Multiply the numerators, and then multiply the denominators.

[latex]\displaystyle \frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}=\frac{35a^{2}}{140a^{3}}[/latex]

Simplify by finding common factors in the numerator and denominator. Simplify the common factors.

[latex]\displaystyle \large\begin{array}{l}\frac{35a^{2}}{140a^{3}}=\frac{5\cdot7\cdot{a}^{2}}{5\cdot7\cdot2\cdot2\cdot{a}^{2}\cdot{a}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\cancel{5}\cdot\cancel{7}\cdot\cancel{{a}^{2}}}{\cancel{5}\cdot\cancel{7}\cdot2\cdot2\cdot\cancel{{a}^{2}}\cdot{a}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\normalsize1\cdot\large\frac{1}{4a}\end{array}[/latex]

Simplify.

[latex]\displaystyle \frac{1}{4a}[/latex]

Answer

[latex]\displaystyle \frac{5{{a}^{2}}}{14}\cdot \frac{7}{10{{a}^{3}}}=\frac{1}{4a}[/latex][latex] \displaystyle [/latex]

Factor the numerators and denominators. Look for the greatest common factors.

[latex] \displaystyle \frac{5\cdot {{a}^{2}}}{7\cdot 2}\cdot \frac{7}{5\cdot 2\cdot {{a}^{2}}\cdot a}[/latex]

Simplify common factors, then multiply.

[latex]\displaystyle \large\begin{array}{c}\frac{5\cdot {{a}^{2}}}{7\cdot 2}\cdot \frac{7}{5\cdot 2\cdot {{a}^{2}}\cdot a}\\\\=\frac{\cancel{5}\cdot\cancel{{a}^{2}}}{\cancel{7}\cdot 2}\cdot \frac{\cancel{7}}{\cancel{5}\cdot 2\cdot\cancel{{a}^{2}}\cdot a}\\\\=\frac{1\cdot1\cdot1}{2\cdot2\cdot{a}}=\frac{1}{4a}\end{array}[/latex]

Answer

[latex]\displaystyle \frac{5a^{2}}{14}\cdot\frac{7}{10a^{3}}=\frac{1}{4a}[/latex]

Factor the numerators and denominators.

[latex]\displaystyle \frac{\left(a-2\right)\left(a+1\right)}{5\cdot{a}}\cdot\frac{5\cdot2\cdot{a}}{\left(a+1\right)}[/latex]

Simplify common factors:

[latex]\large\begin{array}{c}\frac{\left(a-2\right)\cancel{\left(a+1\right)}}{\cancel{5}\cdot{\cancel{a}}}\cdot\frac{\cancel{5}\cdot2\cdot{\cancel{a}}}{\cancel{\left(a+1\right)}}\\\\=\frac{a-2}{1}\cdot\frac{2}{1}\end{array}[/latex]

Multiply simplified rational expressions. This expression can be left with the numerator in factored form or multiplied out.

[latex]\large \begin{array}{c}\frac{\left(a-2\right)}{1}\cdot\frac{2}{1}\\\\=2\left(a-2\right)\end{array}[/latex]

Answer

[latex] \displaystyle \frac{{{a}^{2}}-a-2}{5a}\cdot \frac{10a}{a+1}=2a-4[/latex]

Factor the numerators and denominators.

[latex]\displaystyle \frac{\left(a+2\right)\left(a+2\right)}{\left(2a-5\right)\left(a+2\right)}\cdot\frac{a+5}{a\left(a+2\right)}[/latex]

Simplify common factors.

[latex]\displaystyle \large\frac{\cancel{\left(a+2\right)}\cancel{\left(a+2\right)}}{\left(2a-5\right)\cancel{\left(a+2\right)}}\cdot\frac{a+5}{a\cancel{\left(a+2\right)}}[/latex]

Multiply simplified rational expressions. This expression can be left with the denominator in factored form or multiplied out.

[latex]\displaystyle \frac{1}{\left(2a-5\right)}\cdot\frac{a+5}{a}=\frac{a+5}{a\left(2a-5\right)}[/latex]

Answer

[latex]\displaystyle \frac{a^{2}+4a+4}{2a^{2}-a-10}\cdot\frac{a+5}{a^{2}+2a}=\frac{a+5}{a\left(2a-5\right)}[/latex]

State the Domain:

Find excluded values. 9 and 27 can never equal 0.

Because [latex]15x^{3}[/latex] becomes the denominator in the reciprocal of [latex] \displaystyle \frac{15{{x}^{3}}}{27}[/latex], you must find the values of x that would make [latex]15x^{3}[/latex] equal 0.

[latex]\begin{array}{c}15x^{3}=0\\x=0\,\text{is an excluded value}.\end{array}[/latex]

Divide:

State the quotient in simplest form.  Rewrite division as multiplication by the reciprocal.

