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) To add the fractions \[ \frac{x}{4} \quad \text{and} \quad \frac{3}{7}, \] we first find a common denominator. The least common denominator (LCD) of \(4\) and \(7\) is \(28\). - Multiply the numerator and denominator of \(\frac{x}{4}\) by \(7\): \[ \frac{x}{4} = \frac{7x}{28}. \] - Multiply the numerator and denominator of \(\frac{3}{7}\) by \(4\): \[ \frac{3}{7} = \frac{12}{28}. \] Now, add the fractions: \[ \frac{7x}{28} + \frac{12}{28} = \frac{7x+12}{28}. \] --- (b) To add the fractions \[ \frac{2x}{4+x} \quad \text{and} \quad \frac{x}{3}, \] the common denominator is \(3(4+x)\). - Rewrite \(\frac{2x}{4+x}\) with the denominator \(3(4+x)\) by multiplying numerator and denominator by \(3\): \[ \frac{2x}{4+x} = \frac{6x}{3(4+x)}. \] - Rewrite \(\frac{x}{3}\) with the denominator \(3(4+x)\) by multiplying numerator and denominator by \(4+x\): \[ \frac{x}{3} = \frac{x(4+x)}{3(4+x)}. \] 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: \[ x(4+x) = 4x+x^2. \] Thus, \[ 6x + 4x + x^2 = x^2+10x. \] The simplified expression is: \[ \frac{x^2+10x}{3(4+x)}. \] Alternatively, factor the numerator if desired: \[ \frac{x(x+10)}{3(4+x)}. \] --- (c) Simplify the product \[ \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1}. \] Notice that the denominator \(a^2-1\) factors as: \[ a^2-1 = (a-1)(a+1). \] Substitute the factorization into the expression: \[ \frac{a+1}{b} \times \frac{b^2}{(a-1)(a+1)}. \] Cancel the common factor \((a+1)\): \[ \frac{b^2}{b(a-1)}. \] Simplify by canceling one factor of \(b\): \[ \frac{b}{a-1}. \] --- Final Answers: (a) \(\displaystyle \frac{7x+12}{28}\) (b) \(\displaystyle \frac{x^2+10x}{3(4+x)}\) or \(\displaystyle \frac{x(x+10)}{3(4+x)}\) (c) \(\displaystyle \frac{b}{a-1}\)