SOLUTION: I don't know how to use my computer keyboard to type the correct symbols so I will do it to the best of my ability so you can understand. As you can se this is a division problem.

Question 301901: I don't know how to use my computer keyboard to type the correct symbols so I will do it to the best of my ability so you can understand. As you can se this is a division problem. ^=to the power of
y^2 + 7y - 18 / y^2 + 2y - 63 =
Found 2 solutions by helpnalgebra, london maths tutor:
Answer by helpnalgebra(91) (Show Source): You can put this solution on YOUR website!
y^(2)+7y-(18)/(y^(2))+2y-63
y^(2)+9y-(18)/(y^(2))-63
y^(2)*(y^(2))/(y^(2))+9y*(y^(2))/(y^(2))-(18)/(y^(2))-63*(y^(2))/(y^(2))
(y^(4))/(y^(2))+(9y^(3))/(y^(2))-(18)/(y^(2))-(63y^(2))/(y^(2))
(y^(4)+9y^(3)-18-63y^(2))/(y^(2))
(y^(4)+9y^(3)-63y^(2)-18)/(y^(2)) answer
Answer by london maths tutor(243) (Show Source): You can put this solution on YOUR website!
y^2 + 7y - 18 = (y+9)(y-2)
y^2 + 2y - 63 = (y-7)(y+9)
Therefore,
(y^2 + 7y - 18) / (y^2 + 2y - 63) = (y+9)(y-2) /(y-7)(y+9)
you can cancel (y+9) from the denominator and numerator
therefore the solution becomes :
(y-2)/(y-7)
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