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

To simplify \( \frac{x}{4}+\frac{3}{7} \), you'll want to find a common denominator, which in this case is 28. Thus, you convert \( \frac{x}{4} \) into \( \frac{7x}{28} \) and \( \frac{3}{7} \) into \( \frac{12}{28} \). Now you can combine them: \( \frac{7x+12}{28} \). For \( \frac{2x}{4+x}+\frac{x}{3} \), the common denominator is \( 3(4+x) \). So, you adjust \( \frac{2x}{4+x} \) to \( \frac{6x}{3(4+x)} \) and \( \frac{x}{3} \) to \( \frac{x(4+x)}{3(4+x)} \). Combining gives you \( \frac{6x + x(4 + x)}{3(4 + x)} \), which can be further simplified. In the expression \( \frac{a+1}{b} \times \frac{b^{2}}{a^{2}-1} \), recognize that \( a^{2}-1 \) factors to \( (a-1)(a+1) \). So after canceling, you will end up with \( \frac{b}{a-1} \), as \( b \) remains in the numerator and \( (a+1) \) cancels out.