Question

Verify the following $\dfrac{1}{2} - \left( {\dfrac{3}{4} - \dfrac{5}{4}} \right) = \left( {\dfrac{1}{2} - \dfrac{3}{4}} \right) - \dfrac{5}{4}$

Answer 1

Hint: In this question solve for the L.H.S and the R.H.S part separately. The bracket part of the individual sides are to be evaluated first and then further operation has to be performed. Then prove that the L.H.S part is equal to the R.H.S part.

Complete step-by-step solution -
Verification –
Consider L.H.S
$ \Rightarrow \dfrac{1}{2} - \left( {\dfrac{3}{4} - \dfrac{5}{4}} \right)$
Now simplify it we have,
$ \Rightarrow \dfrac{1}{2} - \left( {\dfrac{{3 - 5}}{4}} \right)$
$ \Rightarrow \dfrac{1}{2} - \left( {\dfrac{{ - 2}}{4}} \right)$
As we know when negative-negative multiplied the sign changes to positive so we have,
 $ \Rightarrow \dfrac{1}{2} + \dfrac{2}{4}$
Now (2/4) is also written as (1/2) so we have,
$ \Rightarrow \dfrac{1}{2} + \dfrac{1}{2}$
$ \Rightarrow \dfrac{{1 + 1}}{2} = \dfrac{2}{2} = 1$
Now consider R.H.S
$ \Rightarrow \left( {\dfrac{1}{2} - \dfrac{3}{4}} \right) + \left( {\dfrac{5}{4}} \right)$
Now in (1/2) multiplied and divide by 2 we have,
$ \Rightarrow \left( {\dfrac{{1 \times 2}}{{2 \times 2}} - \dfrac{3}{4}} \right) + \left( {\dfrac{5}{4}} \right) = \left( {\dfrac{2}{4} - \dfrac{3}{4}} \right) + \left( {\dfrac{5}{4}} \right)$
Now simplify the above equation we have,
$ \Rightarrow \left( {\dfrac{2}{4} - \dfrac{3}{4}} \right) + \left( {\dfrac{5}{4}} \right) = \left( {\dfrac{{2 - 3}}{4}} \right) + \left( {\dfrac{5}{4}} \right)$
\[ \Rightarrow \dfrac{{ - 1}}{4} + \dfrac{5}{4} = \dfrac{{ - 1 + 5}}{4} = \dfrac{4}{4} = 1\]
So as we see that L.H.S = R.H.S
Hence verified.

Note: The concept of L.C.M in the denominator part for evaluation of two fractions is being used to evaluate the given expression. The concept of BODMAS stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. According to BODMAS the brackets have to be solved first followed by powers or roots, then division, Multiplication, Addition and at the end subtraction.