Question #812fe

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Answer 1 · 2017-03-30T13:42:14

See the entire solution process below:

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 $a != 0$