(8.4.1) – Solving rational equations by clearing denominators

[latex]\begin{eqnarray*} 3 x - \frac{1}{2}\,= \frac{1}{x}\,& & \text{ Multiply}\,\text{ each} \text{ term}\,\text{ by}\,\text{ LCD}, (2 x)\\ 3 x (2 x) - \frac{(2 x)}{2}\,= \frac{(2 x)}{x}\,& & \text{ Reduce} \text{ fractions}\\ 6 x^2 - x = 2 & & \text{ Distribute}\\ \underline{- 2 - 2}\,& & \text{ Subtract}\,2 \text{ from}\,\text{ both} \text{ sides}\\ 6 x^2 - x - 2 = 0 & & \text{ Factor}\\ (3 x - 2) (2 x + 1) = 0 & & \text{ Set}\,\text{ each}\,\text{ factor} \text{ equal}\,\text{ to}\,\text{ zero}\\ 3 x - 2 = 0 \text{ or}\,2 x + 1 = 0 & & \text{ Solve}\,\text{ each} \text{ equation}\\ \underline{+ 2 \quad + 2}\, \underline{- 1 - 1}\,& & \\ 3 x = 2 \hspace{3em}\,2 x = - 1 & & \\ x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,& & \text{ Check} \text{ solutions}, \text{ LCD}\,\text{ can}' t \text{ be}\,\text{ zero}\\ 2 \left( \frac{3}{2}\,\right) =2 \left( - \frac{1}{2}\,\right) = - 1 & & \text{ Neither}\,\text{ make}\,\text{ LCD}\,\text{ zero}, \text{ both}\,\text{ are} \text{ solutions}\\ x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,& & \text{ Our} \text{ Solution} \end{eqnarray*}[/latex]

Answer

[latex]\displaystyle x = \frac{2}{3}\,\text{ or}\,x = - \frac{1}{2}\,[/latex]

Clear the denominators by multiplying each side by the common denominator. The common denominator is [latex]3\left(x+1\right)[/latex] since [latex]3\text{ and }x+1[/latex] don’t have any common factors.

[latex]\begin{array}{c}3\left(x+1\right)\left(\frac{8}{x+1}\right)=3\left(x+1\right)\left(\frac{4}{3}\right)\end{array}[/latex]

Simplify common factors.

[latex]\begin{array}{c}3\cancel{\left(x+1\right)}\left(\frac{8}{\cancel{x+1}}\right)=\cancel{3}\left(x+1\right)\left(\frac{4}{\cancel{3}}\right)\\24=4\left(x+1\right)\\24=4x+4\end{array}[/latex]

Now this looks like a linear equation, and we can use the addition and multiplication properties of equality to solve it.

[latex]\begin{array}{c}24=4x+4\\\underline{-4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-4}\\20=4x\,\,\,\,\,\,\,\,\\\\x=5\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Check the solution in the original equation.

[latex]\large \begin{array}{r}\,\,\,\,\,\frac{8}{\left(x+1\right)}=\frac{4}{3}\\\\\frac{8}{\left(5+1\right)}=\frac{4}{3}\\\\\frac{8}{6}=\frac{4}{3}\end{array}[/latex]

Reduce the fraction [latex]\frac{8}{6}[/latex] by simplifying the common factor of 2:

[latex]\displaystyle \frac{\cancel{2}\cdot4}{\cancel{2}\cdot3}\normalsize=\large\frac{4}{3}[/latex]

Answer

[latex]x=5[/latex]

