Find values of a and b that make the following equality into identity:x−1/(x+1)(x−4) = a/x+1 + b/x−4

The value of a and b in given expression must be [tex]\frac{2}{5} \text { and } \frac{3}{5}[/tex] respectively so that given equality becomes identity.Solution:Need to find the value of a and b in following expression so that following equality will become identity.[tex]\frac{(x-1)}{(x+1)(x-4)}=\frac{a}{(x+1)}+\frac{b}{(x-4)}[/tex]   ------- eqn 1Lets Simplify Right hand Side first,[tex]\frac{a}{(x+1)}+\frac{b}{(x-4)}=\frac{a(x-4)+b(x+1)}{(x+1)(x-4)}[/tex][tex]\begin{array}{l}{=\frac{a x-4 a+b x+b}{(x+1)(x-4)}} \\\\ {=\frac{a x+b x-4 a+b}{(x+1)(x-4)}} \\\\ {=\frac{(a+b) x-4 a+b}{(x+1)(x-4)}}\end{array}[/tex][tex]=>\frac{a}{(x+1)}+\frac{b}{(x-4)}=\frac{(a+b) x-4 a+b}{(x+1)(x-4)}[/tex][tex]\text {On substituting } \frac{a}{(x+1)}+\frac{b}{(x-4)}=\frac{(a+b) x-4 a+b}{(x+1)(x-4)} \text { in equation } 1 \text { we get }[/tex][tex]\frac{(x-1)}{(x+1)(x-4)}=\frac{(a+b) x-4 a+b}{(x+1)(x-4)}[/tex]On multiplying both sides by (x+1)(x-4) we get[tex]\frac{(x-1)}{(x+1)(x-4)} \times(x+1)(x-4)=\frac{(a+b) x-4 a+b}{(x+1)(x-4)} \times(x+1)(x-4)[/tex][tex]\Rightarrow x-1=(a+b) x-(4 a-b)[/tex]On comparing coefficient of x and constant term separately, we get a + b = 1 and 4a - b = 1 On adding the two equations we get a + b + 4a - b = 1 + 1 => 5a = 2 => [tex]a = \frac{2}{5}[/tex][tex]\text {Substituting } \mathrm{a}=\frac{2}{5} \text { in equation } a+b=1, \text { we get }[/tex][tex]\begin{array}{l}{\frac{2}{5}+b=1} \\\\ {\Rightarrow b=1-\frac{2}{5}=\frac{3}{5}}\end{array}[/tex]So the value of a and b in given expression must be [tex]\frac{2}{5} \text { and } \frac{3}{5}[/tex] so that given equality becomes identity.