I need help with a math problem =P

Don't we have schools for this?

 

I don't get it

1-1 =/= 1+2

do you want us to substitute in the x's and c's and d's? And if so, where?

 

So, 1/x-1/c=1/d+2/x?

I'm currently trying (and failing) to focus on my own math homework, but the way you'd go about doing that is getting a least common denominator on each side.

 

hmmm...i dont remeber how to do this

 

This the equation, correct?

Code:

1/x - 1/c = 1/d + 2/x

Put the x's on the same side and everything else on the other.

Code:

- 1/c - 1/d = 1/x

or

Code:

1/x = -1/c - 1/d

This is as far as I can get while being sure.
Next is what I think.

Multiply everything by x^2 (x squared).

Code:

x = -1x^2/c - 1x^2/d

That doesn't seem right to me, but that's the best I can do. I'll keep trying.

 

I think it is Undefined.


Or maybe its

X = -(CD/(C+D)

 

Uhm... I am having trouble. Am I confused, or is everyone that far off? To get rid of fractions, you don't square the denominator, and it isn't undefined simply because there are more than 1 variable. It just means the solution for X is not a constant.

Code:

 1/x - 1/c = 1/d + 2/x 

Subtract 2/x from both sides, and add 1/c to both sides.

Code:

 1/x - 2/x = 1/d + 1/c 

Code:

 -1/x = 1/d + 1/c 

Now make 1/d and 1/c have common denominations by multiplying both sides by c/c and d/d (both equal 1, so it won't affect the x side of the equation). Then add the right side.

Code:

 -1/x = c/dc + d/dc 

Code:

 -1/x = c+d/dc 

Now you need to get x out of the fraction. So, multiply both sides of the equation by x (do not multiply x into the right side, this would only be an extra un-needed step).

Code:

 -1 =  (c+d/dc) (x) 

However, x needs to be alone on a single side of the equation. So, multiply both sides by (dc/c+d) [note that this will cancel out with c+d/dc making it 1x or just x on the right side].

Code:

 (-1) (dc/c+d) = x 

Now distribute the negative, and lets flip left side with right side so that x is on the left side of the = sign.

Code:

 x = -dc/c+d 

There you have it, basic algebra (unless I am making some huge mistake here... in which case fuuuuuuuuuu)

Edit: Okay, Jpitty fixed himself.

 

Okay ya, I thought I was alone. I had to go redo it because it just didn't seem right, and if we both got the same answer, it has to be right....right?

 

If I got an answer it is right... because I am right. Always.
/ego

 

Okay, Chuck Norris.

 

The second to last pic is wrong. You divide by (c+d/cd), not multiply by it. If you wanna cancel a variable/number in a multiplication sequence (as is the case with [x] [c+d/cd]) you have to divide both sides by one of those variables/numbers. It should read:

Code:

 (-1) / (c+d/cd) 

Which can be simplified by into.....oh my god......haha wow I just failed miserably. I end up getting the same answer you did sweeny. *facepalm

 

Wow, this is onl;y basic maths. We're all overcomplicating things. What year/grade are you in, OP?

This could probably accomplished at GCSE level.