SOLUTION: Solve. If problem has no solution; write no solution.
1 = 3 over y+2 + 1 over y-2
Here is what I have:
(y+2)(y-2) = 3(y-2)+ y + 2
y^2 + 2y - 2y - 4 = 3y - 6
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Question 69181This question is from textbook Algebra Structure and Method
: Solve. If problem has no solution; write no solution.
1 = 3 over y+2 + 1 over y-2
Here is what I have:
(y+2)(y-2) = 3(y-2)+ y + 2
y^2 + 2y - 2y - 4 = 3y - 6 + y + 2
y^2 - 4 = 4y - 4
and that is as far as I got, I am lost
This question is from textbook Algebra Structure and Method
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
You did fine as far as you went so let's continue from where you left off:
Add 4 to both sides of the equation.
Subtract 4y from both sides.
Factor a y on the left side.
Apply the zero product principle.
and/or
The solutions are:
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