How do you solve $\frac{1}{12} = \frac{3}{2} x - 1 + \frac{2}{3} x$ $\frac{1}{12} = \frac{3}{2} x - 1+ \frac{2}{3} x$ ?

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Answer 1 · 2016-09-20T14:43:21

equalize denominators and then solve the equation

$\frac{1}{12} = \frac{3 x}{2} - \frac{3}{3} + \frac{2 x}{3}$ $1/12 = (3x)/2 - 3/3 + (2x)/3$

$\frac{1}{12} = \frac{9 x}{6} - \frac{6}{6} + \frac{4 x}{6}$ $1/12 = (9x)/6 - 6/6 + (4x)/6$

$\frac{1}{12} = \frac{18 x}{12} - \frac{12}{12} + \frac{8 x}{12}$ $1/12 = (18x)/12 -12/12 + (8x)/12$

$\frac{1}{12} = \frac{26 x - 12}{12}$ $1/12 = (26x-12)/12$

$1 = 26 x - 12$ $1 = 26x - 12$

$26 x = 13$ $26x = 13$

$x = \frac{13}{26}$ $x = 13 / 26$

$x = \frac{1}{2}$ $x = 1 / 2$

How do you solve 112=32x−1+23x\frac{1}{12} = \frac{3}{2} x - 1+ \frac{2}{3} x?