How do you solve $\frac{1}{2} (- 4 + 6 x) = \frac{1}{3} x + \frac{2}{3} (x + 9)$ $\frac { 1 } { 2 } ( - 4 + 6 x ) = \frac { 1 } { 3 } x + \frac { 2 } { 3 } ( x + 9 )$ ?

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How do you solve $\frac{1}{2} (- 4 + 6 x) = \frac{1}{3} x + \frac{2}{3} (x + 9)$ $\frac { 1 } { 2 } ( - 4 + 6 x ) = \frac { 1 } { 3 } x + \frac { 2 } { 3 } ( x + 9 )$ ?

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Answer 1 · 2018-04-01T21:43:30

x=4

Explanation:

First, use distributive property on the $(\frac{1}{2}) (- 4 + 6 x)$ $(1/2)(-4+6x)$ and $(\frac{2}{3}) (x + 9)$ $(2/3)(x+9)$ .

$(\frac{1}{2}) (- 4 + 6 x) = (\frac{1}{2}) \cdot - 4 + (\frac{1}{2}) \cdot 6 x = - 2 + 3 x$ $(1/2)(-4+6x) = (1/2)*-4+(1/2)*6x = -2+3x$
$(\frac{2}{3}) (x + 9) = (\frac{2}{3}) \cdot x + (\frac{2}{3}) \cdot 9 = \frac{2}{3} x + 6$ $(2/3)(x+9) = (2/3)*x+(2/3)*9 = 2/3x+6$

The equation is now $- 2 + 3 x = \frac{1}{3} x + \frac{2}{3} x + 6$ $-2+3x=1/3x+2/3x+6$ .

Combine like terms and make it "prettier".

$3 x - 2 = x + 6$ $3x-2=x+6$ .

Now we've broken it down into something much more simple!

Subtract $6$ $6$ from both sides to get
$3 x - 2 - 6 = x + 6 - 6 \Rightarrow 3 x - 8 = x$ $3x-2-6=x+6-6 => 3x-8=x$

Subtract $x$ $x$ from both sides to get
$3 x - x - 8 = x - x \Rightarrow 2 x - 8 = 0$ $3x-x-8 = x-x => 2x-8=0$

Add $8$ $8$ to both sides to get
$2 x - 8 + 8 = 0 + 8 \Rightarrow 2 x = 8$ $2x-8+8=0+8 => 2x=8$

Final step: Divide $2$ $2$ from both sides to get
$2 \frac{x}{2} = \frac{8}{2} \Rightarrow x = 4$ $2x/2=8/2 => x=4$

Therefore, $x = 4$ $x=4$ .

Yay!

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How do you solve $\frac{1}{2} (- 4 + 6 x) = \frac{1}{3} x + \frac{2}{3} (x + 9)$ $\frac { 1 } { 2 } ( - 4 + 6 x ) = \frac { 1 } { 3 } x + \frac { 2 } { 3 } ( x + 9 )$ ?

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How do you solve 12(−4+6x)=13x+23(x+9)\frac { 1 } { 2 } ( - 4 + 6 x ) = \frac { 1 } { 3 } x + \frac { 2 } { 3 } ( x + 9 )?

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Explanation:

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How do you solve $\frac{1}{2} (- 4 + 6 x) = \frac{1}{3} x + \frac{2}{3} (x + 9)$ $\frac { 1 } { 2 } ( - 4 + 6 x ) = \frac { 1 } { 3 } x + \frac { 2 } { 3 } ( x + 9 )$ ?