[latex]\displaystyle \frac{5x^{2}}{9}\cdot\frac{27}{15x^{3}}[/latex]

Factor the numerators and denominators.

[latex]\displaystyle \frac{5\cdot{x}\cdot{x}}{3\cdot3}\cdot\frac{3\cdot3\cdot3}{5\cdot3\cdot{x}\cdot{x}\cdot{x}}[/latex]

Simplify common factors.

Simplify.

[latex]\displaystyle \large\begin{array}{c}\frac{\cancel{5}\cdot{\cancel{x}}\cdot{\cancel{x}}}{\cancel{3}\cdot\cancel{3}}\cdot\frac{\cancel{3}\cdot\cancel{3}\cdot\cancel{3}}{\cancel{5}\cdot\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}\cdot{x}}\\\\=\frac{1}{x}\end{array}[/latex]

Answer

[latex]\displaystyle \frac{5{{x}^{2}}}{9}\div \frac{15{{x}^{3}}}{27}=\frac{1}{x},x\ne 0[/latex]

Determine the excluded values that make the denominators and the numerator of the divisor equal to 0.

[latex]\begin{array}{r}\left(x+2\right)=0\,\,\,\,\,\\x=-2\\\left({{x}^{2}}+5x+6 \right)=0\,\,\,\,\,\\\left(x+3\right)\left(x+2\right)=0\,\,\,\,\,\\x=-3\,\,\,\,\text{or}\,\,\,\,-2\\6x^{4}=0\,\,\,\,\,\\x=0\,\,\,\,\,\end{array}[/latex]

Domain is all real numbers except [latex]0[/latex], [latex]−2[/latex], and [latex]−3[/latex].

Rewrite division as multiplication by the reciprocal.

[latex]\displaystyle \frac{3x^{2}}{x+2}\cdot\frac{\left(x^{2}+5x+6\right)}{6x^{4}}[/latex]

Factor the numerators and denominators.

[latex]\displaystyle \frac{3\cdot{x}\cdot{x}}{x+2}\cdot\frac{\left(x+2\right)\left(x+3\right)}{2\cdot3\cdot{x}\cdot{x}\cdot{x}\cdot{x}}[/latex]

Simplify common factors

[latex]\displaystyle \frac{\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}}{\cancel{x+2}}\cdot\frac{\cancel{\left(x+2\right)}\left(x+3\right)}{2\cdot\cancel{3}\cdot{\cancel{x}}\cdot{\cancel{x}}\cdot{x}\cdot{x}}[/latex]

Simplify.

[latex]\displaystyle \frac{(x+3)}{2{{x}^{2}}}[/latex]

Answer

[latex]\displaystyle \frac{3{{x}^{2}}}{x+2}\div \frac{6{{x}^{4}}}{({{x}^{2}}+5x+6)}=\frac{x+3}{2{{x}^{2}}}[/latex].

The domain is all real numbers except 0, [latex]−2[/latex], and [latex]−3[/latex].

First, let’s define the domain of each term. Since we have [latex]x[/latex] and [latex]y[/latex] in the denominators, we can say [latex]x\ne0 ,\text{ and }y\ne0[/latex].

Now we have to find the LCD. Since [latex]x[/latex] appears once and [latex]y[/latex] appears once,  the LCD will be [latex]xy[/latex].  We then multiply each expression by the appropriate form of 1 to obtain [latex]xy[/latex] as the denominator for each fraction.

Now that the expressions have the same denominator, we simply add the numerators to find the sum.

The domain is [latex]x\ne-4[/latex]

Answer

[latex] \displaystyle \frac{2{{x}^{2}}}{x+4}+\frac{8x}{x+4}=2x,x\ne -4[/latex], [latex]x\ne0 ,\text{ and }y\ne0[/latex]

Analysis of the Solution

Multiplying by [latex]\displaystyle \frac{y}{y}[/latex] or [latex]\displaystyle \frac{x}{x}[/latex] does not change the value of the original expression because any number divided by itself is 1, and multiplying an expression by 1 gives the original expression.

Note that the denominator of the first expression is a perfect square trinomial, and the denominator of the second expression is a difference of squares so they can be factored using special products.

