Question

simplify: (-3/4*-24/15)+[-11/13*78/55]

Answer 1

Given:

$(\dfrac{-3}{4} \times \dfrac{-24}{15})+(\dfrac{-11}{13} \times \dfrac{78}{55})$

We have to simply $(\dfrac{-3}{4} \times \dfrac{-24}{15})+(\dfrac{-11}{13} \times \dfrac{78}{55})$.

Solution:

∴ $(\dfrac{-3}{4} \times \dfrac{-24}{15})+(\dfrac{-11}{13} \times \dfrac{78}{55})$

= $(\dfrac{1}{1} \times \dfrac{6}{5})+(\dfrac{-1}{1} \times \dfrac{6}{5})$ [ ∵ ( - ) ×  ( - ) = + ]

= $\dfrac{6}{5}+(-\dfrac{6}{5})$

= $\dfrac{6}{5}-\dfrac{6}{5}$

Taking LCM of denominator part, we get

= $\dfrac{6-6}{5}$

= $\dfrac{0}{5}$

= 0

∴ $(\dfrac{-3}{4} \times \dfrac{-24}{15})+(\dfrac{-11}{13} \times \dfrac{78}{55})$ = 0

Thus, the value of $(\dfrac{-3}{4} \times \dfrac{-24}{15})+(\dfrac{-11}{13} \times \dfrac{78}{55})$ is equal to "zero (0)".