Question

1/4 (X - 1/3) = 1/6 (X+1) + 1/12

Answer 1

★ How to do :-

Here, we are given with an equation that has two same variables x. We are said that those two are equal. We are asked to find the value of that variable x. We can find teh answer is this equation by some other concepts such as transition of numbers and variables. In this method, we shift the numbers or variables from one hand side to the other. While we are using this concept, the sign of the particular number or variable changes. We can also verify our answer by the verification method. So, let's solve!!

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➤ Solution :-

${\tt \leadsto \dfrac{1}{4} \bigg( x - \dfrac{1}{3} \bigg) = \dfrac{1}{6} (x + 1) + \dfrac{1}{12}}$

Multiply the numbers outside the bracket with numbers in the bracket.

${\tt \leadsto \dfrac{1}{4} x - \dfrac{1}{12} = \dfrac{1}{6} x + \dfrac{1}{6} + \dfrac{1}{12}}$

Firstly, add the two fractions given in RHS.

${\tt \leadsto \dfrac{1}{4} x - \dfrac{1}{12} = \dfrac{1}{6} x + \dfrac{3}{12}}$

Shift all variable values on LHS and all constant values on RHS.

${\tt \leadsto \dfrac{1}{4} x - \dfrac{1}{6} x = \dfrac{3}{12} + \dfrac{1}{12}}$

Convert the unlike fractions to like fractions on LHS and add the values on RHS.

${\tt \leadsto \dfrac{3}{12} x - \dfrac{2}{12} x = \dfrac{4}{12}}$

Subtract the values on LHS.

${\tt \leadsto \dfrac{1}{12} x = \dfrac{4}{12}}$

Shift the fraction on LHS to RHS, changing it's sign.

${\tt \leadsto x = \dfrac{4}{12} \div \dfrac{1}{12}}$

Take the reciprocal of second fraction and multiply both fractions.

${\tt \leadsto x = \dfrac{4}{12} \times \dfrac{12}{1}}$

Write the numerator and denominator in lowest form by cancellation method.

${\tt \leadsto x = \dfrac{4}{\cancel{12}} \times \dfrac{\cancel{12}}{1} = \dfrac{4 \times 1}{1 \times 1}}$

Simplify the fraction to get the value of x.

${\tt \leadsto \dfrac{4}{1} = 4}$

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${\red{\underline{\boxed{\bf So, \: the \: value \: of \: x \: is \: 4.}}}}$

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Verification :-

${\tt \leadsto \dfrac{1}{4} \bigg( x - \dfrac{1}{3} \bigg) = \dfrac{1}{6} (x + 1) + \dfrac{1}{12}}$

Substitute the value of x.

${\tt \leadsto \dfrac{1}{4} \bigg( 4 - \dfrac{1}{3} \bigg) = \dfrac{1}{6} (4 + 1) + \dfrac{1}{12}}$

Multiply the number outside the bracket with numbers in the bracket.

${\tt \leadsto \dfrac{4}{4} - \dfrac{1}{12} = \dfrac{4}{6} + \dfrac{1}{6} + \dfrac{1}{12}}$

Convert the unlike fractions to like fractions.

${\tt \leadsto \dfrac{12}{12} - \dfrac{1}{12} = \dfrac{8}{12} + \dfrac{2}{12} + \dfrac{1}{12}}$

Subtract the fractions on LHS and add the fractions on RHS.

${\tt \leadsto \dfrac{11}{12} = \dfrac{11}{12}}$

So,

${\sf \leadsto LHS = RHS}$

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Hence verified !!