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Word Problems Involving Rational Equations

Example $1$:

Seven divided by the sum of a number and two is equal to half the difference of the number and three. Find all such numbers.

Let's begin by defining our variable. $$\text{Let }x=\text{the number}$$ Now we can set up our equation. Seven divided by the sum of a number and two $$\frac{7}{x+2}$$ is equal to $$\frac{7}{x+2}=$$ half the difference of the number and three. $$\frac{7}{x+2}=\frac{x-3}{2}$$ Before we solve, take note that there's a variable in the denominator which means we will have a restricted value. So, $x\ne-2$ because that would make the denominator zero which would make the expression undefined. Now we can cross multiply and solve! $$\begin{align} \color{royalblue}{\frac{7}{x+2}}&=\color{magenta}{\frac{x-3}{2}}\\ \color{royalblue}{7}\cdot\color{magenta}{2}&=\color{royalblue}{(x+2)}\color{magenta}{(x-3)}\\ 14&=x^2-x-6\\ \color{#8925AE}{-14}&=\ \ \ \ \ \ \ \ \ \ \color{#8925AE}{-14}\\ \hline 0&=x^2-x-20\\ 0&=(x-5)(x+4)\\ x=5&\text{ o}\text{r }x=-4 \end{align}$$ None of the solutions are our restricted value so we know both these will work. Finally we state our answer. $$\text{The numbers are }5\text{ and}-4\text{.}$$

Distance, Rate, and Time Word Problems

Let's talk about how to solve word problems involving distance, rate, and time. These problems can be tricky, but here is a good guide on how to solve them.

Example:

A boat that can travel fifteen miles per hour in still water can travel thirty-six miles downstream in the same amount of time that it can travel twenty-four miles upstream. Find the speed of the current in the river.

The general formula we use for these problems is:

Overall Rate $=$ speed of the vehicle (with no current/wind) $\pm$ speed of the water or wind

Since these are distance, rate, and time problems, we will be using the distance formula as well: $$\text{Distance}=\text{Rate}\times\text{Time}\\ d=rt$$

Here's the skeleton of the table we make to help solve these problems.

Once we make the outline of our table we begin by filling in the information that was given to us. The problem directly tells us that you travel thirty-six miles downstream and twenty-four miles upstream.

Next we're given that the boat can travel fifteen miles per hour in still water so we have a rate of $15\text{ mph}$. But this is in still water, not going upstream or downstream. So we need to use our formula given above for overall rate.

Overall Rate $=$ speed of the vehicle (with no current/wind) $\pm$ speed of the water or wind

So we know the speed of the vehicle with no current or wind. That's the $15\text{ mph}$. So let's write that in under rate.

Now we need to figure out what gets added or subtracted. When going upstream, you're losing speed because you have to fight the current. So it would be a minus. When going downstream, you're gaining speed because you're moving with the current. So it would be a plus.

So, we're adding or subtracting the speed of the current. The problem is asking us to find the speed of the current. So, we can define our variable here: $$\text{Let }x=\text{speed of current}$$

Next, we need to think about time. Well, we don't know the time. But we do know that our distance equals our rate times our time: $$d=r\cdot t$$ We also have values for distance and rate. So we can solve this equation for time. $$\begin{align} d&=r\cdot t\\ \frac{d}{\color{#8925AE}{r}}&=\frac{r\cdot t}{\color{#8925AE}{r}}\\ t&=\frac{d}r \end{align}$$ So we can just put in the distance divided by the rate into our slot for time.

Time can travel thirty-six miles downstream in the same amount of time that it can travel twenty-four miles upstream. The time it takes to travel upstream is equal to the time it takes to travel downstream. $$\frac{24}{15-x}=\frac{36}{15+x}$$ And now we have a rational equation which we can solve with cross multiplication! $$\begin{align} \color{royalblue}{\frac{24}{15-x}}&=\color{magenta}{\frac{36}{15+x}}\\ \color{royalblue}{24}\color{magenta}{(15+x)}&=\color{magenta}{36}\color{royalblue}{(15-x)}\\ \frac{24(15+x)}{\color{#8925AE}{12}}&=\frac{36(15-x)}{\color{#8925AE}{12}}\\ 2(15+x)&=3(15-x)\\ 30+2x&=45-3x\\ \color{#8925AE}{-30}\ \ \ \ \ \ \ \ \ &=\color{#8925AE}{-30}\\ 2x&=15-3x\\ \color{#8925AE}{+3x}&=\ \ \ \ \color{#8925AE}{+3x}\\ 5x&=15\\ \frac{5x}{\color{#8925AE}{5}}&=\frac{15}{\color{#8925AE}{5}}\\ x&=3 \end{align}$$

Finally, we state our answer. $$\text{The speed of the current is }3\text{ miles per hour.}$$

(Video) Word Problems with Rational Equations

by larryschmidt

Here is a really good video by larryschmidt showing you how to work through some word problems involving rational expressions.

$\large{\text{Summary}}$

One type of problem is when one person can work at a certain rate, and another person can work at a different rate, and you have to calculate how long it would take if these two people worked together.

The first problem he looks at is:

Sandra can paint a kitchen in 6 hours and Roger can paint the same kitchen in 7 hours. How long would it take for both working together to point the kitchen?

We start by setting up some rates per hour. It says that Sandra can paint $1$ kitchen in $6$ hours, or: $$\frac16\text{ kitchen/hour}$$

Roger can paint $1$ kitchen in $7$ hours, or: $$\frac17\text{ kitchen/hour}$$

Finally, let's let $t$ be the time it takes when they work together. So they can paint $1$ kitchen in $t$ hours, or: $$\frac{1}{t}\text{ kitchen/hour}$$

If they work together, this is the same as their rates being added, so we get the equation: $$\frac16+\frac17=\frac1t$$ Let's solve for $t$. $$\begin{align} \frac16+\frac17&=\frac1t\\ \color{#8925AE}{\frac77}\cdot\frac16+\frac17\cdot\color{#8925AE}{\frac66}&=\frac1t\\ \frac{7}{42}+\frac{6}{42}&=\frac1t\\ \frac{13}{42}&=\frac1t\\ \frac{42}{13}&=t\\ 3.23&\approx t \end{align}$$

So it takes approximately $3.23$ hours for them to paint the kitchen together.