Question

$\frac{2x+3}{5}- \frac{1}{2}(x- \frac{3}{4}) = \frac{x}{8} + \frac{6}{5}$

Answer 1

$\frac{2x+3}{5}- \frac{1}{2}x+ \frac{3}{8}= \frac{x}{8} + \frac{6}{5} \\ \\ \frac{2x+3}{5}- \frac{6}{5} - \frac{1}{2} x= \frac{x}{8}- \frac{3}{8} \\ \\ \frac{2x+3-6}{5} - \frac{1}{2}x= \frac{x-3}{8} \\ \\ \frac{2(2x+3)-5x}{10} = \frac{x-3}{8} \\ \\ 4x-6-5x= \frac{10(x-3)}{8} \\ \\ 8(-x-6)=10x-30 \\ -8x-48=10x-30 \\ 18x=-18 \\ x=-1$

Se eu não errei nenhum sinal :)

Answer 2

$\dfrac{2x + 3}{5} - \dfrac{1}{2} (x - \dfrac{3}{4} ) = \dfrac{x}{8} + \dfrac{6}{5} \\ \\ \\ \dfrac{2x + 3}{5} - \dfrac{1}{2} ( \dfrac{4x - 3}{4}) = \dfrac{5x + 48}{40} \\ \\ \\ \dfrac{2x + 3}{5} - \dfrac{4x - 3}{8} - \dfrac{5x + 48}{40} = 0\\ \\ \\ -\dfrac{4x - 39}{40} - \dfrac{5x + 48}{40} = 0 \\ \\ \\ -(4x - 39) - 5x + 48 = 0 \\ \\ \\ -(9x + 9) = 0 (-1) \\ \\ \\ 9x + 9 = 0 \\ \\ 9x = -9 \\ \\ \\ x = -\dfrac{9}{9} \\ \\ \\ => x = -1$