> Simplifying Algebra Question?

Simplifying Algebra Question?

Posted at: 2014-06-09 
I am on a worksheet of complex fractions and the last two problems I just can't solve and need help. I've been trying to work them out for the last couple of hours and I'm coming up with a answer that just isn't right. Thanks! The first one (x-2/2x) +1=x+1/x Second one: 1/x-1+1/x+1=2/x^2-1
On the second one, factor the x^2 - 1 1/x-1+1/x+1=2/(x + 1)(x - 1) so the common denominator is (x + 1)(x - 1). The first fraction needs the (x + 1) so multiply top and bottom by (x + 1). The second fraction needs the (x - 1) so multiply top and bottom by (x - 1). The third has both so it's ok. (x + 1) / (x+1)x-1) + (x - 1) / (x + 1)(x - 1) = 2 / (x+1)(x-1) Now since the denominators are all the same, the numerators must add to be equal. So x + 1 + x - 1 = 2 2x = 2 x = 1 But if x = 1, the first fraction would be 1/0 which is undefined. So, that means there is no solution for this equation. On the first, the common denominator is 2x. The first fraction already has 2x so it's ok. The second, which is 1/1, needs the whole 2x so multiply top and bottom by 2x. I can't tell if the right side has the whole x + 1 over the x, or just the 1 so I'll guess it's the whole x + 1. The denominator needs the 2 so multiply by 2/2 (x - 2) / 2x + 2x / 2x = 2(x + 1) / 2x and now like before, just do the tops x - 2 + 2x = 2(x + 1) 3x - 2 = 2x + 2 x = 4