Question 6 Simplify, by writing as one fraction: a) \( \frac{1-3 x}{4}-\frac{2 x+1}{3} \) b) \( \frac{2 x+7}{4}+\frac{2 x-7}{3} \)

To simplify both expressions into one fraction, we first need to find a common denominator. For part (a), the common denominator for 4 and 3 is 12. We rewrite each fraction: \[ \frac{1-3x}{4} = \frac{3(1-3x)}{12} = \frac{3-9x}{12} \] \[ \frac{2x+1}{3} = \frac{4(2x+1)}{12} = \frac{8x+4}{12} \] Now combining them: \[ \frac{3-9x}{12} - \frac{8x+4}{12} = \frac{(3-9x) - (8x+4)}{12} = \frac{-17x - 1}{12} \] For part (b), using the same approach with a common denominator of 12: \[ \frac{2x+7}{4} = \frac{3(2x+7)}{12} = \frac{6x + 21}{12} \] \[ \frac{2x-7}{3} = \frac{4(2x-7)}{12} = \frac{8x - 28}{12} \] Combining these results: \[ \frac{6x + 21}{12} + \frac{8x - 28}{12} = \frac{(6x + 21) + (8x - 28)}{12} = \frac{14x - 7}{12} \] So, the simplified answers are: a) \( \frac{-17x - 1}{12} \) b) \( \frac{14x - 7}{12} \)