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)** We combine the fractions by finding a common denominator. The denominators are \(4\) and \(7\); hence, the least common denominator is \(28\). \[ \frac{x}{4} + \frac{3}{7} = \frac{7x}{28} + \frac{12}{28} = \frac{7x+12}{28} \] --- **(b)** The given expression is: \[ \frac{2x}{4+x} + \frac{x}{3} \] The denominators are \((4+x)\) and \(3\); hence, the common denominator is \(3(4+x)\). Rewrite each fraction: \[ \frac{2x}{4+x} = \frac{2x \cdot 3}{3(4+x)} = \frac{6x}{3(4+x)} \] \[ \frac{x}{3} = \frac{x(4+x)}{3(4+x)} \] Now add the fractions: \[ \frac{6x}{3(4+x)} + \frac{x(4+x)}{3(4+x)} = \frac{6x + x(4+x)}{3(4+x)} \] Expand the numerator: \[ 6x + x(4+x) = 6x + 4x + x^2 = x^2 + 10x \] Thus, the simplified expression is: \[ \frac{x^2 + 10x}{3(4+x)} \] Optionally, factor \(x\) from the numerator: \[ \frac{x(x+10)}{3(4+x)} \] --- **(3)** We need to simplify: \[ \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} \] First, factor the difference of squares in the denominator: \[ a^{2}-1 = (a-1)(a+1) \] Now substitute: \[ \frac{a+1}{b} \times \frac{b^2}{(a-1)(a+1)} \] Cancel the common factor \(a+1\) (assuming \(a \neq -1\)): \[ \frac{\cancel{a+1}}{b} \times \frac{b^2}{(a-1)\cancel{a+1}} = \frac{b^2}{b(a-1)} \] Cancel a factor of \(b\) (assuming \(b\neq0\)): \[ \frac{b}{a-1} \] --- Final Answers: - **(a)** \(\displaystyle \frac{7x+12}{28}\) - **(b)** \(\displaystyle \frac{x(x+10)}{3(4+x)}\) - **(3)** \(\displaystyle \frac{b}{a-1}\)