Question

Question #812fe

Answer 1

See the entire solution process below:

Explanation:

First, factor and cancel common terms for the fraction on the right:

`(3ab) xx (2 xx 2)/(2a) -> (3ab) xx (color(red)(cancel(color(black)(2))) xx 2)/(color(red)(cancel(color(black)(2)))a) -> (3ab) xx 2/a`

Next, multiply the two terms:

`(3ab xx 2)/a -> (6ab)/a`

Now, we can again cancel a common term in the numerator and denominator:

`(6color(red)(cancel(color(black)(a)))b)/color(red)(cancel(color(black)(a))) -> 6b`

However, from the original expression we cannot divide by `0` therefore the solution is:

`6b` where `2a != 0` or `a != 0`