6. Simplify the ff. (a) \( \frac{x}{4}+\frac{3}{7} \) b) \( \frac{2 x}{4+x}+\frac{x}{3} \) (3) \( \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} \)

**(a) Simplify \( \frac{x}{4} + \frac{3}{7} \)** 1. Find the least common denominator (LCD) of 4 and 7, which is \( 28 \). 2. Rewrite each fraction with the denominator \( 28 \): \[ \frac{x}{4} = \frac{7x}{28}, \quad \frac{3}{7} = \frac{12}{28} \] 3. Combine the fractions: \[ \frac{7x}{28} + \frac{12}{28} = \frac{7x+12}{28} \] --- **(b) Simplify \( \frac{2x}{4+x}+\frac{x}{3} \)** 1. Identify the common denominator for the fractions. The denominators are \( 4+x \) and \( 3 \). Their LCD is \( 3(4+x) \). 2. Rewrite each fraction with the common denominator: \[ \frac{2x}{4+x} = \frac{2x \cdot 3}{3(4+x)} = \frac{6x}{3(4+x)} \] \[ \frac{x}{3} = \frac{x(4+x)}{3(4+x)} \] 3. Combine the numerators: \[ \frac{6x + x(4+x)}{3(4+x)} = \frac{6x + 4x + x^2}{3(4+x)} = \frac{x^2 + 10x}{3(4+x)} \] 4. Factor the numerator if desired: \[ \frac{x(x+10)}{3(4+x)} \] --- **(3) Simplify \( \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} \)** 1. Factor the difference of squares in the denominator: \[ a^2 - 1 = \left(a+1\right)\left(a-1\right) \] 2. Substitute the factorization into the expression: \[ \frac{a+1}{b} \times \frac{b^{2}}{(a+1)(a-1)} \] 3. Cancel the common factor \((a+1)\): \[ \frac{b^{2}}{b(a-1)} \] 4. Simplify by canceling one \( b \): \[ \frac{b}{a-1} \] --- **Final Answers:** (a) \( \frac{7x+12}{28} \) (b) \( \frac{x^2+10x}{3(4+x)} \) or \( \frac{x(x+10)}{3(4+x)} \) (3) \( \frac{b}{a-1} \)