Question 202559
1. (9z^3/16xy)(4x/27z^3) 
2. (3m-9/4m+8)(m^2+5m+6/m^2-9) 
3. (y^2+3y-10/3y+15) / (10-5y) 



Before we do these, we have to agree that {{{4/4 = 1}}}


Further, we have to agree that {{{a^3/a^3 = 1}}}


Those are good facts to have in your mind, so let's begin.



1. (9z^3/16xy)(4x/27z^3)


I read this as:  {{{9z^3/16xy}}}*{{{4x/27z^3}}}


Because you are multiplying, you can factor terms, just like we did above when we said {{{4/4=1}}}



SO - let's factor: 


The 9 in the numerator can be factored into the 27 of the denominator to give you {{{1/3}}}


The 4 in the numerator can be factored into the 16 of the denominator to give you {{{1/4}}}


The {{{z^3}}} in the numerator can be factored into the {{{z^3}}} of the denominator to give you {{{1/1}}}


The "x" in the numerator can be factored into the x of the denominator to again give you {{{1/1}}}


What you have left is: 


{{{1/4(3)y}}}  which is:   {{{1/12y}}}


2. (3m-9/4m+8)(m^2+5m+6/m^2-9)



Let's just break this down piece by piece, because it may be quicker:


Numerator of first fraction:   3m - 9   
Let's factor out 3:    3(m-3)


Denominator of first fraction:  4m + 8
Let's factor out 4:    4(m + 2)


Numerator of 2nd fraction:  {{{m^2+5m+6}}}
Let's factor this to be:  (m+2)(m+3)


Denominator:  {{{m^2 -9)}}}
Let's factor this to be:  (m+3)(m-3)



NOW we have: 


{{{3(m-3)(m+2)(m+3)/4(m+2)(m+3)(m-3)}}}


NOW factor out {{{(m-3)(m+2)(m+3)/(m+2)(m+3)(m-3)}}}



What you have left is {{{3/4}}}


3.(y^2+3y-10/3y+15) / (10-5y) 


Numerator of first fraction:{{{y^2+3y-10}}}
Factor to:  (y+5)(y-2)


Denominator of first fraction:3y+15
Let's factor that to:  3(y+5)


Second fraction:  10-5y
Factor to:  5(2-y)



Now we have:   {{{(y+5)(y-2)5(2-y)/3(y+5)}}}



(y+5) can factor out of the numerator and denominator so we have left....


{{{5(2-y)(y-2)/3}}}



I hope this helps. :-)