How do you solve #((2a)/3)*(6/a)=4#?

1 Answer
Sep 17, 2015

Multiplication of fractions is very easy, we just multiply everything above the bar together, multiply everything below the bar together, and if there's any common factors we can cut it out.

For example:
#((2a)/3)*(6/a) = 4#

Everything on top is multiplied and so is everything on the bottom

#(2a*6)/(3*a) = 4#

We have an #a# both on top and on bottom, so we can cut it out! (Because any number divided by itself is 1. Think for example, of you having 2 pieces of candy to divide to 2 friends, they both get one each. The same applies to 3 pieces and 3 friends, 4 pieces and 4 friends, and so on)

#(2cancel(a)*6)/(3*cancel(a)) = 4#

Now we just have numbers left, we can just multiply.
#(2*6)/3 = 12/3 = 4#

Or

We know that #6 = 3*2#, so we can rewrite that as
#(2*3*2)/3#, and use the same logic above. There's a 3 above and below so we can cut them out.

#(2*cancel(3)*2)/cancel(3) = 2*2#

And #2*2# as we know is #4#