# Thread: Help a soldja out.

1. ## Help a soldja out.

Ahh, need assistance with this stuff, help is like always extremely appreciated. Thanx tons guys.

2. Originally Posted by Soldja Soul3ss
Ahh, need assistance with this stuff, help is like always extremely appreciated. Thanx tons guys.

Wow, soulja needs a lot o' helpin'

1) the first is not anything to worry about, just replace a and b with the numbers and work it out using your calculator, that is:

what is: [(-1)^3 - 3]/[-1 + 3]

2) a fraction is undefined when the denominator is zero, so the function is undefined for x + 2 = 0

3) there is a single term in the bottom, just divide it into each

(8x - 8)/4 = (8/4)x - (8/4)

or if you want to be elegant, factor a 4 out the top:

(8x - 8)/4 = 4(2x - 2)/4

now the answer is blatantly staring you in the face

4) factorize the top, something should cancel

when you get those tell me, and if i'm still online i'll help you with the rest

3. Okay i'll try a few out, but algebra is the most intemidating subject for me to even try to comprehend lol.

4. Originally Posted by Soldja Soul3ss
Okay i'll try a few out, but algebra is the most intemidating subject for me to even try to comprehend lol.
it's like that for everyone, but it becomes easier with practice

5. for 1 i got -1

for 2 im not even sure how to go about

and for 3 i THINK its E if i had to guess.

6. ## Re:

RE:
1. A
2. C
3. C
4. B
5. A
6. See Attachment!
7.D
8.A
9.B
10. B

7. woaw #1 was A?!?! i thought for sure i had that right with B

8. For 2 and 3 i pretty much gave you the answers

Originally Posted by Soldja Soul3ss
Ahh, need assistance with this stuff, help is like always extremely appreciated. Thanx tons guys.

1) if a = -1 and b = 3

then (a^3 - b)/(a + b) = [(-1)^3 - 3]/(-1 + 3) = (-1 - 3)/(2) = -4/2 = -2

2) as i said, a fraction is undefined where the denominator is zero, so the values of x for which our fraction is undefined comes from the equation:

x + 2 = 0 ............i would think this is not a hard problem to solve, no matter what your limitations with algebra is, anyway...

=> x + 2 - 2 = 0 - 2 .................subtract 2 from both sides since we want x
=> x = -2

3) (8x - 8)/4 = 4(2x - 2)/4 = 2x - 2 ...........the 4's cancel

4) (x^2 + 3x - 10)/(x + 5) = [(x + 5)(x - 2)]/(x + 5) = x-2 ...the (x + 5)'s cancel

9. ## Re:

Originally Posted by Soldja Soul3ss
woaw #1 was A?!?! i thought for sure i had that right with B
RE:

Defiantly not!! It's A