Question

What should be added to $\left(\frac{-13}{4}+\frac{-3}{8}\right)$ to get 1?

Answer 1

$\begin{array}{l}\text{Let the required}\text{number that is to be added be}\mathit{\text{x.}}\\ \left(\frac{-13}{4}+\frac{-3}{8}\right)+x=1\\ =>x=1-\left(\frac{-13}{4}+\frac{(-3)}{8}\right)\\ \mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}4\mathrm{and}8\mathrm{is}8.\\ \begin{array}{l}=1-\left(\frac{-26-3}{8}\right)\\ =1-\left(\frac{-29}{8}\right)\\ =1+\frac{29}{8}\\ =\frac{8+29}{8}\end{array}\\ =\frac{37}{8}\\ \text{Hence, the required number is}\frac{37}{8}\text{.}\end{array}$