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Solve the following equations and verify your answer:
(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$
(ii) $\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$
Given:
The given equations are:
(i) $\frac{2}{3x}-\frac{3}{2x}=\frac{1}{12}$
(ii) $\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$
To do:
We have to solve the given equations and verify the answers.
Solution:
To verify the answer 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 $6x$
$\frac{2(2)-3(3)}{6x}=\frac{1}{12}$
$\frac{4-9}{6x}=\frac{1}{12}$
$\frac{-5}{6x}=\frac{1}{12}$
On cross multiplication, we get,
$12(-5)=(1)(6x)$
$-60=6x$
$6x=-60$
$x=\frac{-60}{6}$
$x=-10$
Verification:
LHS $=\frac{2}{3x}-\frac{3}{2x}$
$=\frac{2}{3(-10)}-\frac{3}{2(-10)}$
$=\frac{2}{-30}-\frac{3}{-20}$
$=\frac{-1}{15}+\frac{3}{20}$
$=\frac{-1\times4+3\times3}{60}$ (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 $\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$
$\frac{3x+5}{4x+2}=\frac{3x+4}{4x+7}$
On cross multiplication, we get,
$(3x+5)(4x+7)=(3x+4)(4x+2)$
$3x(4x+7)+5(4x+7)=3x(4x+2)+4(4x+2)$
$12x^2+21x+20x+35=12x^2+6x+16x+8$
On rearranging, we get,
$12x^2-12x^2+41x-22x=8-35$
$19x=-27$
$x=\frac{-27}{19}$
Verification:
LHS $=\frac{3x+5}{4x+2}$
$=\frac{3(\frac{-27}{19})+5}{4(\frac{-27}{19})+2}$
$=\frac{\frac{3\times(-27)}{19}+5}{\frac{4\times(-27)}{19}+2}$
$=\frac{\frac{-81+5\times19}{19}}{\frac{-108+2\times19}{19}}$
$=\frac{\frac{-81+95}{19}}{\frac{-108+38}{19}}$
$=\frac{\frac{14}{19}}{\frac{-70}{19}}$
$=\frac{14}{19}\times\frac{19}{-70}$
$=\frac{1}{1}\times\frac{1}{-5}$
$=\frac{-1}{5}$
RHS $=\frac{3x+4}{4x+7}$
$=\frac{3(\frac{-27}{19})+4}{4(\frac{-27}{19})+7}$
$=\frac{\frac{3\times(-27)}{19})+4}{\frac{4\times(-27)}{19})+7}$
$=\frac{\frac{-81+19\times4}{19}}{\frac{-108+19\times7}{19}}$
$=\frac{\frac{-81+76}{19}}{\frac{-108+133}{19}}$
$=\frac{\frac{-5}{19}}{\frac{25}{19}}$
$=\frac{-5}{19}\times\frac{19}{25}$
$=\frac{-1}{1}\times{1}{5}$
$=\frac{-1}{5}$
LHS $=$ RHS
Hence verified.
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