Simplifying expressions, lets consider this one:
1/(2y-1) + 1/(1+2y) - 1/(y+2) = 0
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Okay, so the problem is the multiple denominators here; so lets get rid of those,
but first we rearrange it so that we do not work with a 0 on the rhs:
As we add 1/(y+2) to both sides:
(1/(2y-1) + 1/(1+2y) - 1/(y+2)) + 1/(y+2) = (0) + 1/(y+2) <=>
1/(2y-1) + 1/(1+2y) = 1/(y+2) <=>
Getting rid of the left hand denominator by multiplying both sides with (y+2) gives:
(y+2)(1/(2y-1) + 1/(1+2y)) = (y+2)(1/(y+2)) <=>
(y+2)/(2y-1) + (y+2)/(1+2y) = 1 <=>
Now lets get rid of another by directly multiplying with (1+2y) on both sides:
(1+2y)((y+2)/(2y-1) + (y+2)/(1+2y)) = (1+2y)1 <=>
(1+2y)(y+2)/(2y-1) + (y+2) = (1+2y) <=>
Same story with the (2y-1):
(2y-1) ((1+2y)(y+2)/(2y-1) + (y+2)) = (2y-1)(1+2y) <=>
(1+2y)(y+2) + (y+2)(2y-1) = (1+2y)(2y-1) <=>
Lets have it equal 0 and make teachers happy!
(1+2y)(y+2) + (y+2)(2y-1) - (1+2y)(2y-1) = 0 <=>
Expanding it:
(y + 2 + 2yy + 4y) + (2yy - y + 4y - 2) - (2y - 1 + 4yy - 2y) = 0 <=>
Cancelling out terms/factors within parentheses:
(2yy + 5y + 2 ) + (2yy + 3y) - (4yy - 1) = 0 <=>
Getting rid of parentheses:
2yy + 5y + 2 + 2yy + 3y - 4yy + 1 = 0 <=>
Cancelling out terms/factors:
8y + 1 = 0 <=>
Rearranging the constants:
8y = -1 <=>
Finally dividing by the factor infront of our variable, in this case y:
y = -1/8
Tada!
And yes, I'm late to the party.
Last edited by Smogard49; Nov 27, 2013 at 12:17 AM.
Reason: Damn, k6 were fast as hell to understand that reference...
