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

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Answer 1 · 2015-06-01T20:00:19

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

We are going to put it all on six by multiplying both denominators and numerators

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

We substract $\frac{2}{6}$ $2/6$ on both sides

$\frac{3 x}{6} - \frac{9}{6} - \frac{2}{6} = \frac{4 x}{6}$ $(3x)/6-9/6-2/6=(4x)/6$

$\frac{3 x}{6} - \frac{11}{6} = \frac{4 x}{6}$ $(3x)/6-11/6=(4x)/6$

We substract $\frac{3 x}{6}$ $(3x)/6$ on both side

$- \frac{11}{6} = \frac{4 x}{6} - \frac{3 x}{6}$ $-11/6=(4x)/6-(3x)/6$

$- \frac{11}{6} = \frac{x}{6}$ $-11/6=x/6$

We multiply both side by six to get $x$ $x$

$x = - 11$ $x=-11$

How do you solve 12(x−3)=13(2x+1)1/2 (x-3) = 1/3 (2x+1)?