Question

resolva por favor essas duas equações ​

Answer 1

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

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

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

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

$\frac{6+2x-4+4x}{4}= \frac{x-1}{4}$

$6+2x-4+4x=x-1$

$2x+4x-x=-1-6+4$

$5x=-3$

$x=-\frac{3}{5}$

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

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

$\frac{4x-4}{6}+\frac{9+9x}{6}=\frac{3}{6}-\frac{2x-2}{6}$

$(4x-4)+(9+9x)=3-(2x-2)$

$4x-4+9+9x=3-2x+2$

$4x+9x+2x=3+2+4-9$

$15x=0$

$x=0$