In this topic we use the methods discussed of adding, subtracting, multiplying and dividing REs to solve an equation. Since our equation will have REs in it, any value that makes a denominator equal zero cannot be allowed as a solution to the equation. In other words, all solutions must be in the domain of the rational expressions listed in the equation. It is a good idea to check your solutions back into the original equation.
The process shown here to solve is called Clearing the Fractions. First find the CD, then multiply each term by the factor that is missing from the CD, the remaining factors will cancel out and there will not be any fractions left. Proceed to solve the resulting equation and check proposed solutions in the original equation.
Solve the following equation: \(\large{\frac{4}{x-5}}\normalsize{-}\large{\frac{2}{x+5}}\normalsize{=}\large{\frac{1}{x^2-25}}\).
| Get a CD: the denominators of \(x-5, x+5\) and \(x^2-25=(x-5)(x+5)\) create a CD \(=(x-5)(x+5)\). ***NOTE: when finding a CD, the restricted values come from where each denominator equals zero: $$\begin{array}{ccc} x-5=0 & \implies & x=5 \\ x+5=0 & \implies & x=-5\end{array}$$ |
| Multiply through by the CD: $$\color{blue}{(x-5)(x+5)}\left(\frac{4}{x-5}\right)-\color{blue}{(x-5)(x+5)}\left(\frac{2}{x+5}\right)=\color{blue}{(x-5)(x+5)}\left(\frac{1}{x^2-25}\right)$$ Notice factors cancel out within each multiplication. For example, the factor \(x-5\) cancels out in the first multiplication, \(x+5\) cancels out in the second, and \((x-5)(x+5)\) cancels out in the third: $$\color{red}{(x-5)}\color{blue}{(x+5)}\left(\frac{4}{\color{red}{x-5}}\right)-\color{blue}{(x-5)}\color{red}{(x+5)}\left(\frac{2}{\color{red}{x+5}}\right)=\color{red}{(x-5)(x+5)}\left(\frac{1}{\color{red}{(x-5)(x+5)}}\right)$$ |
| After multiplying, a new equation results with NO fractions: $$\color{blue}{(x+5)}(4)-\color{blue}{(x-5)}(2)=1$$ |
| Solve the new equation: \(4x+20-2x+10=1\implies 2x+30=1\implies 2x=-29 \implies x=\frac{-29}{2}\;\color{green}{\checkmark}\) |
| This proposed solution \(x=-\large{\frac{29}{2}}\) is not one of the restricted values of \(-5\) or \(5\). Upon checking in the original equation using Desmos, or a calculator, it is determined to be the solution. |
This next example will show a shortcut, just multiplying through by what is missing instead of the entire CD. This is the preferred method and all examples to follow will show this shortcut technique.
Solve the following equation: \(\large{\frac{2}{x-3}}\normalsize{=}\large{\frac{12}{x^2-9}}\normalsize{+}\large{\frac{3}{x+3}}\).
| Factor to get CD: | \(x-3, x+3\) and \(x^2-9=(x+3)(x-3)\implies\) CD = \((x-3)(x+3)\) Restricted values \(x=\pm3\). |
| Apply the SHORTCUT: Multiply only by the factor missing from the CD: |
\(2\color{blue}{(x+3)}=12+3\color{blue}{(x-3)}\) |
| Solve resulting equation: | \(2x+6=12+3x-9\implies 15=12+x\implies 3=x\;\color{green}{\checkmark}\) |
| The proposed solution \(x=3\) is one of the restricted values found during the CD, so this equation has NO solution. | |
Solve the following equation: \(\large{\frac{4}{w^2+w-6}}\normalsize{-}\large{\frac{1}{x^2-4}}\normalsize{=}\large{\frac{2}{x^2+5w+6}}\).
| Factor to find CD: | \((w+3)(w-2)\), \((w+2)(w-2)\) and \((w+3)(w+2)\implies\) CD \(=(w+3)(w-2)(w+2)\) The restricted values are \(w=\pm2,\; w=-3\) |
| Apply the SHORTCUT: | \(4(w+2)-1(w+3)=2(w-2)\) |
| Solve the resulting equation: | \(4w+8-w-3=2w-4\implies 3w+5=2w-4\implies w=-9\;\color{green}{\checkmark}\) |
| The proposed solution is not one of the restricted values of \(w=\pm2,\; w=-3\). Upon checking in the original equation using Desmos, or a calculator, it is determined to be the solution. | |