1. ## sum of solutions?

Solve:

5/(x+4) = 4 + [3/(x-2)]

What is the sum of your solutions?

the sum of your solutions is 6x + 10x - 4x squared..

if you were to solve this you get x equal to 2.5 or -1

hope that helps

Solve:

5/(x+4) = 4 + [3/(x-2)]

What is the sum of your solutions?
Make a common denominator by multiplying all the denominators to make one. So (x+4)(1)(x-2) will give you a denominator of (x+4)(x-2)

now you'll have
(something)/(x+4)(x-2) = (something)/(x+4)(x-2) + (something)/(x+4)(x-2)

to find out what that something is, make each term so that it denominator is this common one we just found.
1) term 5/(x+4) ---> [5(x-2)]/[(x+4)(x-2)]
2) term 4 ---> [4(x+4)(x-2)]/[(x+4)(x-2)]
3) term 3/(x-2) ---> [3(x+4)]/(x+4)(x-2)]
so now you have your terms, where their numerators can be added or subtracted with eachother since they all have like denominators.

[5(x-2)]/[(x+4)(x-2)] = [4(x+4)(x-2) + 3(x+4)]/[(x+4)(x-2)]
expand the numerator on each side. don't worry about the denominator since it's the same everywhere.

so, 5x-10 = 4(x^2+4x-2x-8) + 3x+12
5x-10 = 4x^2+16x-8x-32+3x+12
5x-10 = 4x^2+11x-10

So the question is... how do we make these two sides equal. Well, there's a number x which makes this possible. What's this number x? Let's find out. Bring all terms to one side, to start:
0 = 4x^2+11x-10-5x+10
0 = 4x^2+6x
0 = 2x(2x+3)
0 = 2x and 0 = 2x+3
So x = 0 and x = -3/2

There are two numbers that when substitued in for x, will make your original equation true.

Did that help, because I'm not too sure what you were specifically asking.

4. Yes! thanks a lot

6. No problem!

7. ## Sum of solutions.

Solve:

5/(x+4) = 4 + [3/(x-2)]

What is the sum of your solutions?
The sum of the solutions is -1.5 or -3/2.

LCM = (x+4)(x-2) , x not equal to -4; 2

5(x-2) = 4(x+4)(x-2) + 3(x+4)
5x-10 = 4(x^2 +2x-8)+3x+12
0 = 4x^2 +8x-32+3x+12-5x+10
0 = 4x^2 +6x-10
0 = 2x^2 +3x-5
0 = (2x+5)(x-1)

x=-5/2 or x=1 (these are the solutions)

the sum of the solutions
= -5/2 + 1
= -2.5 + 1
= -1.5

(I hope there aren't any mistakes...nice question. Cheers)