How to solve abstract fractional (rational) equations

How do we solve rational equations?

At times we’d like to take an equation that has at least one fraction with a variable in its denominator and write the equation in a different way.

We’ll call an equation like this an abstract fractional equation. This lesson will look at how to do that.

There are a few things we want to remember about rational functions in general.

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1. Multiplying a fraction by its reciprocal will always give you a value of $1$.

For example $x/y$ has a reciprocal of $y/x$ because

$\frac{x}{y}\cdot\frac{y}{x}=1$

Remember that you can’t divide by $0$ so some instructors might want you to include that $x$ and $y$ can’t equal $0$ like this:

$\frac{x}{y}\cdot\frac{y}{x}=1$ where $x,y\neq0$

2. To clear a fraction from an equation, multiply all of the terms on both sides of the equation by the fraction’s denominator.

For example, to clear the $b$ from the fraction in

$ax+\frac{m}{b}=c$

multiply the equation by $b$ on both sides.

$ax+\frac{m}{b}=c$

$b\left(ax+\frac{m}{b}=c\right)$

$abx+m=bc$

And in case your instructor wants to see it, remember you can’t divide by $0$, so that means your new equation is true only where $b\neq0$.

How to solve rational (abstract fractional) equations by multiplying by the least common multiple of all the denominators

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Solving the equation by clearing the denominators

Example

Solve the fractional expression for $n$, if $n\neq0$.

$\frac{m}{n}+x+ab=c$

In order to get rid of the fraction, we have to multiply every term on both sides of the equation by the denominator of $m/n$.

$\frac{m}{n}+x+ab=c$

$n\left(\frac{m}{n}+x+ab=c\right)$

$n\cdot\frac{m}{n}+n(x)+n(ab)=n(c)$

$m+nx+nab=nc$

To solve for $n$, we’ll need to collect all terms containing $n$ on one side of the equation, and then factor out the $n$.

Let’s move $m$ to the right hand side and $nc$ to the left.

$nx+nab-nc=-m$

Now factor out $n$.

$n(x+ab-c)=-m$

Divide both sides by $(x+ab-c)$.

$n=\frac{-m}{x+ab-c}$

Write the negative sign in front.

$n=-\frac{m}{x+ab-c}$

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Let’s try another one.

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To clear a fraction from an equation, multiply all of the terms on both sides of the equation by the fraction’s denominator.

Example

Solve for $x$ if $x\neq0$ and $y\neq0$.

$\frac{1}{x}-\frac{m}{y}=p$

In order to get rid of the fractions, we have to multiply every term on both sides of the equation by both denominators, $x$ and $y$.

$\frac{1}{x}-\frac{m}{y}=p$

$xy\left(\frac{1}{x}-\frac{m}{y}=p\right)$

$xy\left(\frac{1}{x}\right)-xy\left(\frac{m}{y}\right)=xy(p)$

$1y-mx=xyp$

$y-mx=xyp$

To solve for $x$ we’ll need to collect all terms containing $x$ on one side of the equation, and then factor out the $x$.

Let’s move $mx$ to the right to get

$y=mx+xyp$

Now factor out the $x$.

$y=x(m+yp)$

Divide both sides by $m+yp$.

$\frac{y}{m+yp}=\frac{x(m+yp)}{m+yp}$

$\frac{y}{m+yp}=x$

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