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

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Answer 1 · 2016-12-26T05:08:26

$x = - \frac{7}{15}$ $x=-7/15$

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

Multiply the equation by the LCD 12.

$12 \cdot [\frac{- 1 - 3 x}{4} - \frac{2 x + x}{6} = \frac{1}{3}]$ $12*[frac(-1-3x)(4)-frac(2x+x)(6)=1/3]$

$\frac{12}{4} \cdot (- 1 - 3 x) - \frac{12}{6} \cdot (2 x + x) = \frac{12}{3}$ $12/4 * (-1-3x)-12/6*(2x+x)=12/3$

$3 (- 1 - 3 x) - 2 (2 x + x) = 4$ $3(-1-3x) -2(2x+x)=4$

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

$- 15 x - 3 = a 4$ $-15x-3=color(white)a4$
$a a a a a + 3 a + 3$ $color(white)(aaaaa)+3color(white)(a)+3$

$- 15 x = 7$ $-15x=7$

$\frac{- 15 x}{- 15} = \frac{7}{- 15}$ $(-15x)/-15=7/-15$

$x = - \frac{7}{15}$ $x=-7/15$

How do you solve \frac { - 1- 3x } { 4} - \frac { 2x + x } { 6} = \frac { 1} { 3}−1−3x4−2x+x6=13\frac { - 1- 3x } { 4} - \frac { 2x + x } { 6} = \frac { 1} { 3}?