Solve the following:
$ \frac{3}{4}(7 x-1)-\left(2 x-\frac{1-x}{2}\right)=x+\frac{3}{2} $

Given:

Given equation is $\frac{3}{4}(7x-1)-(2x-\frac{1-x}{2})=x+\frac{3}{2}$.

To do:

We have to find the value of $x$.

Solution:

$\frac{3}{4}(7x-1)-(2x-\frac{1-x}{2})=x+\frac{3}{2}$

$\frac{3}{4}(7x-1)-(\frac{2(2x)-(1-x)}{2})=\frac{2(x)+3}{2}$

$\frac{3}{4}(7x-1)-(\frac{4x-1+x}{2})=\frac{2x+3}{2}$

$\frac{3}{4}(7x-1)-(\frac{5x-1}{2})=\frac{2x+3}{2}$

$\frac{3}{2}(7x-1)-(5x-1)=2x+3$

$\frac{3(7x-1)-2(5x-1)}{2}=2x+3$

$21x-3-10x+2=2(2x+3)$    (On cross multiplication)

$11x-1=4x+6$

$11x-4x=6+1$

$7x=7$

$x=1$

The value of $x$ is $1$.

Tutorialspoint

Updated on: 10-Oct-2022

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