Why does this pre-algebra problem come out this way

(2/3) ^ -2 = 1/(2/3) ^ 2 = 1/ (4/9) = 9/4

Why can the nine just move to the numerator of the equation at the end? What are the actual steps for that to happen?

Hi there. The easiest way to explain it is to consider that you can multiply the numerator and denominator by the same number and keep the fraction correct. In this case, you multiply top and bottom by 9. So the numerator becomes 9 x 1 = 9 and the numerator becomes 9 x (4/9) = 4.

Hope that helps :-)

That last step is actually a law of algebra. It's really simple. (1/x)x=1, the inverse property

When you are dividing by a number you are essentially multiplying by the same number, but flipped.

If you had the number 20 and you divided that by 5 you are doing the same operation as 20(1/5). You are multiply 20 by one fifth.

So 1 divided by that fraction of 9/4 is the same as multiply 1 by 4/9.

I assume you understand up to this 1/ (4/9) = 9/4
remember if one fraction divide by another then the result is the top fraction Multiply the reciprocal of the bottom.
so
1/(4/9) = (1/1)/(4/9) = 1/1 * 9/4 = 9/4
Hope it helps

A number divided by a fraction is equal to that same number multiplied by its reciprocal. 1/(4/9) is equal to 1*(9/4). So the answer is 9/4.

because (1/1)/(4/9) is9/4 fartimes far neartimes near

Youre ght