Answer:
Here, the given expression,
\(\frac{3}{x^2+5x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}\),
\(\frac{3}{x^2+3x+2x+6}+\frac{x-1}{x+2}=\frac{7}{x+3}\)
\(\frac{3}{x(x+3)+2(x+3)}+\frac{x-1}{x+2}=\frac{7}{x+3}\)
\(\frac{3}{(x+3)(x+2)}+\frac{x-1}{x+2}=\frac{7}{x+3}\)
\(\frac{3+(x-1)(x+3)}{(x+3)(x+2)} = \frac{7}{x+3}\)
\(3+(x-1)(x+3)=7(x+2)\)
\(3+x^2-x+3x-3=7x+14\)
\(x^2+2x=7x+14\)
\(x^2+2x-7x-14=0\)
\(x^2-5x-14=0\)
\(x^2-7x+2x-14=0\)
\(x(x-7)+2(x-7)=0\)
\((x+2)(x-7)=0\)
\(\implies x = -2\text{ or } x = 7\)
For x = -2,
The right side of the given equation is not defined,
⇒ x = -2 can not be the solution of the given equation.
While for x = 7,
\(\frac{3}{7^2+5\times 7+6}+\frac{7-1}{7+2}=\frac{7}{7+3}\),
Hence, the solution of the given equation is,
x = 7.
