Question

Find the value of x, if : $m=\frac{1-3x}{5}$, $n=\frac{1-2x}{3}$ and $3(n-2m)+1=0$.

• A. $-\frac{1}{8}$
• B. $-\frac{1}{3}$
• C. $-\frac{1}{2}$
• D. $-\frac{1}{9}$

Answer 1

The correct option is C $-\frac{1}{2}$

$3(n-2m)+1=0$
$=>3\left[\right(\frac{1-2x}{3})-2(\frac{1-3x}{5}\left)\right]+1=0$
$=>\frac{3-6x}{3}-\frac{2+6x}{5}+1=0$
$=>\frac{15-30x-6-18x+15}{15}=0$
$=>-48x-24=0$
$=>-48x=24$
$=>x=-\frac{24}{48}$
$=>x=-\frac{1}{2}$