Rational Functions: n+2/-3n=n+5/-3n-1, 5/(x-3)/(-6/x)+3,

[whitneyla13]

1. n+2/-3n=n+5/-3n-1
Times both sides by what they don't have.
(n+2)(3n-1)=n+5(-3n)
n+6n-2=n-15n
22n-2=n
25=n
Is this correct! What am I doing wrong?

2. 5/(x-3)/(-6/x)+3
5/(x-3)/-3/2x
-3*5/(x-3)*2x*(x-3)
-15x+2x^2-6x
What am I doing wrong?

3.s^2+3s/(s-3)(s-4)*(s-4)/(s+3)
s^2+3s^2+9s+s^2-7s+12+s-4/(s-3)(s+3)(s-4)
3s^2+3s+8
What am I doing wrong?

4. x/x^2-9+3/x-3=1/x+3
A.4 B.-4 C.2 D.No Solution
D. I plugged each of the other numbers in and got no solution!

5.(7S)/(S+4)(S-4)-(4)/(S+3)(S+4)
7S(S+3)-4(S-4)/(S+4)(S-4)(S+3)
7S^2-4X+37/(S+4)(S-4)(S+3)
iS THIS CORRECT?

 

[Loren]

>1. n+2/-3n=n+5/-3n-1
Times both sides by what they don't have.
(n+2)(3n-1)=n+5(-3n)
n+6n-2=n-15n
22n-2=n
25=n
Is this correct! What am I doing wrong?

First, I would suggest that you need to use parenthesis correctly to clarify what is meant. As written, the problem means \(\displaystyle n+\frac{2}{-3}n=n+\frac{5}{-3}n-1\).
The first step of your solution implies that you meant (n+2)/(-3n)=(n+5)/(-3n-1) which means \(\displaystyle \frac{n+2}{-3n}=\frac{n+5}{-3n-1}\).
If that is what is meant, your first step should be (n+2)(-3n-1)=(n+5)(-3n) which will yield...

-3n[sup:1tnhpaj0]2[/sup:1tnhpaj0]-7n-2=-3n[sup:1tnhpaj0]2[/sup:1tnhpaj0]-15n.

You can take it from there?

 

[mmm4444bot]

1. n+2/-3n=n+5/-3n-1

(n+2)(3n-1)=n+5(-3n)

You dropped the negative sign.

n+6n-2=n-15n

This result makes no sense. Where are the n-squared terms?

Is this correct! What am I doing wrong?

It appears to me that you do not know how to use FOIL to multiply two binomials.

Without this skill, you are not ready to work with rational functions. Please speak with your instructor because you've fallen too far behind for me to provide in this forum the missing lessons that you need.