Question

Simplify:

$\frac{4}{8}-\frac{1}{2}+\frac{3}{4}$

Answer 1

Given, $\frac{4}{8}-\frac{1}{2}+\frac{3}{4}$

Consider the denominator of $\frac{4}{8}-\frac{1}{2}+\frac{3}{4}$i.e, $8,2,4$

The LCM of $8,2,4$is $8$

Now the fraction becomes

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{4}{8}}\times 8-{\displaystyle \frac{1}{2}}\times 8+{\displaystyle \frac{3}{4}}\times 8}{8} \\ \Rightarrow \frac{4\times 1-1\times 4+3\times 2}{8} \\ \Rightarrow \frac{4-4+6}{8} \\ =\frac{6}{8} \\ =\frac{3}{4} \end{gathered}$$

Hence, the required fraction is $\frac{3}{4}$.