[latex] \begin{eqnarray*}\frac{5 x + 5}{x + 2}\,+ 3 x = \frac{x^2}{x + 2}\,& & \text{ Multiply} \text{ each}\,\text{ term}\,\text{ by}\,\text{ LCD}, (x + 2)\\ & & \\ \frac{(5 x + 5) (x + 2)}{x + 2}\,+ 3 x (x + 2) = \frac{x^2 (x + 2)}{x + 2} & & \text{ Reduce}\,\text{ fractions}\\ & & \\ 5 x + 5 + 3 x (x + 2) = x^2 & & \text{ Distribute}\\ 5 x + 5 + 3 x^2 + 6 x = x^2 & & \text{ Combine}\,\text{ like}\,\text{ terms}\\ 3 x^2 + 11 x + 5 = x^2 & & \text{ Make}\,\text{ equation}\,\text{ equal} \text{ zero}\\ \underline{- x^2 - x^2}\,& & \text{ Subtract}\,x^2 \text{ from}\,\text{ both} \text{ sides}\\ 2 x^2 + 11 x + 5 = 0 & & \text{ Factor}\\ (2 x + 1) (x + 5) = 0 & & \text{ Set}\,\text{ each}\,\text{ factor} \text{ equal}\,\text{ to}\,\text{ zero}\\ 2 x + 1 = 0 \text{ or}\,x + 5 = 0 & & \text{ Solve}\,\text{ each} \text{ equation}\\ 2 x = - 1 \text{ or}\,x = - 5 & & \\ x = - \frac{1}{2}\,\text{ or}\,- 5 & & \text{ Check}\,\text{ solutions}, \text{ LCD}\,\text{ can}' t \text{ be}\,\text{ zero}\\ - \frac{1}{2}\,+ 2 = \frac{3}{2}\,- 5 + 2 = - 3 & & \text{ Neither} \text{ make}\,\text{ LCD}\,\text{ zero}, \text{ both}\,\text{ are} \text{ solutions}\\ x = - \frac{1}{2}\,\text{ or}\,- 5 & & \text{ Our}\,\text{ Solution} \end{eqnarray*}[/latex]

Answer

[latex]x = - \frac{1}{2}\,\text{ or}\,x = - 5[/latex]

Determine any values for [latex]x[/latex] that would make the denominator 0.

[latex]\displaystyle \frac{2x-5}{x-5}=\frac{15}{x-5}[/latex]

5 is an excluded value because it makes the denominator [latex]x-5[/latex] equal to 0.

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for [latex]x[/latex].

[latex]\large \begin{array}{r}2x-5=15\\2x=20\\x=10\end{array}[/latex]

Check the solution in the original equation.

[latex]\large \begin{array}{r}\frac{2x-5}{x-5}=\frac{15}{x-5}\,\,\\\\\frac{2(10)-5}{10-5}=\frac{15}{10-5}\\\\\frac{20-5}{10-5}=\frac{15}{10-5}\\\\\frac{15}{5}=\frac{15}{5}\,\,\,\,\,\,\,\,\,\end{array}[/latex]

Answer

[latex]x=10[/latex]

Determine any values for [latex]m[/latex] that would make the denominator 0. [latex]−4[/latex] is an excluded value because it makes [latex]m+4[/latex] equal to 0.

Since the denominator of each expression in the equation is the same, the numerators must be equal. Set the numerators equal to one another and solve for [latex]m[/latex].

[latex]\large \begin{array}{l}16=m^{2}\\\,\,\,0={{m}^{2}}-16\\\,\,\,0=\left( m+4 \right)\left(m-4 \right)\end{array}[/latex]

[latex]\large \begin{array}{c}0=m+4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,0=m-4\\m=-4\,\,\,\,\,\,\text{or}\,\,\,\,\,\,m=4\\m=4,-4\end{array}[/latex]

Check the solutions in the original equation.

Since [latex]m=−4[/latex] leads to division by 0, it is an extraneous solution.

[latex]\large \begin{array}{c}\frac{16}{m+4}=\frac{{{m}^{2}}}{m+4}\\\\\frac{16}{-4+4}=\frac{{{(-4)}^{2}}}{-4+4}\\\\\frac{16}{0}=\frac{16}{0}\end{array}[/latex]

[latex]-4[/latex] is excluded because it leads to division by 0.

[latex]\large \begin{array}{c}\frac{16}{4+4}=\frac{{{(4)}^{2}}}{4+4}\\\\\frac{16}{8}=\frac{16}{8}\end{array}[/latex]

Answer

[latex]m=4[/latex]