[latex]\large \begin{array}{cc}\frac{6}{{\left(x+2\right)}^{2}}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\hfill & \text{Factor}.\hfill \\ \frac{6}{{\left(x+2\right)}^{2}}\cdot \frac{x - 2}{x - 2}-\frac{2}{\left(x+2\right)\left(x - 2\right)}\cdot \frac{x+2}{x+2}\hfill & \text{Multiply each fraction to get LCD as denominator}.\hfill \\ \frac{6\left(x - 2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}-\frac{2\left(x+2\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Multiply}.\hfill \\ \frac{6x - 12-\left(2x+4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Apply distributive property}.\hfill \\ \frac{4x - 16}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Subtract}.\hfill \\ \frac{4\left(x - 4\right)}{{\left(x+2\right)}^{2}\left(x - 2\right)}\hfill & \text{Simplify}.\hfill \end{array}[/latex]

The domain is [latex]x\ne-6[/latex]

Answer

[latex]\displaystyle \frac{4x+7}{x+6}-\frac{2x+8}{x+6}=\frac{2x-1}{x+6},\text{}x\ne-6[/latex]

Analysis of the solution

In the last example, the LCD was  [latex]\displaystyle \left(x+2\right)^2\left(x-2\right)[/latex].  The reason we need to include [latex]\left(x+2\right)[/latex] two times is because it appears two times in the expression [latex]\displaystyle \frac{6}{{x}^{2}+4x+4}[/latex].

Rewrite the complex fraction as a division problem.

[latex]\displaystyle\large \frac{\,\frac{12}{35}\,}{\,\frac{6}{7}\,}=\normalsize\frac{12}{35}\div \frac{6}{7}[/latex]

Rewrite the division as multiplication, using the reciprocal of the divisor.

[latex]\displaystyle =\frac{12}{35}\cdot \frac{7}{6}[/latex]

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

[latex]\large \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}[/latex]

Answer

[latex]\displaystyle \frac{\,\frac{12}{35}\,}{\,\frac{6}{7}\,}=\normalsize\frac{2}{5}[/latex]

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 if you are confused.

[latex]\displaystyle \frac{\,\frac{3}{4}+\frac{1}{2}\,}{\,\frac{4}{5}-\frac{1}{10}\,}=\frac{\,\frac{5}{4}\,}{\,\frac{7}{10}\,}[/latex]

Rewrite the complex fraction as a division problem.

[latex]\displaystyle \frac{\,\,\frac{5}{4}\,\,}{\,\,\frac{7}{10}\,\,}=\normalsize\frac{5}{4}\div \frac{7}{10}[/latex]

Rewrite the division as multiplication, using the reciprocal of the divisor.

[latex]\displaystyle =\frac{5}{4}\cdot \frac{10}{7}[/latex]

Multiply and simplify as needed.

[latex]\displaystyle \frac{5}{4}\cdot \frac{10}{7}=\frac{5\cdot5\cdot2}{2\cdot2\cdot7}=\frac{25}{14}[/latex]

Answer

[latex]\displaystyle \frac{\,\frac{3}{4}+\frac{1}{2}\,}{\,\frac{4}{5}-\frac{1}{10}\,}=\normalsize\frac{25}{14}[/latex]

Rewrite the complex rational expression as a division problem.

[latex]\displaystyle =\frac{x+5}{{{x}^{2}}-16}\div \frac{{{x}^{2}}-25}{x-4}[/latex]

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

[latex]\displaystyle =\frac{x+5}{{{x}^{2}}-16}\cdot \frac{x-4}{{{x}^{2}}-25}[/latex]

Factor the numerator and denominator, looking for common factors. In this case, [latex]x+5[/latex] and [latex]x–4[/latex] are common factors of the numerator and denominator. Notice that [latex] \frac{(x+5)(x-4)}{(x+5)(x-4)}[/latex] is equal to 1.

[latex]\large \begin{array}{l}=\frac{(x+5)(x-4)}{(x+4)(x-4)(x+5)(x-5)}\\\\=\frac{(x+5)(x-4)}{(x+5)(x-4)}\cdot \frac{1}{(x+4)(x-5)}\end{array}[/latex]

Answer

[latex]\displaystyle \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[/latex]

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.

[latex]\Large \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}[/latex]

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.)

[latex]\displaystyle \large =\frac{{{x}^{2}}-9}{{{x}^{2}}}\div \frac{{{x}^{2}}+5x+6}{{{x}^{2}}}[/latex]

Rewrite the division as multiplication, using the reciprocal of the divisor.

[latex]\displaystyle \large =\frac{{{x}^{2}}-9}{{{x}^{2}}}\cdot \frac{{{x}^{2}}}{{{x}^{2}}+5x+6}[/latex]

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

[latex]\Large \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}[/latex]

Answer

[latex]\displaystyle \large \frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}=\frac{x-3}{x+2},x\ne -3,-2,0[/latex]

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

[latex]\Large \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}[/latex]

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, [latex]-2[/latex] and [latex]-3[/latex].

[latex]\Large \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}[/latex]

Answer

[latex]\displaystyle \large \frac{1-\frac{9}{{{x}^{2}}}}{1+\frac{5}{x}+\frac{6}{{{x}^{2}}}}=\frac{x-3}{x+2},x\ne -3,-2,0[/latex]