Multiply the numerators, and then multiply the denominators.

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

Factor the numerators and denominators.

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

Answer

[latex]\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]

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]

Answer

The domain is all real numbers except 0.

Rewrite division as multiplication by the reciprocal.

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

Factor the numerators and denominators.

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

Factor the numerators and denominators.

[latex]\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]\large\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]\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].

Since the denominators are the same, add the numerators.

[latex]\frac{2{{x}^{2}}+8x}{x+4}[/latex]

Factor the numerator.

[latex]\frac{2x(x+4)}{x+4}[/latex]

Simplify common factors and.

[latex]\large\begin{array}{c}\frac{2x\cancel{(x+4)}}{\cancel{x+4}}\\\\=\frac{2x}{1}\end{array}[/latex]

The domain is found by setting the denominators in the original sum equal to zero.

[latex]\begin{array}{l}x+4=0\\x=-4\end{array}[/latex]

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]

Subtract the second numerator from the first and keep the denominator the same.

[latex]\frac{4x+7-(2x+8)}{x+6}[/latex]

Be careful to distribute the negative to both terms of the second numerator.

[latex]\frac{4x+7-2x-8}{x+6}[/latex]

Combine like terms. This rational expression cannot be simplified any further.

[latex]\frac{2x-1}{x+6}[/latex]

The domain is found from the denominators of original expression.

[latex]\begin{array}{l}x+6=0\\x=-6\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]

[latex]\begin{array}{r}6=3\cdot2\\10=5\cdot2\\4=2\cdot2\end{array}[/latex]

The LCM will contain factors of 2, 3, and 5. Multiply each number the maximum number of times it appears in a single factorization.

In this case, 3 appears once, 5 appears once, and 2 is used twice because it appears twice in the prime factorization of 4.

Therefore, the LCM of 6, 10, and 4 is [latex]3\cdot5\cdot2\cdot2[/latex], or 60.

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,6=3\cdot2\\\,\,\,\,\,\,\,\,10=5\cdot2\\\,\,\,\,\,\,\,\,\,\,\,4=2\cdot2\\\text{LCM}=3\cdot5\cdot2\cdot2\end{array}[/latex]

Answer

The least common multiple of 6, 10, and 4 is 60.

Find the prime factorization of each denominator.

[latex]\begin{array}{l}15m^{2}\,=\,3\cdot5\cdot{m}\cdot{m}\\\,\,\,21m\,=\,3\cdot7\cdot{m}\end{array}[/latex]

Find the least common multiple. 3 appears exactly once in both of the expressions, so it will appear once in the least common multiple. Both 5 and 7 appear at most once. For the variables, the most m appears is twice.

Use the least common multiple for your new common denominator, it will be the LCD.

[latex]\begin{array}{l}15m^{2}\,=\,3\cdot5\cdot{m}\cdot{m}\\21m\,\,\,=\,3\cdot7\cdot{m}\\\text{LCM}:3\cdot5\cdot7\cdot{m}\cdot{m}\\\text{LCM}:105m^{2}\end{array}[/latex]

Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]105m^{2}[/latex]. Remember that m cannot be 0 because the denominators would be 0.

The first denominator is [latex]15m^{2}[/latex] and the LCD is [latex]105m^{2}[/latex]. You need to multiply [latex]15m^{2}[/latex] by 7 to get the LCD, so multiply the entire rational expression by [latex]\frac{7}{7}[/latex].

The second denominator is [latex]21m[/latex] and the LCD is [latex]105m^{2}[/latex]. You need to multiply [latex]21m[/latex] by [latex]5m[/latex] to get the LCD, so multiply the entire rational expression by [latex]\frac{5m}{5m}[/latex].

[latex]\begin{array}{c}\frac{2n}{15m^{2}}\cdot\frac{7}{7}=\frac{14n}{105m^{2}}\\\\\frac{3n}{21m}\cdot\frac{5m}{5m}=\frac{15mn}{105m^{2}}\end{array}[/latex]

Add the numerators and keep the denominator the same.

[latex]\frac{14n}{105{{m}^{2}}}+\frac{15mn}{105{{m}^{2}}}=\frac{14n+15mn}{105{{m}^{2}}}[/latex]

If possible, simplify by finding common factors in the numerator and denominator. This rational expression is already in simplest form because the numerator and denominator have no factors in common.

[latex]\displaystyle \frac{n(14+15m)}{105{{m}^{2}}}[/latex]

Answer

[latex]\displaystyle \frac{2n}{15{{m}^{2}}}+\frac{3n}{21m}=\frac{n(14+15m)}{105{{m}^{2}}},m\ne 0[/latex]

Find the prime factorization of each denominator. [latex]t+1[/latex] cannot be factored any further, but [latex]{{t}^{2}}-t-2[/latex] can be. Remember that t cannot be [latex]-1[/latex] or [latex]2[/latex] because the denominators would be 0.

