Some of the work in solving the problem is shown below. 3(x+ a)-1(x+b) (x + 1)(x+2) 3 x+1 What are the values of 'a', 'b', 'c', and 'd'? a= b= C= 1 x+2 d= cx + d (x + 1)(x+2)

Given that $\frac{3}{x+1}-\frac{1}{x+2}=\frac{3\left(x+a\right)-1\left(x+b\right)}{\left(x+1\right)\left(x+2\right)}=\frac{cx+d}{\left(x+1\right)\left(x+2\right)}......\left(1\right)$.

We have to find the value of $a, b, c and d$.

Simplify the expression $\frac{3}{x+1}-\frac{1}{x+2}$

$\begin{array}{c}\frac{3}{x+1}-\frac{1}{x+2}=\frac{3\left(x+2\right)-1\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}\\ =\frac{3x+6-x-1}{\left(x+1\right)\left(x+2\right)}\\ =\frac{2x+5}{\left(x+1\right)\left(x+2\right)}\end{array}$

Hence, $\frac{3}{x+1}-\frac{1}{x+2}=\frac{2x+5}{\left(x+1\right)\left(x+2\right)}......\left(2\right)$.