How do you solve $- 3 v - \frac{5}{2} = \frac{7}{2} v - \frac{4}{3}$ $-3v - \frac { 5} { 2} = \frac { 7} { 2} v - \frac { 4} { 3}$ ?

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Answer 1 · 2017-06-22T16:54:56

$v = \frac{23}{39}$ $v=23/39$

First, get rid of the fractions by multiplying everything by a number which a multiple of both $2$ $2$ and $3$ $3$ , say $6$ $6$ :

$6 \times (- 3 v - \frac{5}{2}) = 6 \times (\frac{7}{2} v - \frac{4}{3})$ $color(red)(6)xx(-3v-5/2)=color(red)(6)xx(7/2v-4/3)$

Distribute the $6$ $6$ through the parentheses on both sides

$6 \times (- 3 v) - 6 \times (\frac{5}{2}) = 6 \times (\frac{7}{2} v) - 6 \times (\frac{4}{3})$ $6xx(-3v)-6xx(5/2)=6xx(7/2v)-6xx(4/3)$

$- 18 v - 3 \times 5 = 3 \times 7 v - 2 \times 4$ $-18v-3xx5=3xx7v-2xx4$

$- 18 v - 15 = 21 v - 8$ $-18v-15=21v-8$

Add $18 v$ $18v$ to both sides

$- 18 v + 18 v - 15 = 21 v + 18 v - 8$ $-18vcolor(red)(+18v)-15=21vcolor(red)(+18v)-8$

$15 = 39 v - 8$ $15=39v-8$

Add $8$ $8$ to both sides, to the variable $v$ $v$ "more alone"

$15 + 8 = 39 v - 8 + 8$ $15color(red)(+8)=39v-8color(red)(+8)$

$23 = 39 v$ $23=39v$

Divide both sides by $39$ $39$ to isolate the $v$ $v$

$23/color(red)(39)=(cancel(39)v)/color(red)(cancel(39))$

This leaves

$v=23/39$

Because $23$ is a prime number, this fraction is reduced to lowest terms.

How do you solve -3v - \frac { 5} { 2} = \frac { 7} { 2} v - \frac { 4} { 3}−3v−52=72v−43-3v - \frac { 5} { 2} = \frac { 7} { 2} v - \frac { 4} { 3}?