Lesson: February 10, 2025. Quiz: February 13, 2025.

  • Solve rational equations in one variable. Identify extraneous solutions to rational equations and give examples of how they arise.

Assignment

Solving a Rational Equation

The general strategy is to eliminate your denominators so that you are back in polynomial territory.

\[\begin{align} \frac{1}{x+4} &= 2 \\[1em] 1 &= 2(x+4) \\ 0 &= 2x + 7 \\ x &= -\frac{7}{2} \end{align}\]

The catch is that you have to watch out for extraneous solutions. Your original equation will have restrictions on the domain, so any solution that falls outside the domain is not actually a solution, despite appearances.

\[\begin{align} \frac{1}{x-5} + \frac{x}{x-3} &= \frac{2}{x^2-8x+15} \\[1em] \frac{(x-3)+(x^2-5x)}{(x-5)(x-3)} &= \frac{2}{(x-5)(x-3)} \\[1em] x^2 - 4x - 3 &= 2 \\ x^2 - 4x - 5 &= 0 \\ (x-5)(x+1) &= 0 \\ x &=\{-1,5\} \end{align}\]

Based on the original domain, $x\neq5$ since it’s an extraneous solution and we are left only with $x=-1$.

Here’s another example from the book (4B on page 227), but worked out a different way. Pay careful attention to what happens from $(12)$ to $(13)$.

\[\begin{align} \frac{3}{x-3} &= \frac{x}{x-3} - \frac{x}{4} \\[1em] \frac{3}{x-3} - \frac{x}{x-3} &= - \frac{x}{4} \\[1em] \frac{3 - x}{x-3} &= -\frac{x}{4} \\[1em] -\frac{x-3}{x-3} &= -\frac{x}{4} \\[1em] -1 &= -\frac{x}{4} \\[1em] x&=4 \end{align}\]

The way that problem was worked out above didn’t lead to any extraneous solutions, but will if you don’t move your expressions with like denominators to the same side.

Rate Problems

Whoops. Added to the notes late since I forgot about these.

This section also throws work-rate problems at you. These types of problems involve two or more people working are varying speeds to accomplish a single task. You can add rates together to find a combined speed, but setting it up correctly is important.

\[\begin{align} \frac{\text{Work completed}}{\text{Time}} \end{align}\]

Time as the denominator is the key. If you are ever unsure, remember that miles per hour is also a rate and there, hours is in the denominator.

Let’s set up the Try-It from example 2. A pool is fed by two pipes and fills in 12 hours, and one pipe flows three times faster than the other. We don’t know either pipe’s rate, but we can express their combined rate in terms of the just the slower pipe.

\[\begin{align} \frac{x \text{ pool}}{\text{hour}} + \frac{3x\text{ pool}}{\text{hour}} \end{align}\]

We also weren’t given an explicit unit for the pool, like gallons or liters, but you can think of $x$ as a part of the pool (i.e. $x$ amount of the pool is filled after an hour by the slower of the two pipes). Let’s go ahead and compare that to what we know about filling the whole pool.

\[\begin{align} \frac{x \text{ pool}}{\text{hour}} + \frac{3x\text{ pool}}{\text{hour}} = \frac{1 \text{ pool}}{12 \text{ hours}} \end{align}\]

Now we have something we can solve.

\[\begin{align} \frac{x \text{ pool}}{\text{hour}} + \frac{3x\text{ pool}}{\text{hour}} &= \frac{1 \text{ pool}}{12 \text{ hours}} \\[1em] \frac{4x \text{ pool}}{1 \text{ hour}} &= \frac{1 \text{ pool}}{12 \text{ hours}} \\[1em] 48x \text{ pool hours} &= 1 \text{ pool hours} \\[0.5em] x &= \frac{1}{48} \end{align}\]

Be careful with your solution and make sure you interpret it correctly. I never presented the question asked, but it was looking for how long it would take the slower pipe to fill the pool by itself. What we found was that the slower pipe could fill $1/48$ of the pool in an hour, meaning it would take $48$ hours to fill the whole thing.

Extra Example

I’m going to redo the book’s example as well, since

Arthur and Cheyenne paint a wall in 6 hours, with Cheyenne working twice as fast as Arthur. Here’s the setup, with $a$ representing the portion of the wall painted by Arthur. Cheyenne’s portion is expressed in terms of Arthur’s contribution to the project.

\[\begin{align} \frac{a \text{ wall}}{1 \text{ hour}} + \frac{2a \text{ wall}}{1 \text{ hour}} = \frac{1 \text{ wall}}{6 \text{ hours}} \end{align}\]

I’ll do this one a bit different just to show off other strategies. This time I’ll get a common denominator first, saving us some fraction work.

\[\begin{align} \frac{a \text{ wall}}{1 \text{ hour}} + \frac{2a \text{ wall}}{1 \text{ hour}} &= \frac{1 \text{ wall}}{6 \text{ hours}} \\[1em] \frac{6a \text{ wall}}{6 \text{ hours}} + \frac{12a \text{ wall}}{6 \text{ hours}} &= \frac{1 \text{ wall}}{6 \text{ hours}} \\[1em] 18a \text{ wall} &= 1 \text{ wall} \\[0.5em] a &= \frac{1}{18} \end{align}\]

Arthur can paint $1/18$ of the wall in an hour, meaning it would take him 18 hours to do the whole thing himself. Cheyenne on the other hand, can do $2/18$ or $1/9$, so it would only take her 9 hours by herself.

But the Book Does it Different

I wrote this whole thing up based on how I worked out the problems, realizing after the fact that the book did it from the point of view of the hours. Variables were used for the time it took to accomplish the job, rather than how much was accomplished in an hour. Arthur can paint a wall in $x$ hours, while Cheyenne could do two in the same amount of time.

\[\begin{align} \frac{1 \text{ wall}}{x \text{ hour}} + \frac{2 \text{ walls}}{x \text{ hour}} &= \frac{1 \text{ wall}}{6 \text{ hours}} \\[1em] \frac{3 \text{ walls}}{x \text{ hours}} &= \frac{1 \text{ wall}}{6 \text{ hours}} \\[1em] x &= 18 \end{align}\]

You can pick your poison here as it only matters which version makes the most sense to you. I like not having variables in the denominator and unit rates. On the other hand, most of the questions focus on finding time spent, so you might prefer the more direct approach.