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  [latex]2[/latex],[latex]3[/latex], and [latex]5[/latex]. Multiply each number the maximum number of times it appears in a single factorization.

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

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

[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  [latex]6, 10[/latex], and [latex]4[/latex] is [latex]60[/latex].

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. [latex]3[/latex] appears exactly once in both of the expressions, so it will appear once in the least common multiple. Both [latex]5[/latex] and [latex]7[/latex] 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 [latex]0[/latex].

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 [latex]7[/latex] 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]

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

Now we have to find the LCD. Since x appears once and y 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.

[latex]\frac{5}{x}+\frac{6}{y}=\frac{(6x+5y)}{xy}[/latex], [latex]x\ne0 ,\text{ and }y\ne0[/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 [latex]0[/latex].

[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]

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]\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-2,2[/latex]

[latex]\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}=\frac{4(x-4)}{(x+2)^2(x-2)}[/latex],   [latex]x\ne-2,2[/latex]