Question

Solve each of the following equations and also check your results in each case:​
$\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$

Answer 1

Given: $\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}$

$\Rightarrow \frac{7x-1}{4}-(\frac{2}{3}x-\frac{1-x}{6})=\frac{10}{3}$

$\Rightarrow \frac{7x-1}{4}-\frac{4x-1+x}{6}=\frac{10}{3}$

$\Rightarrow \frac{7x-1}{4}-\frac{5x-1}{6}=\frac{10}{3}$

$\Rightarrow \frac{(7x-1)6-(5x-1)4}{24}=\frac{10}{3}$

$\Rightarrow \frac{42x-6-20x+4}{24}=\frac{10}{3}$

$\Rightarrow \frac{22x-2}{24}=\frac{10}{3}$

$\Rightarrow 22x-2=10\left(8\right)$

$\Rightarrow 22x-2=80$

$\Rightarrow 22x=82$

$\Rightarrow x=\frac{82}{22}$

$\therefore x=\frac{41}{11}$

To check : $\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})=\frac{10}{3}\text{for}x=\frac{41}{11}$

$L.H.S=\frac{7x-1}{4}-\frac{1}{3}(2x-\frac{1-x}{2})$

$=\frac{7x-1}{4}-\frac{5x-1}{6}$

$=\frac{7\left(\frac{41}{11}\right)-1}{4}-\frac{5\left(\frac{41}{11}\right)-1}{6}$

$=\frac{\frac{287}{11}-1}{4}-\frac{\frac{205}{11}-1}{6}$

$=\frac{287-11}{44}-\frac{205-11}{66}$

$=\frac{276}{44}-\frac{194}{66}$

$=\frac{69}{11}-\frac{97}{33}$

$=\frac{(69\times 3)-(97\times 1)}{33}$

$=\frac{207-97}{33}$

$=\frac{110}{33}$

$=\frac{10}{3}=R.H.S$

$\therefore L.H.S=R.H.S$

Hence, the given equation is verified for $x=\frac{41}{11}$.
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