Question

$\frac{1}{3x + y} + \frac{1}{3x - y } = \frac{3}{4} \\ \\ \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = - \frac{1}{8}\ \textless \ br /\ \textgreater \ ​$
​

Answer 1

Step-by-step explanation:

$\frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \\ \frac{3x - y + 3x + y}{(3x + y)(3x - y)} = \frac{3}{4} \\ \frac{6x}{( {3x)}^{2} - {y}^{2} } = \frac{3}{4} \\ 12x = 3(9 {x}^{2} - {y}^{2} ) \\ 27 {x}^{2} - 3 {y}^{2} - 12x = 0 \\ 9 {x }^{2} - {y}^{2} - 4x = 0$

2.

$\frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = - \frac{1}{8} \\ \frac{1}{2} ( \frac{3x - y - 3x - y}{(3x + y)(3x - y)} = - \frac{1}{8} \\ \frac{2y}{9 {x}^{2} - {y}^{2} } = - \frac{1}{4} \\ - 8y = 9 {x}^{2} - {y}^{2} \\ 9 {x}^{2} - {y}^{2} + 8y = 0$