Question

Can someone help me solve for x?

1/(x-1) + 1/2 = 2/ (x²-1)

Answer 1

Have common denominator (x^2-1)

Answer 2

How does the x²-1 go in to the denominator of 2?

Answer 3

its already the denominator!

Answer 4

I'm talking about the 1/2 . How can x²-1 go into 2?

Answer 5

Multiply both sides by \((x^{2}-1)\) to get\[\frac{(x^{2}-1)}{(x-1)} +\frac{(x^{2}-1)}{2} = 2\]
Then recall that \((x^{2}-1)=(x-1)(x+1)\) which gives \[\frac{(x-1)(x+1)}{(x-1)} +\frac{(x^{2}-1)}{2} = 2\].
The first term simplifies and you are left with (after some rearranging) \[x^{2} -2x -3 = 0\]. Then just factorise or use the quadratic equation method to solve for x.

Answer 6

Thank you JonnyMcA, your answer seems to be pretty good so far. :)

Answer 7

Another method you could use is to multiply by the LCM of the denominators.

Answer 8

Remember that for a quadratic equation \(ax^{2}+bx+c=0\), the solutions for x will be \[x=-b\pm\sqrt{\frac{(b^{2}-4ac)}{2a}}\]

Answer 9

jonnymca, I still don't see how you got to the x² - 2x - 3. I understand the factoring part and the rearranging, but it seems when you simplify the (x-1)(x+1)/(x-1), you would cancel out the x-1 part.

Answer 10

Let me see if I can put this in a better way...
When I have \[\chi^2-1/2\]
I don't see how I would divide to simplify it. This is where I keep getting stuck.

Answer 11

no probs, ill go through the rearranging part. So we have \[\frac{(x−1)(x+1)}{(x-1)}+\frac{(x^{2}-1)}{2}=2\]. This becomes \[(x+1)+\frac{(x^{2}-1)}{2}=2\], which then is teh same as \[2(x+1)+(x^{2}-1)=4\] or \[2x+2+x^{2}-1 - 4=0\] Leaving \[2x+2+x^{2}-3=0\]. and it is here i notice that I made a mistake in teh original in that the expression should be \(x^{2}+2x-3=0\) not \(x^{2}-2x-3=0\)

Answer 12

But it is not \(x^{2}-\frac{1}{2}\) it is \[\frac{x^{2}-1}{2}\]. To get rid of teh 2 on teh bottom line, just multiply all terms by 2

Answer 13

Ah ok

Answer 14

Your simplifying is very helpful! Thank you. :D

Answer 15

Your very welcome. Glad I could help.