Please show me how to solve this! 2/y^2-3y+2 + 7/y^2-1?

Answer 1

2/(y^2-3y+2) + 7/(y^2-1) <-- I think you mean this.
2/[(y-1)(y-2)] + 7/[(y-1)(y+1)]
= 2(y+1)/[(y-1)(y-2)(y+1)] + 7(y-2)/[(y-1)(y-2)(y+1)]
= (2y+2+7y-14)/[(y-1)(y-2)(y+1)]
= (9y-12)/[(y-1)(y-2)(y+1)]

Answer 2

I am assuming the equation should be written as such:

[2/(y^2 - 3y + 2)] + [7/ (y^2 - 1)]

The first step is to factor the bottom to produce the following:

[2/(y-2)(y-1)] + [7/(y-1)(y+2)]

find the least common denominator and add the two together to get the following:

[2(y+1) + 7(y-2)] / [(y-2)(y+1)(y-1)]

multiply out the top to get the final answer:

[3(3y-4)]/[[(y-2)(y+1)(y-1)]

Answer 3

2 / y^2 - 3y +2 + 7 / y^2 -1
9 / y^2 - 3y +1

Answer 4

2/y^2-3y+2 + 7/y^2-1
9/y^2 - 3y + 1

Answer 5

2/( y^2-3y+2)+ 7(/y^2-1)
= 2/ [(y -2)(y -1)] + 7 / [(y+1)(y - 1)]
= [ 2(y+1) + 7(y - 2) ] / [ (y - 2)(y + 1)(y - 1)]
= ( 9y - 12 ) / [ (y - 2)(y + 1)(y - 1)]
= 3( 3y - 4) / [ (y - 2)(y + 1)(y - 1)]

Answer 6

I can screw up this answer twice as fas as you can
Therefore : 2( 2/y^2-3y+27y^2-1) = normal screw up time:)

Answer 7

To solve this expression, you must transform it into an equation because you can only solve equations, not expressions. Here's how.

First find an equal sign. Place your given expression on the left side of the equal sign. Then find another term or number that goes with this expression albeit on the OTHER side of the equal sign. Voila! You now have an equation, which you can solve.

Answer 8

should put some parathesis in this to help clarify
i don't know if this is ;
(2/y^2) -3y +2
or 2/(y^2-3y+2)

Answer 9

http://www-math.cudenver.edu/~rbyrne/2423q1f98an.htm