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

Question

Savannah Anderson · Accepted Answer

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#

Julia Adams · Answer

undefined To solve the equation ((2a)/3)*(6/a)=4, you can start by simplifying the expression on the left side. By canceling out the common factor of "a" in the numerator and denominator, the equation becomes (2/3)*6=4. Multiplying 2/3 by 6 gives you 12/3, which simplifies to 4. Therefore, the equation simplifies to 4=4. Since both sides are equal, this equation is true for all values of "a".