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 [latex]1[/latex] 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 [latex]0[/latex].

[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 [latex]1[/latex] 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 [latex]1[/latex] 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]