[latex]\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\end{array}[/latex]

Find the least common multiple. [latex]t+1[/latex] appears exactly once in both of the expressions, so it will appear once in the least common denominator. [latex]t–2[/latex] also appears once.

This means that [latex]\left(t-2\right)\left(t+1\right)[/latex] is the least common multiple. In this case, it is easier to leave the common multiple in terms of the factors, so you will not multiply it out.

Use the least common multiple for your new common denominator, it will be the LCD.

[latex]\begin{array}{c}t+1=t+1\\t^{2}-t-2=\left(t-2\right)\left(t+1\right)\\\text{LCM}:\left(t+1\right)\left(t-1\right)\end{array}[/latex]

Compare each original denominator and the new common denominator. Now rewrite the rational expressions to each have the common denominator of [latex]\left(t+1\right)\left(t–2\right)[/latex].

You need to multiply [latex]t+1[/latex] by [latex]t–2[/latex] to get the LCD, so multiply the entire rational expression by [latex]\displaystyle \frac{t-2}{t-2}[/latex].

The second expression already has a denominator of [latex]\left(t+1\right)\left(t–2\right)[/latex], so you do not need to multiply it by anything.

[latex]\begin{array}{c}\frac{2}{t+1}\cdot \frac{t-2}{t-2}=\frac{2(t-2)}{(t+1)(t-2)}\\\\\,\,\,\frac{t-2}{{{t}^{2}}-t-2}=\frac{t-2}{(t+1)(t-2)}\end{array}[/latex]

Then rewrite the subtraction problem with the common denominator.

[latex]\frac{2\left(t-2\right)}{\left(t+1\right)\left(t-2\right)}-\frac{t-2}{\left(t+1\right)\left(t-2\right)}[/latex]

Subtract the numerators and simplify. Remember that parentheses need to be included around the second [latex]\left(t–2\right)[/latex] in the numerator because the whole quantity is subtracted. Otherwise you would be subtracting just the t.

[latex]\begin{array}{c}\frac{2(t-2)-(t-2)}{(t+1)(t-2)}\\\\\frac{2t-4-t+2}{(t+1)(t-2)}\\\\\frac{t-2}{(t+1)(t-2)}\end{array}[/latex]

The numerator and denominator have a common factor of [latex]t–2[/latex], so the rational expression can be simplified.

[latex]\large\begin{array}{c}\frac{\cancel{t-2}}{(t+1)\cancel{(t-2)}}\\\\=\frac{1}{t+1}\end{array}[/latex]

Answer

[latex]\displaystyle \frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}=\frac{1}{t+1},t\ne -1,2[/latex]

Neither [latex]2y–1[/latex] nor [latex]y–5[/latex] can be factored. Because they have no common factors, the least common multiple, which will become the least common denominator, is the product of these denominators.

[latex]\text{LCM}=\left(2y-1\right)\left(y-5\right)[/latex]

Multiply each expression by the equivalent of 1 that will give it the common denominator.

Then rewrite the subtraction problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.

[latex]\begin{array}{c}\frac{3y}{2y-1}\cdot \frac{y-5}{y-5}=\frac{3y(y-5)}{(2y-1)(y-5)}\\\\\frac{4}{y-5}\cdot \frac{2y-1}{2y-1}=\frac{4(2y-1)}{(2y-1)(y-5)}\\\\\frac{3y(y-5)}{(2y-1)(y-5)}-\frac{4(2y-1)}{(2y-1)(y-5)}\end{array}[/latex]

Subtract and simplify.

[latex]\begin{array}{c}\frac{3{{y}^{2}}-15y}{(2y-1)(y-5)}-\frac{8y-4}{(2y-1)(y-5)}\\\\\frac{3{{y}^{2}}-15y-(8y-4)}{(2y-1)(y-5)}\\\\\frac{3{{y}^{2}}-15y-8y+4}{(2y-1)(y-5)}\end{array}[/latex]

The domain is found by setting the original denominators equal to zero.

[latex]\begin{array}{l}2y-1=0\text{ and }y-5=0\\\,\,\,y=\frac{1}{2}\,\,\,\,\,\,\,\text{ and }y=5\end{array}[/latex]

The domain is [latex]y\ne\frac{1}{2}, y\ne5[/latex]

Answer

[latex]\displaystyle \frac{3y}{2y-1}-\frac{4}{y-5}=\frac{3{{y}^{2}}-23y+4}{2{{y}^{2}}-11y+5},y\ne \frac{1}{2},5[/latex]

Find the least common multiple by factoring each denominator. Multiply each factor the maximum number of times it appears in a single factorization. Remember that x cannot be [latex]2[/latex] or [latex]-2[/latex] because the denominators would be 0.

