Question

1. \\(\frac{1}{b^2} + \frac{4b + 3}{b^2} = \frac{1}{b}\\)

Explanation:

Step1: Combine fractions on left

Since denominators are same ($b^2$), add numerators: $\frac{1 + 4b + 3}{b^2} = \frac{1}{b}$ → $\frac{4b + 4}{b^2} = \frac{1}{b}$

Step2: Simplify left fraction

Factor numerator: $\frac{4(b + 1)}{b^2} = \frac{1}{b}$

Step3: Cross - multiply (b ≠ 0)

Multiply both sides by $b^2$: $4(b + 1) = b$

Step4: Solve for b

Expand: $4b + 4 = b$ → Subtract $b$: $3b + 4 = 0$ → Subtract 4: $3b = - 4$ → $b = -\frac{4}{3}$
Check: Denominators $b^2$ and $b$ not zero when $b = -\frac{4}{3}$.

Answer:

$b = -\frac{4}{3}$