Question

Add the following -3a2b2, (–5/2) a2b2, 4a2b2, and (⅔) a2b2​

Answer 1

Given:

The terms are $-3a^2b^2,-\dfrac{5}{2}a^2b^2,4a^2b^2,\dfrac{2}{3}a^2b^2$.

To find:

The sum of given terms.

Solution:

Sum of given terms is

$-3a^2b^2+(-\dfrac{5}{2}a^2b^2)+4a^2b^2+\dfrac{2}{3}a^2b^2$

$=-3a^2b^2-\dfrac{5}{2}a^2b^2+4a^2b^2+\dfrac{2}{3}a^2b^2$

Taking out common factors, we get

$=\left(-3-\dfrac{5}{2}+4+\dfrac{2}{3}\right)a^2b^2$

$=\left(\dfrac{-18-15+24+4}{6}\right)a^2b^2$

$=\left(\dfrac{-5}{6}\right)a^2b^2$

$=\dfrac{-5}{6}a^2b^2$

Therefore, the sum of given terms is $\dfrac{-5}{6}a^2b^2$.