[latex]\left(x+2\right)[/latex] appears a maximum of one time, as does [latex]\left(x–2\right)[/latex]. This means the LCM is [latex]\left(x+2\right)\left(x–2\right)[/latex].

[latex]\begin{array}{l}x^{2}-4=\left(x+2\right)\left(x-2\right)\\\,\,x-2=x-2\\\,\,x+2=x+2\\\,\,\text{LCM}=\left(x+2\right)\left(x-2\right)\end{array}[/latex]

The LCM becomes the common denominator.

Multiply each expression by the equivalent of 1 that will give it the common denominator.

[latex]\begin{array}{r}\frac{2{{x}^{2}}}{{{x}^{2}}-4}=\frac{2{{x}^{2}}}{(x+2)(x-2)}\\\frac{x}{x-2}\cdot \frac{x+2}{x+2}=\frac{x(x+2)}{(x+2)(x-2)}\\\frac{1}{x+2}\cdot \frac{x-2}{x-2}=\frac{1(x-2)}{(x+2)(x-2)}\end{array}[/latex]

Rewrite the original problem with the common denominator. It makes sense to keep the denominator in factored form in order to check for common factors.

[latex]\displaystyle \frac{2{{x}^{2}}}{(x+2)(x-2)}+\frac{x(x+2)}{(x+2)(x-2)}-\frac{1(x-2)}{(x+2)(x-2)}[/latex]

Combine the numerators.

[latex]\begin{array}{c}\frac{2{{x}^{2}}+x(x+2)-1(x-2)}{(x+2)(x-2)}\\\\\frac{2{{x}^{2}}+{{x}^{2}}+2x-x+2}{(x+2)(x-2)}\end{array}[/latex]

Check for simplest form. Since neither [latex]\left(x+2\right)[/latex] nor [latex]\left(x-2\right)[/latex] is a factor of [latex]3{{x}^{2}}+x+2[/latex], this expression is in simplest form.

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

Answer

[latex]\displaystyle \frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}-\frac{1}{x+2}=\frac{3{{x}^{2}}+x+2}{(x+2)(x-2)}[/latex][latex]\displaystyle x\ne 2,-2[/latex]

Find the least common multiple by factoring each denominator. Multiply each factor the maximum number of times it appears in a single factorization.

[latex]\begin{array}{l}\,\,\,\,\,\,\,\,3y=3\cdot{y}\\\,\,\,\,\,\,\,\,\,\,x=x\\\,\,\,\,\,\,\,\,\,\,9=3\cdot3\\\text{LCM}=3\cdot3\cdot{x}\cdot{y}\\\text{LCM}=9xy\end{array}[/latex]

The LCM becomes the common denominator. Multiply each expression by the equivalent of 1 that will give it the common denominator.

[latex]\begin{array}{r}\frac{{{y}^{2}}}{3y}\cdot \frac{3x}{3x}=\frac{3x{{y}^{2}}}{9xy}\\\\\frac{2}{x}\cdot \frac{9y}{9y}=\frac{18y}{9xy}\,\,\\\\\frac{15}{9}\cdot \frac{xy}{xy}=\frac{15xy}{9xy}\end{array}[/latex]

Rewrite the original problem with the common denominator.

[latex]\frac{3x{{y}^{2}}}{9xy}-\frac{18y}{9xy}-\frac{15xy}{9xy}[/latex]

Combine the numerators.

[latex]\frac{3x{{y}^{2}}-18y-15xy}{9xy}[/latex]

Check for simplest form.

[latex]\large\begin{array}{c}\frac{3y(xy-6-5x)}{9xy}\\\\=\frac{3y(xy-6-5x)}{3y(3x)}\\\\=\frac{\cancel{3y}(xy-6-5x)}{\cancel{3y}(3x)}\\\\=\frac{xy-6-5x}{3x}\end{array}[/latex]

The domain is found by setting the denominators equal to zero. [latex]9=0[/latex] is nonsense, so we don’t need to worry about that denominator.

[latex]\begin{array}{l}3y=0\text{ and }x=0\\y=0\text{ and }x=0\end{array}[/latex]

The domain is [latex]y\ne0, x\ne0[/latex]

Answer

[latex]\frac{{{y}^{2}}}{3y}-\frac{2}{x}-\frac{15}{9}=\frac{xy-5x-6}{3x},y\ne 0,x\ne 0[/latex]