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Question #ad88d

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Jun 27, 2017

$x = \frac{32}{7}$ $x=32/7$

Explanation:

The first step is to multiply the $\frac{1}{2}$ $1/2$ term through the $(x - 6)$ $(x-6)$ term.

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

Because one half of six is three, we can rewrite it as

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

Next, multiply the $\frac{1}{4}$ $1/4$ term through the $(x - 12)$ $(x-12)$ term.

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

Because one quarter of 12 is 3, we can rewrite it as

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

You can combine the $\frac{1}{2} x$ $1/2x$ and the $\frac{1}{4} x$ $1/4x$ . Because

$\frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$ $1/2+1/4=2/4 + 1/4=3/4$ , we can rewrite our equation as

$\frac{3}{4} x - 3 - 3 = 2 - x$ $3/4x - 3 -3=2-x$

Because negative 3 minus 3 is negative six, we can rewrite

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

Add $6$ $6$ to both sides

$\frac{3}{4} x = 8 - x$ $3/4x=8-x$

Add $x$ $x$ to both sides

$\frac{3}{4} x + x = 8$ $3/4x+x=8$

Because $x$ $x$ is really just $1 \times x$ $1xxx$ , we can rewrite as

$\frac{3}{4} x + 1 \times x = 8$ $3/4x+1xxx=8$

$\frac{3}{4} x + \frac{4}{4} x = 8$ $3/4x+4/4x=8$

$\frac{7}{4} x = 8$ $7/4x=8$

Multiply both sides by $4$ $4$

$7 x = 32$ $7x=32$

Divide both sides by 7

$x = \frac{32}{7}$ $x=32/7$

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