What is #(4x^2)/(y) * (xy^2)/(12)#?

3 Answers
Sep 10, 2016

Answer:

#(4x^2)/y*(xy^2)/12=(x^3y)/3#

Explanation:

#(4x^2)/y*(xy^2)/12#

= #(2xx2xx x xx x)/y*(x xxyxxy)/(2xx2xx3)#

= #(cancel2xxcancel2xx x xx x)/cancely*(x xxcancelyxxy)/(cancel2xxcancel2xx3)#

= #(x xx x xx x xxy)/3#

= #(x^3y)/3#

Sep 10, 2016

Answer:

#(x^3y)/3#

Explanation:

I always use the following approach in simplifying the multiplication and division of fractions like these in algebra.

Step 1 #rarr# Determine the final sign of the answer.

Done once, you do not need to look at it again.
An EVEN number of negative signs will give a POSITIVE
An ODD number of negative signs will give a NEGATIVE

Step 2 #rarr# sort out any negative indices by moving the bases to or from the numerator or denominator.

Step 3 simplify the numbers, cancel first if possible.

Step 4 combine all the variables to give one numerator and one denominator.

Step 5 Simplify the indices of like bases.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#(4x^2)/y xx (xy^2)/12 " "larr# no negative signs or negative indices.

=#(cancel4x^2)/y xx (xy^2)/cancel12^3 " "larr# cancel numbers

=#(x^3y^2)/(3y) " "larr# make one numerator and one denominator.

=#(x^3y)/3" "larr# subtract indices of like bases

Sep 15, 2016

Answer:

#(x^3y)/3#

Explanation:

Using the property called commutative. (can travel#-># commute)

Using an example:

#color(blue)(2/3xx1/56)color(green)( =(2xx1)/(3xx56))color(brown)( = (2xx1)/(56xx3))color(magenta)( = 2/56xx1/3)#

Notice the way the denominators are able to swap round without changing the final value
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Write as:

#(4x^2)/12xx(xy^2)/y#

#4/12 xx x^2 xx x xx y^2/y#

#" "1/3xx x^3 xx y" " =" " (x^3y)/3#