Question

Me ajudem por favor


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

resolva as equações do primeiro grau​:

Answer 1

B)

$\frac{x + 3}{2} + \frac{x + 2}{3} = - \frac{1}{2} \\ \\ \frac{3(x + 3)}{3 \times 2} + \frac{2(x + 2)}{2 \times 3} = \frac{ - 1}{2} \\ \\ \frac{3x + 9}{6} + \frac{2x + 4}{6} = \frac{ - 1}{2} \\ \\ \frac{5x + 13}{6} = \frac{ - 1}{2} \\ \\ 2(5x + 13) = 6( - 1) \\ 10x + 26 = - 6 \\ 10x = - 32 \\ x = - \frac{32}{10}$

A)

$\frac{3 + x}{2} - (1 - x) = \frac{ x - 1}{4} = \\ \\ \frac{3 + x}{2} - 1 + x = \frac{x - 1}{4} = \\ \\ \frac{3 + x}{2} - \frac{2}{2} + \frac{2x}{2} = \frac{x - 1}{4} \\ \\ \frac{3 - 2 + x + 2x}{2} = \frac{ x - 1}{4} \\ \\ \frac{3x + 1}{2} = \frac{x - 1}{4} \\ \\ 4(3x + 1) = 2(x - 1) \\ \\ 12x + 4 = 2x - 2 \\ 10x = - 2 - 4 \\ x = \frac{ - 6}{10}$