Ask your own question, for FREE!
Algebra 19 Online
OpenStudy (anonymous):

Find the Quotient of 20x^2y-88xy/-4xy

jimthompson5910 (jim_thompson5910):

The problem is this, right? \[\Large \frac{20x^2y-88xy}{-4xy}\]

OpenStudy (anonymous):

Yes

jimthompson5910 (jim_thompson5910):

we can factor out -4xy from the numerator \[\Large \frac{20x^2y-88xy}{-4xy} = \frac{-4xy(-5x+22)}{-4xy} \] what comes next?

OpenStudy (anonymous):

I'm having a very hard time understanding even from the beginning, if you can i would love for you to help explain it to me.

jimthompson5910 (jim_thompson5910):

well I factored out -4xy because that's what is in the denominator I want to cancel out the denominator

jimthompson5910 (jim_thompson5910):

did you learn about factoring yet?

OpenStudy (anonymous):

i believe so but I'm not positive

jimthompson5910 (jim_thompson5910):

the idea of factoring is to rewrite a number or expression as a product of two smaller things

jimthompson5910 (jim_thompson5910):

ex: 27 = 9*3 16 = 8*2

OpenStudy (anonymous):

could possibly be 22x^2y^2+20x^2y

jimthompson5910 (jim_thompson5910):

what do you mean?

OpenStudy (anonymous):

for the awnser i asked my mom and she said she thinks that is what it is

jimthompson5910 (jim_thompson5910):

have you learned about the distributive property?

OpenStudy (anonymous):

is that not with (X)^Y

jimthompson5910 (jim_thompson5910):

it's something like 2*(x+y) = 2*x + 2*y

OpenStudy (anonymous):

yeah i have

jimthompson5910 (jim_thompson5910):

so when I factored back up at the top, I used the distributive property (in reverse)

jimthompson5910 (jim_thompson5910):

if you have 2*x + 2*y you can factor out 2 to get 2*(x+y)

jimthompson5910 (jim_thompson5910):

same idea with your problem, but I pulled out -4xy

OpenStudy (anonymous):

okay im some what following lol

jimthompson5910 (jim_thompson5910):

it turns out that -4xy is a common factor of 20x^2y and -88xy

jimthompson5910 (jim_thompson5910):

|dw:1436404481447:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!