7.5 Solve Applications with Rational Equations

Solving Applications with Rational Equations

Rational equations show up whenever a real-world problem involves ratios, rates, or quantities that vary together. Setting up these equations correctly is often the hardest part, so this section focuses on recognizing problem types, translating words into equations, and checking that your answers actually make sense.

Proportions in Real-World Contexts

A proportion states that two ratios are equal. You'll use these whenever two quantities have a consistent relationship, like cost per item or miles per gallon.

How to solve a proportion problem:

1. Identify the two quantities being compared and their units
2. Set up two ratios with matching units: same units in both numerators, same units in both denominators
3. Cross-multiply and solve the resulting equation for the unknown variable
4. Check that your answer is reasonable in context (a negative cost or 500 mpg should raise a red flag)

Example: If 5 pounds of apples cost $8.50, how much do 12 pounds cost?

$\frac{8.50}{5} = \frac{x}{12}$

Cross-multiply: $8.50 \times 12 = 5x$, so $102 = 5x$, giving $x = 20.40$. That's $20.40 for 12 pounds, which seems reasonable.

Rational Equations for Similar Figures

Similar figures have the same shape but different sizes. Their corresponding sides are proportional, meaning the ratios of matching sides are all equal.

1. Match up corresponding sides between the two figures
2. Write a proportion using one pair of known corresponding sides and one pair that includes the unknown
3. Cross-multiply and solve for the unknown side length
4. Verify your answer is positive and consistent with the scale between the figures

Example: Two similar triangles have sides where the first triangle's base is 6 and the second triangle's base is 15. If the first triangle has a height of 4, the second triangle's height $h$ satisfies:

$\frac{6}{15} = \frac{4}{h}$

Cross-multiplying gives $6h = 60$, so $h = 10$.

Uniform Motion with Rational Equations

Motion problems use the relationship $d = rt$ (distance = rate × time), which you can rearrange to $r = \frac{d}{t}$ or $t = \frac{d}{r}$ depending on what's unknown.

These problems often involve two legs of a trip (going and returning, or two travelers). The key is writing a separate expression for each leg, then connecting them with whatever relationship the problem gives you.

Steps:

1. Identify what's given and what's unknown for each leg (distance, rate, time)
2. Write expressions using $t = \frac{d}{r}$ for each leg
3. Use the connecting relationship (total time, same time, etc.) to build one equation
4. Solve for the unknown, then find any remaining quantities

Example: A boat travels 30 miles upstream and 30 miles downstream. The current is 5 mph, and the total trip takes 4 hours. What's the boat's speed in still water?

Let $b$ = boat speed in still water. Upstream rate is $b - 5$, downstream rate is $b + 5$.

$\frac{30}{b - 5} + \frac{30}{b + 5} = 4$

Multiply both sides by $(b - 5)(b + 5)$:

$30(b + 5) + 30(b - 5) = 4(b^2 - 25)$

$60b = 4b^2 - 100$

$4b^2 - 60b - 100 = 0 \implies b^2 - 15b - 25 = 0$

Use the quadratic formula to solve, then discard any negative answer since speed must be positive.

Work-Rate Problems and Rational Expressions

Work-rate problems ask how long it takes people or machines to complete a task together. The core idea: if someone finishes a job in $t$ hours, their rate is $\frac{1}{t}$ of the job per hour.

The combined work formula:

$\frac{1}{t_1} + \frac{1}{t_2} = \frac{1}{t_{\text{together}}}$

where $t_1$ and $t_2$ are individual completion times and $t_{\text{together}}$ is the combined time.

Steps:

1. Find each worker's rate as a fraction of the job per unit time
2. Add the rates and set equal to $\frac{1}{t_{\text{together}}}$
3. Solve for the unknown time
4. Check: the combined time should always be less than the smallest individual time

Example: Pump A fills a tank in 6 hours. Pump B fills it in 4 hours. Together:

$\frac{1}{6} + \frac{1}{4} = \frac{1}{t}$

$\frac{5}{12} = \frac{1}{t}$, so $t = \frac{12}{5} = 2.4$ hours. That's less than 4 hours, so it checks out.

Direct Variation in Practical Situations

Two quantities vary directly when one is a constant multiple of the other. As one increases, the other increases proportionally.

Formula: $y = kx$, where $k$ is the constant of variation

Steps:

1. Confirm the relationship is direct variation (doubling one quantity doubles the other)
2. Use a known pair of values to find $k$: plug in $x$ and $y$, then solve $k = \frac{y}{x}$
3. Use your equation $y = kx$ with the new value to solve for the unknown

Example: The cost of gasoline varies directly with the number of gallons purchased. If 8 gallons cost $28, how much do 14 gallons cost?

Find $k$: $k = \frac{28}{8} = 3.50$. So $y = 3.50x$. For 14 gallons: $y = 3.50(14) = 49$. The cost is $49.

Inverse Variation from Given Data

Two quantities vary inversely when their product is constant. As one increases, the other decreases.

Formula: $xy = k$, or equivalently $y = \frac{k}{x}$

Steps:

1. Confirm the relationship is inverse variation (doubling one quantity halves the other)
2. Use a known pair of values to find $k$: multiply $x \times y = k$
3. Substitute the new value and solve for the unknown

Example: Boyle's Law says gas pressure and volume vary inversely at constant temperature. If a gas has pressure 150 kPa at volume 4 L, what's the pressure at 10 L?

Find $k$: $150 \times 4 = 600$. So $P \times V = 600$. At 10 L: $P = \frac{600}{10} = 60$ kPa.

Interpreting and Checking Solutions

Always interpret your answer in the context of the original problem. A mathematically correct solution can still be wrong if it doesn't make sense for the situation.

1. Verify units — Does your answer have the right units (dollars, hours, miles)?
2. Check the range — Is the value realistic? Negative time or a speed of 3,000 mph should signal an error.
3. Substitute back — Plug your solution into the original equation to confirm it works. This also catches extraneous solutions introduced when you multiplied both sides by a variable expression.

Problem-Solving Strategies for Rational Equations

• Identify the problem type first. Is it a proportion, motion, work-rate, or variation problem? This tells you which formula or setup to use.
• Define your variable clearly. Write down what $x$ (or whatever letter you choose) represents, including units.
• Translate carefully. Convert the word problem into algebraic fractions one phrase at a time. Rushing this step causes most errors.
• Convert units if needed. If one quantity is in minutes and another in hours, make them match before setting up the equation.
• Follow the four-step cycle: understand the problem, plan your equation, solve it, then check your answer.