Solve the following equation and also verify your solution:
$\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$

Given:

The given equation is $\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$.

To do:

We have to solve the given equation and verify the solution.

Solution:

To verify the solution we have to find the value of the variable and substitute it in the equation. Find the value of LHS and the value of RHS and check whether both are equal.

The given equation is $\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$.

$\frac{2(x-5)}{3}-\frac{1(x-2)}{4}=\frac{9}{2}$

$\frac{2x-10}{3}-\frac{x-2}{4}=\frac{9}{2}$

LCM of denominators $3$ and $4$ is $12$.

$\frac{(2x-10)\times4-(x-2)\times3}{12}=\frac{9}{2}$

$\frac{8x-40-3x+6}{12}=\frac{9}{2}$

$\frac{5x-34}{12}=\frac{9}{2}$

On cross multiplication, we get,

$(5x-34=\frac{9\times12}{2}$

$5x-34=9\times6$

$5x-34=54$

$5x=54+34$

$5x=88$

$x=\frac{88}{5}$

Verification:

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

$=\frac{2}{3}(\frac{88}{5}-5)-\frac{1}{4}(\frac{88}{5}-2)$

$=\frac{2}{3}(\frac{88-5\times5}{5})-\frac{1}{4}(\frac{88-2\times5}{5})$

$=\frac{2}{3}(\frac{88-25}{5})-\frac{1}{4}(\frac{88-10}{5})$

$=\frac{2}{3}(\frac{63}{5})-\frac{1}{4}(\frac{78}{5})$

$=\frac{2}{1}(\frac{21}{5})-\frac{1}{2}(\frac{39}{5})$

$=\frac{42}{5}-\frac{39}{10}$

$=\frac{42\times2-39}{10}$

$=\frac{84-39}{10}$

$=\frac{45}{10}$

$=\frac{9}{2}$

RHS $=\frac{9}{2}$

LHS $=$ RHS

Hence verified.

Updated on: 13-Apr-2023

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