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(Videos) Solving Rational Equations

by mathman1024

Here's a video by mathman1024 where he shows you how to solve rational equations. Rational equations are equations that contain rational expressions.

The first thing we need to learn about is proportions. A proportion is two fractions that are equal. $$\frac{a}b=\frac{c}d$$ If we have a proportion, we also know the cross products are equal. $$a\cdot d=b\cdot c$$

The first problem he looks at is, $\dfrac{x+5}7=\dfrac{2x-3}{10}$.

$$\begin{align} \color{royalblue}{\dfrac{x+5}7}&=\color{magenta}{\dfrac{2x-3}{10}}\\ \color{magenta}{10}\color{royalblue}{(x+5)}&=\color{royalblue}{7}\color{magenta}{(2x-3)}\\ 10x+50&=14x-21\\ \color{#8925AE}{+21}&=\ \ \ \ \ \ \color{#8925AE}{+21}\\ \hline 10x+71&=14x\\ \color{#8925AE}{-10x}\ \ \ \ \ \ \ \ \ &=\color{#8925AE}{-10x}\\ \hline 71&=4x\\ x&=\frac{71}4 \end{align}$$

(Videos) Solving with Multiple Rational Expressions

by mathman1024

Here's a video by mathman1024 showing you how to solve rational equations. In this example there's multiple rational expressions and you can't use cross multiplication.

The problem he looks at is, $\dfrac{x}{x^2+7x+6}+\dfrac{2}{x+6}=\dfrac{4}{x+1}$. The first step is to factor the quadratic. $$\begin{align} \dfrac{x}{x^2+7x+6}+\dfrac{2}{x+6}&=\dfrac{4}{x+1}\\ \dfrac{x}{(x+6)(x+1)}+\dfrac{2}{x+6}&=\dfrac{4}{x+1} \end{align}$$ So now we multiply each term by the missing factors so that we get common denominators. $$\begin{align} \dfrac{x}{x^2+7x+6}+\dfrac{2}{x+6}&=\dfrac{4}{x+1}\\ \dfrac{x}{(x+6)(x+1)}+\dfrac{2}{x+6}&=\dfrac{4}{x+1}\\ \dfrac{x}{(x+6)(x+1)}+\dfrac{2}{x+6}\cdot\color{#8925AE}{\frac{x+1}{x+1}}&=\dfrac{4}{x+1}\cdot\color{#8925AE}{\frac{x+6}{x+6}}\\ \dfrac{x}{(x+6)(x+1)}+\dfrac{2(x+1)}{(x+6)(x+1)}&=\dfrac{4(x+6)}{(x+6)(x+1)} \end{align}$$ Now that all the terms have a common denominator we can just remove the denominator and focus on only the numerator terms. $$\begin{align} x+2(x+1)&=4(x+6)\\ x+2x+2&=4x+24\\ 3x+2&=4x+24\\ \color{#8925AE}{-24}&=\ \ \ \ \color{#8925AE}{-24}\\ \hline 3x-22&=4x\\ \color{#8925AE}{-3x}\ \ \ \ \ \ \ \ \ &=\color{#8925AE}{-3x}\\ \hline -22&=x\\ x&=-22 \end{align}$$

Before we accept this as an answer we have to make sure that it's not one of the terms that make any of the denominators $0$. If we look at the denominators we can see that $x\ne-6,-1$. Therefore, $x=-22$ is the solution.

Background

In the previous lessons, we learned how to simplify rational expressions, whether they're added/subtracted or multiplied/divided together. We can extend these concepts into solving rational equations. The main difference is that because these are equalities, we can solve for the value of a variable.

Solving Rational Equations

When solving a rational equation, our goal is the same as always. We want to find the value of the variable.

But rational equations are written as fractions, so we have to be careful. The denominator can never equal $0$.

If the denominator were $0$, the fraction would not make sense. You can’t divide by zero since there are no “parts of no whole.”

So before and after solving, always check your answers to make sure they do not make the denominator equal $0$.

You've seen this concept before - think back to proportions, when we used the cross-product property to solve proportions!

$$\frac{4x+1}{3x-1} = \frac{x-5}{2x-7}$$

That is an example of rational equations! We know how to solve this: with cross multiplication. We do some cross multiplying of both sides.

