How do you simplify #\frac { 9x ^ { 4} } { 35y ^ { 4} } \cdot \frac { 15x y } { 3x ^ { 4} y ^ { 2} }#?

1 Answer
Nov 26, 2017

#(9x)/(7y^5)#

Explanation:

Instead of multiplying than simplifying, we could simplify the fractions, than cross-simplify.

Since #(9x^4)/(35y^4)# is already in its simplest form, we leave it alone.

#(15xy)/(3x^4y^2)# is NOT in its simplest form. The numerator and the denominator shares #xy# as their common factor. We multiply #(15xy)/(3x^4y^2)# by #(1/(xy))/(1/(xy))# to get #(15)/(3x^3y)#

Now, we cross-simplify #(9x^4)/(35y^4)# and #(15)/(3x^3y)#. The main question here is this: What is the GCF between #9x^4# and #3x^3y#? What about #35y^4# and #15#?

The GCF between #9x^4# and #3x^3y# is #3x^3#, so divide #9x^4# and #3x^3y# by #3x^3# to get #3x# and #y#

We now have #(3x)/(35y^4)xx15/y#

Our next task is to find the GCF between #35y^4# and #15# ,which is #5#. Divide #35y^4# and #15# by #5# to get #7y^4# and #3#.

Now, the original expression becomes #(3x)/(7y^4)xx3/y#

Multiply like we would multiply any other fraction to get #(9x)/(7y^5)#. That is the answer!