How do you simplify #(3x^2+6y)/(3x^2+9x)#?

1 Answer
Pythagoras
Sep 25, 2015

#(x^2 + 2y)/(x*(x + 3)#

Explanation:

#(3x^2 +6y)/(3x^2 +9x)#

Factor out 3 from numerator, and 3x from the denominator:
#=(3*x^2 + (3*2)y)/(3x*x + 3*3x)#
#=(3*(x^2 + 2y))/(3x*(x + 3))#

Then cancel out the 3 from the numerator and the denominator:
#=(cancel 3*(x^2 + 2y))/(cancel 3 x*(x + 3))#

So you have:
#=(x^2 + 2y)/(x*(x + 3)#

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