J JW86 New member Joined Apr 22, 2009 Messages 1 Apr 22, 2009 #1 Equation: (2/x^2 - 1) - 1 = 1/x - 1 Next: 2/(x + 1)(x - 1) - 1 = 1/x - 1 ^^Is that correct? I need help solving the rest. I think it is the minus one that is throwing me off. Thank you! Jenn
Equation: (2/x^2 - 1) - 1 = 1/x - 1 Next: 2/(x + 1)(x - 1) - 1 = 1/x - 1 ^^Is that correct? I need help solving the rest. I think it is the minus one that is throwing me off. Thank you! Jenn
D DrMike Full Member Joined Mar 31, 2009 Messages 251 Apr 22, 2009 #2 JW86 said: Equation: (2/x^2 - 1) - 1 = 1/x - 1 Next: 2/(x + 1)(x - 1) - 1 = 1/x - 1 ^^Is that correct? I need help solving the rest. I think it is the minus one that is throwing me off. Thank you! Jenn Click to expand... If your equation was \(\displaystyle \frac{2}{x^2-1}-1=1/x-1\) then it is correct that \(\displaystyle \frac{2}{(x+1)(x-1)}-1=1/x-1\) However, the way you wrote it looks more like \(\displaystyle \frac{2}{x^2}-1=\frac{1}{x}-1\) Perhaps you need to add some brackets to your post?
JW86 said: Equation: (2/x^2 - 1) - 1 = 1/x - 1 Next: 2/(x + 1)(x - 1) - 1 = 1/x - 1 ^^Is that correct? I need help solving the rest. I think it is the minus one that is throwing me off. Thank you! Jenn Click to expand... If your equation was \(\displaystyle \frac{2}{x^2-1}-1=1/x-1\) then it is correct that \(\displaystyle \frac{2}{(x+1)(x-1)}-1=1/x-1\) However, the way you wrote it looks more like \(\displaystyle \frac{2}{x^2}-1=\frac{1}{x}-1\) Perhaps you need to add some brackets to your post?
