $s$. At 60 mph, the trip takes 4 hours. Find the equation. $t = k/s \rightarrow 4 = k/60 \rightarrow k=240$. The equation is $t = 240/s$.
• Using the equation $t = 240/s$, how long would the trip take at a speed of 75 mph? $t = 240/75 = 3.2$ hours.
• The number of people, $N$, needed to complete a project varies inversely with the time, $t$, allowed. If 5 people can do it in 12 days, how many people are needed to do it in 10 days? First, find $k$: $N=k/t \rightarrow 5=k/12 \rightarrow k=60$. Then, find the new number of people: $N = 60/10 = 6$ people.

Explanation

Inverse variation describes a relationship where one quantity goes up as the other goes down. Think of travel time and speed: the faster you drive, the less time a trip takes. They move in opposite directions.