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)

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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_
**Posts:**3**Joined:**Thu Oct 03, 2013 3:27 pm

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.

- theshadow
**Posts:**91**Joined:**Sun Feb 22, 2009 11:12 pm

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_
**Posts:**3**Joined:**Thu Oct 03, 2013 3:27 pm

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.

- anonmeans
**Posts:**56**Joined:**Sat Jan 24, 2009 7:18 pm

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

- canadia_
**Posts:**3**Joined:**Thu Oct 03, 2013 3:27 pm

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