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Spartan_Of_AresBest ResponseYou've already chosen the best response.0
dw:1356272011895:dw
 one year ago

math>philosophyBest ResponseYou've already chosen the best response.0
Find a common denominator
 one year ago

math>philosophyBest ResponseYou've already chosen the best response.0
I think you need to either review the distributive property or learn FOIL method
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
\[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] use \(a=8,b=x+2,c=4,d=x+3\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
use parentheses and make a direct substitution
 one year ago

Spartan_Of_AresBest ResponseYou've already chosen the best response.0
wait so i cross multiply?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
there is no such thing as "cross multiply" to add fractions, you compute \[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] that is how it is always done in this case \(a=8\) so replace \(a\) by \(8\) \(b=x+2\) so replace \(b\) by \(x+2\) \(c=4\) replace \(c\) by \(4\) and \(d=x+3\) replace \(d\) by \(x+3\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
\[\frac{\overbrace{8}^a\times \overbrace{(x+3)}^d+\overbrace{(x+2)}^b\times \overbrace{4}^c}{\overbrace{(x+2)}^b\times \overbrace{(x+3)}^d}\]
 one year ago

Spartan_Of_AresBest ResponseYou've already chosen the best response.0
do i combine the denominators?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
you have a choice, you can either leave it in factored form as \[(x+2)(x+3)\] or you can multiply out using the distributive law \[(x+2)(x+3)=x^2+2x+3x+6=x^2+5x+6\] your choice
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
your real job is to use the distributive property in the numerator, and then combine like terms, because no one want to look at \[8(x+3)+4(x+2)\]
 one year ago

Spartan_Of_AresBest ResponseYou've already chosen the best response.0
well to get a final answer i would need to use \[x ^{2}5x+6\] wouldnt I?
 one year ago

Spartan_Of_AresBest ResponseYou've already chosen the best response.0
so 8x+24 + 4x+8 = 12x+32 ?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
your numerator is correct, i would be \(12x+3 2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
and your denominator is \(x^2+5x+6\)
 one year ago

Spartan_Of_AresBest ResponseYou've already chosen the best response.0
would i have any restrictions?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
yes, the same restrictions you had to begin with the denominator cannot be zero, so you know \(x\neq 2\) and also \(x\neq 3\)
 one year ago

Spartan_Of_AresBest ResponseYou've already chosen the best response.0
so it cant be the 2 and 3 from the original denominators?
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
it cannot be \(2\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
one denominator is \(x+2\) so if \(x=2\) then you get \(2+2=0\) and the denominator cannot be 0
 one year ago

Spartan_Of_AresBest ResponseYou've already chosen the best response.0
ok that makes sense thanks :)
 one year ago
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