# Solve each of the following equations and also verify your solution:(i) $\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$(ii) $13(y-4)-3(y-9)-5(y+4)=0$

Given:

The given equations are:

(i) $\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$

(ii) $13(y-4)-3(y-9)-5(y+4)=0$

To do:

We have to solve the given equations and verify the solutions.

Solution:

To verify the solutions we have to find the values of the variables and substitute them in the equation. Find the value of LHS and the value of RHS and check whether both are equal.

(i) The given equation is $\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$.

$\frac{(2x-1)}{3}-\frac{(6x-2)}{5}=\frac{1}{3}$

LCM of denominators $3$ and $5$ is $15$

$\frac{(2x-1)\times5-(6x-2) \times3}{15}=\frac{1}{3}$

$\frac{10x-5-18x+6}{15}=\frac{1}{3}$

$\frac{-8x+1}{15}=\frac{1}{3}$

On cross multiplication, we get,

$-8x+1=\frac{1\times15}{3}$

$-8x+1=5$

$8x=1-5$

$8x=-4$

$x=\frac{-4}{8}$

$x=\frac{-1}{2}$

Verification:

LHS $=\frac{(2x-1)}{3}-\frac{(6x-2)}{5}$

$=\frac{(2\times\frac{-1}{2}-1)}{3}-\frac{(6\times\frac{-1}{2}-2)}{5}$

$=\frac{-1-1}{3}-\frac{-3-2}{5}$

$=\frac{-2}{3}-\frac{-5}{5}$

$=\frac{-2}{3}+1$

$=\frac{-2+1\times3}{3}$

$=\frac{-2+3}{3}$

$=\frac{1}{3}$

RHS $=\frac{1}{3}$

LHS $=$ RHS

Hence verified.

(ii) The given equation is $13(y-4)-3(y-9)-5(y+4)=0$.

$13(y-4)-3(y-9)-5(y+4)=0$

$13y-52-3y+27-5y-20=0$

$13y-8y-72+27=0$

$5y-45=0$

$5y=45$

$y=\frac{45}{5}$

$y=9$

Verification:

LHS $=13(y-4)-3(y-9)-5(y+4)$

$=13(9-4)-3(9-9)-5(9+4)$

$=13(5)-3(0)-5(13)$

$=65-0-65$

$=0$

RHS $=0$

LHS $=$ RHS

Hence verified.

Updated on: 13-Apr-2023

129 Views 