Step-by-Step Solution

Solve the equation: $\frac{1}{x+1} + \frac{2}{x+2} = \frac{7}{x+5} \quad (x \ne -1, -2, -5)$

Step-by-Step Solution

Step 1: Combine LHS fractions

$\frac{1(x+2) + 2(x+1)}{(x+1)(x+2)} = \frac{7}{x+5}$

$\frac{3x + 4}{x^2 + 3x + 2} = \frac{7}{x+5}$

Step 2: Cross-multiply and expand

$(3x + 4)(x + 5) = 7(x^2 + 3x + 2)$

$3x^2 + 19x + 20 = 7x^2 + 21x + 14$

$4x^2 + 2x - 6 = 0$

Divide by $2$: $2x^2 + x - 3 = 0$

Step 3: Factorize

$(2x + 3)(x - 1) = 0$

$2x + 3 = 0 \implies x = -\frac{3}{2}$ $x - 1 = 0 \implies x = 1$

Step 4: Check solutions

• For $x = 1$: $\dfrac{1}{2} + \dfrac{2}{3} = \dfrac{7}{6} = \text{RHS}$ $\checkmark$
• For $x = -\dfrac{3}{2}$: $\dfrac{1}{-0.5} + \dfrac{2}{0.5} = -2 + 4 = 2$, and $\text{RHS} = \dfrac{7}{3.5} = 2$ $\checkmark$

$\boxed{\text{Solution Set: } \left\{1, \; -\frac{3}{2}\right\}}$