This is actually in the Linear Equations section of my book. I've looked everywhere for a similar problem. please help me break it down.

x/(x^2-1) - (x+3)/(x^2-x) = -3/(x^2+x)

This is actually in the Linear Equations section of my book. I've looked everywhere for a similar problem. please help me break it down.

x/(x^2-1) - (x+3)/(x^2-x) = -3/(x^2+x)

x/(x^2-1) - (x+3)/(x^2-x) = -3/(x^2+x)

canadia_ wrote:This is actually in the Linear Equations section of my book. I've looked everywhere for a similar problem. please help me break it down.

x/(x^2-1) - (x+3)/(x^2-x) = -3/(x^2+x)

They show how to do this kind of equation here. First you factor x^2 - x and x^2 + x to be x(x - 1) and x(x + 1). Then you factor the difference of squares to get x^2 - 1 to be (x - 1)(x + 1). Then multiply through by the common denom x(x - 1)(x + 1). Then solve the quadratic.

Please post back if you get stuck. Thanks.

at what point do i solve the quadratic. after i multiply by common denominators I got: (x)(x) - (x+1)(x+3) = (x-1)(-3) ...

this is how i set up the cancellations: (x)(x+1)(x-1)*(x/(x-1)(x+1) - (x)(x+1)(x-1)*(x+3)/(x(x-1)) = (x)(x+1)(x-1)* -3/(x(x+1))

this is how i set up the cancellations: (x)(x+1)(x-1)*(x/(x-1)(x+1) - (x)(x+1)(x-1)*(x+3)/(x(x-1)) = (x)(x+1)(x-1)* -3/(x(x+1))

canadia_ wrote:at what point do i solve the quadratic. after i multiply by common denominators I got: (x)(x) - (x+1)(x+3) = (x-1)(-3) ...

You multiply out what you got. Then you put everything on one side of the equals, with zero on the other side. Then you solve.

Thanks guys. you made my day. be back soon!