Very easy algebraic problem that I'm struggling with

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It's just been so long that I've done this, I dont remember how

How do you solve the following for Y?


1/2y(y+2)=24?


No matter what I do, I screw it up!!!

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by scoutkb » Tue Apr 10, 2007 5:51 am
Ok..here is my take.

1/2Y(Y+2)=24

1) Notice the fraction...i always try to get rid of the fractions. So here i would multiply both sides of the equation by "2".

2 * 1/2Y(Y+2) = 24 * 2

2) The above step got rid of the 1/2, now solve for Y.

Y(Y+2)= 48

Multiply by Y

Y^2 + 2Y=48

Now subtract 48 from the right side and move to the left and set EQ to 0

Y^2 + 2Y - 48= 0

reverse FOIL

(Y-6) (Y+ 8 ) = 0

set Y-6=0 ---> Y=6
set Y+8=0 ---> Y=-8


So Y= 6 & -8

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by Stacey Koprince » Mon Apr 16, 2007 10:40 pm
Not sure if your problem should be read as:

(1/2y)(y+2)=24
or
(1/2)(y)(y+2) = 24

For the first version:
distribute the 1/2y to get: 1/2 + 1/y = 24
subtract the 1/2 to get: 1/y = 23.5
then take reciprocal of both sides to get: y = 1/23.5 = 2/47

For the second version:
multiply by 1/2 to remove fraction: y(y+2) = 48
distribute the left side: y^2 + 2y = 48
get everything on one side: y^2 + 2y - 48 = 0
factor: (y+8)(y-6) = 0
y = -8, y = 6
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