Find the least common denominator of [latex]4[/latex] and [latex]8[/latex]. Remember, to find the LCD, identify the greatest number of times each factor appears in each factorization. Here, [latex]2[/latex] appears [latex]3[/latex] times, so [latex]2\cdot2\cdot2[/latex], or [latex]8[/latex], will be the LCD.

Multiply both sides of the equation by the common denominator, [latex]8[/latex], to keep the equation balanced and to eliminate the denominators.

[latex]\begin{array}{r}8\cdot \frac{x+5}{8}=\frac{7}{4}\cdot 8\,\,\,\,\,\,\,\\\\\frac{8(x+5)}{8}=\frac{7(8)}{4}\,\,\,\,\,\,\\\\\frac{8}{8}\cdot (x+5)=\frac{7(4\cdot 2)}{4}\\\\\frac{8}{8}\cdot (x+5)=7\cdot 2\cdot \frac{4}{4}\\\\1\cdot (x+5)=14\cdot 1\,\,\,\end{array}[/latex]

Simplify and solve for x.

[latex]\begin{array}{r}x+5=14\\x=9\,\,\,\end{array}[/latex]

Check the solution by substituting [latex]9[/latex] for x in the original equation.

[latex]\begin{array}{r}\frac{x+5}{8}=\frac{7}{4}\\\\\frac{9+5}{8}=\frac{7}{4}\\\\\frac{14}{8}=\frac{7}{4}\\\\\frac{7}{4}=\frac{7}{4}\end{array}[/latex]

Therefore, [latex]x=9[/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] do not 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]\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}\\\frac{4}{3}=\frac{4}{3}\end{array}[/latex]

Therefore, [latex]x=5[/latex].

Both fractions in the equation have a denominator of [latex]3[/latex]. Multiply every term on both sides of the equation (not just the fractions!) by [latex]3[/latex].

[latex] 3\left( \frac{x}{3}+1 \right)=3\left( \frac{4}{3} \right)[/latex]

Apply the distributive property and multiply [latex]3[/latex] by each term within the parentheses. Then simplify and solve for x.

[latex]\begin{array}{r}3\left( \frac{x}{3} \right)+3\left( 1 \right)=3\left( \frac{4}{3} \right)\\\\\cancel{3}\left( \frac{x}{\cancel{3}} \right)+3\left( 1 \right)=\cancel{3}\left( \frac{4}{\cancel{3}} \right)\\\\ x+3=4\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{-3}\,\,\,\,\,\underline{-3}\\\\x=1\end{array}[/latex]

Therefore, [latex]x=1[/latex].

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

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

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

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 x.

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

Check the solution in the original equation.

[latex]\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]

Therefore, [latex]x=10[/latex].

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

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 m.

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

[latex]\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 [latex]0[/latex], it is an extraneous solution.

Notice, however, that latex]m=4[/latex] is a solution that results in a true statement.

[latex]\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 [latex]0[/latex].

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

Therefore, [latex]m=4[/latex].