Given:
[tex]$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}[/tex]
To find:
The simplified rational expression by subtraction.
Solution:
Let us factor [tex]x^2-1[/tex]. It can be written as [tex]x^2-1^2[/tex].
[tex]x^2-1^2=(x-1)(x+1)[/tex] using algebraic identity.
[tex]$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}=\frac{x+4}{x-1}-\frac{5}{(x+1)(x-1)}[/tex]
LCM of [tex]x-1,(x+1)(x-1)=(x+1)(x-1)[/tex]
Make the denominators same using LCM.
Multiply and divide the first term by (x + 1) to make the denominator same.
[tex]$=\frac{(x+4)(x+1)}{(x-1)(x+1)}-\frac{5}{(x-1)(x+1)}[/tex]
Now, denominators are same, you can subtract the fractions.
[tex]$=\frac{(x+4)(x+1)-5}{(x-1)(x+1)}[/tex]
Expand [tex](x+4)(x+1)-5[/tex].
[tex]$=\frac{x^2+4x+x+4-5}{(x-1)(x+1)}[/tex]
[tex]$=\frac{x^{2}+5 x-1}{(x-1)(x+1)}[/tex]
[tex]$\frac{x+4}{x-1}-\frac{5}{x^{2}-1}=\frac{x^{2}+5 x-1}{(x-1)(x+1)}[/tex]
