What is the simplified form of the quantity of x plus 9, all over the quantity of 2x plus 3 + the quantity of x plus 4, all over the quantity of x plus 2?

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What is the simplified form of the quantity of x plus 9, all over the quantity of 2x plus 3 + the quantity of x plus 4, all over the quantity of x plus 2?

Mathematics
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|dw:1438784510372:dw|
to combine those, multiply both of them by a ratio of the common denominator, (2x + 3)(x+2)
... im confused

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|dw:1438785227000:dw|
THe common denominator is (2x+3)(x+3), Just multiplied the original expression by 1, the common denominator over itself
What does that do to add them together?
When you multiply that out, both will be over the same denominator, so you can add them together
\[\frac{ (x+9)(x+2) }{ (2x+3)(x+2) }+\frac{ (x+4)(2x+3) }{ (2x+3)(x+2) }\]
okay, so when you have the same denominator you just add the numerators together,correct?
right, you need the same denominators
will the denominator for this one be 2x^2+7x+6?
yes, but i would leave that as it is before
\[\frac{ (x+9)(x+2)+(x+4)(2x+3) }{ (2x+3)(x+2) }\]
the answer choice has it out how i wrote it
oh k, yeah you can try expanding everything out and combining like terms in the numerator too.
would the numerator be 2x+13?
Reason i said leave the denominator, sometimes the numerator will give a factor that will cancel
the numerator goes to \[x^2+11x+18+2x^2+3x+8x+12\] then just combine
\[\frac{ 3x^2+22x+30 }{ (2x+3)(x+2) }\]
thank you!
you get how to do those probs now?
welcome
yes, one question though.. If this were subtraction how would it change what you did?
\[\frac{ (x+9)(x+2)-(x+4)(2x+3) }{ (2x+3)(x+2) }\] it would change the sign on all the terms that are expanded (-1)(x+4)(2x+3)
- (2x^2 +11x + 12) -2x^2 - 11x - 12
it wouldn't change what you did to the denominator?
nope, same thing
okay, thank you !

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