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anonymous
 5 years ago
Find the solution set to each equation. x+7/x+4 = x+1/x2
Please help!!
anonymous
 5 years ago
Find the solution set to each equation. x+7/x+4 = x+1/x2 Please help!!

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There are no real solutions for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh this equation is no solution? cause it'd be x+7=0/x+4=0 right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes. well, which class are u and where are u studying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0West Coast University. It's called Algebra for college students. I have this huge test at 7am :(

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have another question, have time? & thanks!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Please ask. If I acn, I will help u.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's really easy but I keep messing up on one part, & just need someone to explain it. Reduce to the lowest terms. (2x)(x+1)(x+3) I know for the (x+1)(x+3) it is x^2+3x+1x+3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0After that I just get stuck and mess up my #'s so do I distribute the (2x) to the original or to x^2+3x+1x+3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Answer= x^3  2x^2 + 5x  6 (After u get x^2+4x+3, multiply this by 2x)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(2x)(x+1)(x+3)=(2x)(x^2+4x+3)=2x^2+8x+6x^34x^23x=x^32x^2+5x+6\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh okay so my question do I start off with 2 for example 2(x^2)+2(4x)+2(5x)2(6) get what I'm saying, that part totally throws me off, I know it's easy but I'm stuck :/
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