Algebra II/Exponents & Rational Functions/Modeling with Rational Equations

Algebra II Regents in 15 days

Algebra II · Lesson 1

Modeling with Rational Equations

Setting up and solving equations where the unknown lives in a denominator.

A garden hose fills a pool in 12 hours. A wider hose fills the same pool in 8 hours. A homeowner turns on both hoses at the same time and wants to know how long it takes together. That question cannot be answered with a linear equation. The unknowns end up in denominators, and that is exactly what rational equations are built for.

The key idea is rates. A rate describes how much of a job gets done per unit of time. If one hose fills the pool in 12 hours, it completes $\frac$ of the job each hour. If the other fills it in 8 hours, it completes $\frac$ of the job each hour. When both run simultaneously, their rates add. Let $t$ be the number of hours to fill the pool together. In $t$ hours, the first hose does $\frac$ of the job and the second does $\frac$ . Together they complete exactly one whole job:

\frac, + \frac, = 1

This is a rational equation. To solve it, multiply every term by the least common denominator, which clears all fractions and leaves a linear equation you can solve directly.

Two hoses filling a pool

\frac{t}{12} + \frac{t}{8} = 1

Each fraction represents the portion of the job done by one hose in t hours.

24 \cdot \frac{t}{12} + 24 \cdot \frac{t}{8} = 24 \cdot 1

The LCD of 12 and 8 is 24. Multiply every term by 24 to clear the denominators.

2t + 3t = 24

24 divided by 12 is 2, and 24 divided by 8 is 3.

5t = 24

Combine like terms on the left.

t = \frac{24}{5} = 4.8 \text{ hours}

Divide both sides by 5. Together the hoses take 4 hours and 48 minutes — less than either alone, which makes sense.

Work problems always follow this structure. If person A completes a job in $a$ hours and person B completes the same job in $b$ hours, then working together for $t$ hours:

\frac, + \frac, = 1

Some problems give you the combined time and ask for one of the individual times. The setup is the same — the algebra just looks different because the unknown is in a denominator from the start.

Finding an unknown individual rate

\frac{1}{6} + \frac{1}{x} = \frac{1}{4}

Printer A takes 6 hours alone. Together they take 4 hours. This asks how long Printer B takes alone. Each fraction is a rate: jobs per hour.

12x \cdot \frac{1}{6} + 12x \cdot \frac{1}{x} = 12x \cdot \frac{1}{4}

The LCD of 6, x, and 4 is 12x. Multiply every term by 12x.

2x + 12 = 3x

12x divided by 6 is 2x, 12x divided by x is 12, 12x divided by 4 is 3x.

12 = x

Subtract 2x from both sides. Printer B takes 12 hours alone.

x = 12 \checkmark

Check: 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4. The combined rate is 1/4 job per hour, so together they finish in 4 hours. Correct.

Rational equations also model situations where two objects travel at different speeds and the unknown is in the denominator because time equals distance divided by rate. A boat travels 18 miles upstream and 18 miles back. The current flows at 3 mph. The whole trip takes 4 hours. The boat's still-water speed $r$ satisfies:

\frac, + \frac, = 4

Upstream the effective speed is $r - 3$ , downstream it is $r + 3$ , and time equals distance over speed. Setting up that equation correctly is half the problem. The other half is solving it.

Boat traveling upstream and downstream

\frac{18}{r-3} + \frac{18}{r+3} = 4

Distance divided by speed gives time. The upstream and downstream times must add up to 4 hours.

(r-3)(r+3) \cdot \frac{18}{r-3} + (r-3)(r+3) \cdot \frac{18}{r+3} = 4(r-3)(r+3)

Multiply every term by (r-3)(r+3), the LCD. This cancels one factor from each fraction's denominator.

18(r+3) + 18(r-3) = 4(r^2 - 9)

Each denominator cancels with its matching factor. The right side uses the difference of squares pattern.

18r + 54 + 18r - 54 = 4r^2 - 36

Distribute on both sides.

36r = 4r^2 - 36

The +54 and -54 cancel. Combine the r terms on the left.

0 = 4r^2 - 36r - 36

Move everything to one side by subtracting 36r from both sides.

0 = r^2 - 9r - 9

Divide every term by 4 to simplify before using the quadratic formula.

r = \frac{9 \pm \sqrt{81 + 36}}{2} = \frac{9 \pm \sqrt{117}}{2}

Apply the quadratic formula with a=1, b=-9, c=-9.

r \approx \frac{9 + 10.82}{2} \approx 9.91 \text{ mph}

Take the positive root only. A negative speed has no physical meaning, and r must be greater than 3 or the upstream speed would be zero or negative.

Interactive graph — scroll to zoom, drag to pan

The graph shows the total trip time as a function of the boat's still-water speed. The horizontal line at $y = 4$ marks the 4-hour target. Their intersection at roughly $r \approx 9.91$ is the answer from the algebra. Notice how the curve approaches the vertical asymptote at $r = 3$ — as the boat's speed gets closer to the current's speed, the upstream leg takes longer and longer until the trip becomes impossible.

Practice Questions

\frac{t}{10} + \frac{t}{15} = 1

\frac{1}{x} + \frac{1}{x+2} = \frac{5}{12}

\frac{20}{r+4} + \frac{20}{r-4} = 3

Regents Corner

On Part II and Part III of the Algebra II Regents, modeling problems with rational equations are worth 2 to 4 credits. Full credit requires setting up the equation correctly, showing all algebraic steps, and stating your answer with appropriate units. A numerical answer with no supporting work earns at most one credit even if it is correct.

After multiplying through by the LCD and solving a quadratic, students find two solutions and report both without checking them. One solution often makes a denominator equal to zero or produces a negative time or speed. That solution must be rejected explicitly in your work. Writing "x = 4 or x = -6/5" with no further comment will cost you a point. Write "x = -6/5 is rejected because time cannot be negative" and circle x = 4.

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