Rewrite the complex fraction as a division problem.

$\displaystyle\large \frac{\,\frac{12}{35}\,}{\,\frac{6}{7}\,}=\normalsize\frac{12}{35}\div \frac{6}{7}$

Rewrite the division as multiplication and take the reciprocal of the divisor.

$=\frac{12}{35}\cdot \frac{7}{6}$

Factor the numerator and denominator looking for common factors before multiplying numbers together.

$\begin{array}{l}=\frac{2\cdot 6\cdot 7}{5\cdot 7\cdot 6}\\\\=\frac{2}{5}\cdot \frac{6\cdot 7}{6\cdot 7}\\\\=\frac{2}{5}\cdot 1\end{array}$

$\displaystyle\Large \frac{\,\frac{12}{35}\,}{\,\frac{6}{7}\,}=\normalsize\frac{2}{5}$

First combine the numerator and denominator by adding or subtracting. You may need to find a common denominator first. Note that we do not show the steps for finding a common denominator, so please review that in the previous section if you are confused.

$\displaystyle\Large \frac{\,\frac{3}{4}+\frac{1}{2}\,}{\,\frac{4}{5}-\frac{1}{10}\,}=\frac{\,\frac{5}{4}\,}{\,\frac{7}{10}\,}$

Rewrite the complex fraction as a division problem.

$\displaystyle\Large \frac{\,\,\frac{5}{4}\,\,}{\,\,\frac{7}{10}\,\,}=\normalsize\frac{5}{4}\div \frac{7}{10}$

Rewrite the division as multiplication and take the reciprocal of the divisor.

$=\dfrac{5}{4}\cdot \frac{10}{7}$

Multiply and simplify as needed.

$\dfrac{5}{4}\cdot \frac{10}{7}=\frac{5\cdot5\cdot2}{2\cdot2\cdot7}=\frac{25}{14}$

$\displaystyle\Large \frac{\,\frac{3}{4}+\frac{1}{2}\,}{\,\frac{4}{5}-\frac{1}{10}\,}=\normalsize\frac{25}{14}$

Rewrite the complex rational expression as a division problem.

$=\frac{x+5}{{{x}^{2}}-16}\div \frac{{{x}^{2}}-25}{x-4}$

Rewrite the division as multiplication and take the reciprocal of the divisor. Note that the excluded values for this are $-4$, $4$, $-5$, and $5$, because those values make the denominators of one of the fractions zero.

$=\frac{x+5}{{{x}^{2}}-16}\cdot \frac{x-4}{{{x}^{2}}-25}$

Factor the numerator and denominator, looking for common factors. In this case, $x+5$ and $x–4$ are common factors of the numerator and denominator.

$\begin{array}{l}=\frac{(x+5)(x-4)}{(x+4)(x-4)(x+5)(x-5)}\\\\=\frac{1}{(x+4)(x-5)}\end{array}$

$\displaystyle\Large \frac{\,\,\frac{x+5}{{{x}^{2}}-16}\,}{\,\,\frac{{{x}^{2}}-25}{x-4}\,}\normalsize=\frac{1}{(x+4)(x-5)},x\ne -4,4,-5,5$

Combine the expressions in the numerator and denominator. To do this, rewrite the expressions using a common denominator. There is an excluded value of $0$ because this makes the denominators of the fractions zero.

$\begin{array}{l}=\frac{\frac{{{x}^{2}}}{{{x}^{2}}}-\frac{9}{{{x}^{2}}}}{\frac{{{x}^{2}}}{{{x}^{2}}}+\frac{5x}{{{x}^{2}}}+\frac{6}{{{x}^{2}}}}\\\\=\frac{\frac{{{x}^{2}}-9}{{{x}^{2}}}}{\frac{{{x}^{2}}+5x+6}{{{x}^{2}}}}\end{array}$

Rewrite the complex rational expression as a division problem. (When you are comfortable with the step of rewriting the complex rational fraction as a division problem, you might skip this step and go straight to rewriting it as multiplication.)

$=\frac{{{x}^{2}}-9}{{{x}^{2}}}\div \frac{{{x}^{2}}+5x+6}{{{x}^{2}}}$

Rewrite the division as multiplication and take the reciprocal of the divisor.

$=\frac{{{x}^{2}}-9}{{{x}^{2}}}\cdot \frac{{{x}^{2}}}{{{x}^{2}}+5x+6}$

Factor the numerator and denominator looking for common factors. In this case, $x+3$ and $x^{2}$ are common factors. We can now see there are two additional excluded values, $-2$ and $-3$.

$\begin{array}{l}=\frac{(x+3)(x-3){{x}^{2}}}{{{x}^{2}}(x+3)(x+2)}\\\\=\frac{(x-3)}{(x+2)}\cdot \frac{{{x}^{2}}(x+3)}{{{x}^{2}}(x+3)}\end{array}$

$\frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}=\frac{x-3}{x+2},x\ne -3,-2,0$

Before combining the expressions, find a common denominator for all of the rational expressions. In this case, $x^{2}$ is a common denominator. Multiply by $1$ in the form of a fraction with the common denominator in both the numerator and denominator. In this case, multiply by $\frac{{{x}^{2}}}{{{x}^{2}}}$. There is an excluded value of $0$ because this makes the denominators of the fractions zero.

$\begin{array}{l}=\frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}\cdot \frac{{{x}^{2}}}{{{x}^{2}}}\\\\=\frac{\left( 1-\frac{9}{{{x}^{2}}} \right){{x}^{2}}}{\left( 1+\frac{5}{x}+\frac{6}{{{x}^{2}}} \right){{x}^{2}}}\\\\=\frac{{{x}^{2}}-9}{{{x}^{2}}+5x+6}\end{array}$

Notice that the expression is no longer complex! You can simplify by factoring and identifying common factors. We can now see there are two additional excluded values, $-2$ and $-3$.

$\begin{array}{l}=\frac{(x+3)(x-3)}{(x+3)(x+2)}\\\\=\frac{x+3}{x+3}\cdot \frac{x-3}{x+2}\\\\=1\cdot \frac{x-3}{x+2}\end{array}$

$\frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}=\frac{x-3}{x+2},x\ne -3,-2,0$