Question

Question 4
$x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}$

Answer 1

$x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}\Rightarrow \frac{x}{1}-\frac{8x}{3}+\frac{5x}{2}=\frac{17}{6}-\frac{7}{1}$
$\Rightarrow \frac{6x-16x+15x}{6}=\frac{17-42}{6}\Rightarrow \frac{5x}{6}=\frac{-25}{6}$
$\Rightarrow x=\frac{-25\times 6}{6\times 5}\Rightarrow x=-5$
To check
$x+7-\frac{8x}{3}=\frac{17}{6}-\frac{5x}{2}\Rightarrow -5+7-\frac{8\times (-5)}{3}=\frac{17}{6}-\frac{5\times (-5)}{2}$
$\Rightarrow 2+\frac{40}{3}=\frac{17}{6}+\frac{25}{2}\Rightarrow \frac{6+40}{3}=\frac{17+75}{6}$
$\Rightarrow \frac{46}{3}=\frac{92}{6}\Rightarrow \frac{46}{3}=\frac{46}{3}$
$\Rightarrow L.H.S=R.H.S$
Therefore, it is correct.