#Algebra please help
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wheat flare
Yes, that's right.
You can conclude that 2xy-x-3y-11<0 because d>0 and LHS is -5 (<0)
That underlined thing will also be an integer.
Basically you can use the concept of diophantine equations to proceed after that.
quartz horizon
Is it? I'm not sure if it is.
We can perform a change of variables to get rid of linear terms: x = X + a, y = Y + b.
2xy - x - 3y - 11 = 2(X + a)(Y + b) - (X + a) - 3(Y + b) - 11 = 2XY + (2b - 1)X + (2a - 3)Y + 2ab - a - 3b - 11
So, a = 3/2 and b = 1/2. So, for x = X + 3/2, y = Y + 1/2 we get:
2xy - x - 3y - 11 = 2XY - 25/2
Thus:
2xy - x - 3y - 11 = 2(x - 3/2)(y - 1/2) - 25/2
This function doesn't have constant sign. It's a hyperbolic paraboloid, after all.
wheat flare
See, we have -5=d(2xy-x-3y-11)
Both d and 2xy-x-3y-11 are integers
-5= 5.(-1) or -5.1
But since d>0
You will only get negative value for 2xy-x-3y-11
Using little diophantine eqn here
quartz horizon
Oh, I didn't notice that this was in integers, sorry.
Then yeah, it can only be negative.
bleak sableBOT
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