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Spartan_Of_Ares
 3 years ago
can some one help me add these rational expressions?
Spartan_Of_Ares
 3 years ago
can some one help me add these rational expressions?

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Spartan_Of_Ares
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1356272011895:dw

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0Find a common denominator

math>philosophy
 3 years ago
Best ResponseYou've already chosen the best response.0I think you need to either review the distributive property or learn FOIL method

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2\[\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\] use \(a=8,b=x+2,c=4,d=x+3\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2use parentheses and make a direct substitution

Spartan_Of_Ares
 3 years ago
Best ResponseYou've already chosen the best response.0wait so i cross multiply?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2there 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 \(x+3\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2\[\frac{\overbrace{8}^a\times \overbrace{(x+3)}^d+\overbrace{(x+2)}^b\times \overbrace{4}^c}{\overbrace{(x+2)}^b\times \overbrace{(x+3)}^d}\]

Spartan_Of_Ares
 3 years ago
Best ResponseYou've already chosen the best response.0do i combine the denominators?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2you 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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2your real job is to use the distributive property in the numerator, and then combine like terms, because no one want to look at \[8(x+3)+4(x+2)\]

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

Spartan_Of_Ares
 3 years ago
Best ResponseYou've already chosen the best response.0so 8x+24 + 4x+8 = 12x+32 ?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2your numerator is correct, i would be \(12x+3 2\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2and your denominator is \(x^2+5x+6\)

Spartan_Of_Ares
 3 years ago
Best ResponseYou've already chosen the best response.0would i have any restrictions?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2yes, the same restrictions you had to begin with the denominator cannot be zero, so you know \(x\neq 2\) and also \(x\neq 3\)

Spartan_Of_Ares
 3 years ago
Best ResponseYou've already chosen the best response.0so it cant be the 2 and 3 from the original denominators?

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2it cannot be \(2\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.2one denominator is \(x+2\) so if \(x=2\) then you get \(2+2=0\) and the denominator cannot be 0

Spartan_Of_Ares
 3 years ago
Best ResponseYou've already chosen the best response.0ok that makes sense thanks :)
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