# Solve each of the following equations and also check your results in each case:(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$(ii) $\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$

Given:

The given equations are:

(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$

(ii) $\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$

To do:

We have to solve the given equations and check the results.

Solution:

To check the results 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 $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$

$\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$

LCM of denominators $3x$ and $2x$ is $6x$

$\frac{2\times2-3\times3}{6x}=\frac{1}{12}$

$\frac{4-9}{6x}=\frac{1}{12}$

$\frac{-5}{6x}=\frac{1}{12}$

On cross multiplication, we get,

$-5\times12=1\times6x$

$6x=-60$

$x=\frac{-60}{6}$

$x=-10$

Verification:

LHS $=\frac{2}{3x}-\frac{3}{2x}$

$=\frac{2}{3(-10)}-\frac{3}{2(-10)}$

$=\frac{2}{-30}-\frac{3}{-20}$

$=\frac{-1}{15}-(\frac{-3}{20}$

$=\frac{-1}{15}+\frac{3}{20}$

$=\frac{-1\times4+3\times3}{60}$          (LCM of $15$ and $20$ is $60$)

$=\frac{-4+9}{60}$

$=\frac{5}{60}$

$=\frac{1}{12}$

RHS $=\frac{1}{12}$

LHS $=$ RHS

Hence verified.

(ii) The given equation is $\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$

$\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}=\frac{8x+19}{18}$

On rearranging, we get,

$\frac{4x}{9}+\frac{13x}{108}-\frac{8x+19}{18}=-\frac{1}{3}$

LCM of $9, 108$ and $18$ is $108$

$\frac{4x \times 12+13x \times1- (8x+19)\times6}{108}=-\frac{1}{3}$

$\frac{48x+13x-48x-114}{108}=-\frac{1}{3}$

$\frac{13x-114}{108}=-\frac{1}{3}$

On cross multiplication, we get,

$13x-114=\frac{-1\times108}{3}$

$13x-114=-36$

$13x=-36+114$

$13x=78$

$x=\frac{78}{13}$

$x=6$

Verification:

LHS $=\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}$

$=\frac{4x}{9}+\frac{1}{3}+\frac{13x}{108}$

$=\frac{4(6)}{9}+\frac{1}{3}+\frac{13\times6}{108}$

$=\frac{8}{3}+\frac{1}{3}+\frac{13\times1}{18}$

$=\frac{8}{3}+\frac{1}{3}+\frac{13}{18}$

$=\frac{8\times6+1\times6+13}{18}$                   (LCM of $3$ and $18$ is $18$)

$=\frac{48+6+13}{18}$

$=\frac{67}{18}$

RHS $=\frac{8x+19}{18}$

$=\frac{8(6)+19}{18}$

$=\frac{48+19}{18}$

$=\frac{67}{18}$

LHS $=$ RHS

Hence verified.

Updated on: 13-Apr-2023

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