But let's take this a step further. Think about what would happen if we had some kind of operation on one side that required some extra work!

$$\frac{4x+1}{3x-1} + \frac{x-2}{x+1} = \frac{x-5}{2x-7}$$

We can use the techniques we learned in the previous lessons to add the two rational expressions on the left.

We'll start with finding a common denominator.

$$\begin{align} \frac{4x+1}{3x-1} \cdot \frac{x+1}{x+1} \\ &= \frac{(4x+1) \cdot (x+1)}{(3x-1) \cdot (x+1)} \\ &= \frac{4x^2+4x+x+1}{3x^2+3x-x-1} \\ &= \frac{4x^2+5x+1}{3x^2+2x-1} \\ \end{align}$$ $$\begin{align} \frac{x-2}{x+1} \cdot \frac{3x-1}{3x-1} \\ &= \frac{(x-2) \cdot (3x-1)}{(x+1) \cdot (3x-1)} \\ &= \frac{3x^2-x-6x+2}{3x^2-x+3x-1} \\ &= \frac{3x^2-7x+2}{3x^2+2x-1} \\ \end{align}$$

So, our new equation looks like:

$$\frac{4x^2+5x+1}{3x^2+2x-1} + \frac{3x^2-7x+2}{3x^2+2x-1} = \frac{x-5}{2x-7}$$

This looks daunting! Let's simplify.

$$\begin{align} \frac{(4x^2+5x+1) + (3x^2-7x+2)}{3x^2+2x-1} = \frac{x-5}{2x-7} \\ \frac{7x^2-2x+3}{3x^2+2x-1} = \frac{x-5}{2x-7} \\ \end{align}$$

We can now use cross-multiplication to solve even more.

$$\begin{align} (7x^2-2x+3)(2x-7) &= (3x^2+2x-1)(x-5) \\ 14x^3 -49x^2 -4x^2+14x +6x-21 &= 3x^3-15x^2+2x^2-10x-x+5 \\ 14x^3 -53x^2 +20x -21 &= 3x^3 -13x^2 -11x+5 \\ 11x^3 -40x^2 + 31x -26 &= 0\\ \end{align}$$

If we solve the resulting equation for $x$, we'll find that $x=2.95$.

After you solve the equation, it’s important to check your work.

To do this, plug your answer back into the original equation. If the equation is still true, your solution is correct. If it is not true, there was a mistake somewhere in your steps.

In more advanced math, you might get something called an extraneous solution. This is an answer that shows up during your work but does not actually solve the original equation.

Extraneous solutions happen more often with quadratic equations than with linear equations. That’s why it’s always a good idea to plug your answers back in and check them.

Solving Rational Equations

In this lesson, we will learn how to solve rational equations.

A rational equation is similar to a rational expression, but now there is an equal sign. This means we are solving for a variable instead of just simplifying.

You can think of rational equations as the next step after working with rational expressions.

To solve this type of equation, we need to remove the denominators first.

We can do this by multiplying every term in the equation by the Least Common Denominator (LCD). When we multiply by the LCD, the fractions cancel out. This leaves us with an equation that is easier to solve.

Let's look at an example:

$$\frac{2}{x} + \frac{3}{x+1} = \frac{5x+1}{x(x+1)}$$

This equation looks really tricky - but we can solve it step-by-step. First, we need to find the LCD. In the above example, the LCD is $x(x+1)$. Multiply each term in the equation by the LCD to get rid of the fractions:

$$x(x+1) \cdot \frac{2}{x} + \frac{3}{x+1} = x(x+1) \cdot \frac{5x+1}{x(x+1)}$$

Now, we can simplify each term:

$$2(x+1) + 3x = 5x + 1$$

Distribute and combine like terms:

$$2x+2+3x = 5x+1 \rightarrow 5x+2=5x+1 \rightarrow 2=1$$

Wait...that doesn't make sense!

This means that there is no solution because the solved equation leads to a contradiction. Always check for restrictions, like values that make the denominator zero, and watch for extraneous solutions.

Looking Further

In the next lessons, you will extend this idea to systems of rational equations and solving rational inequalities. Once you've mastered those, you'll move onto partial fraction decomposition!