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)** Combine the fractions by finding a common denominator. The denominators are \(4\) and \(7\), so the common denominator is \(28\): \[ \frac{x}{4} = \frac{7x}{28},\quad \frac{3}{7} = \frac{12}{28} \] Thus, the sum becomes: \[ \frac{x}{4} + \frac{3}{7} = \frac{7x}{28} + \frac{12}{28} = \frac{7x+12}{28} \] --- **(b)** To add \(\frac{2x}{4+x}\) and \(\frac{x}{3}\), find a common denominator which is \(3(4+x)\): \[ \frac{2x}{4+x} = \frac{2x \cdot 3}{3(4+x)} = \frac{6x}{3(4+x)} \] \[ \frac{x}{3} = \frac{x(4+x)}{3(4+x)} \] Now add the fractions: \[ \frac{2x}{4+x} + \frac{x}{3} = \frac{6x + x(4+x)}{3(4+x)} \] Simplify the numerator: \[ 6x + x(4+x) = 6x + 4x + x^2 = x^2 + 10x \] So the expression becomes: \[ \frac{x^2+10x}{3(4+x)} \] Factor \(x\) from the numerator: \[ \frac{x^2+10x}{3(4+x)} = \frac{x(x+10)}{3(4+x)} \] --- **(3)** Simplify the product: \[ \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} \] Factor the difference of squares in the denominator: \[ a^{2}-1 = (a+1)(a-1) \] Substitute back into the expression: \[ \frac{a+1}{b} \times \frac{b^{2}}{(a+1)(a-1)} \] Cancel the common factor \(a+1\) (assuming \(a \neq -1\)): \[ = \frac{b^{2}}{b(a-1)} = \frac{b}{a-1} \]