Respuesta :
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 1
Lets 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.
