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Q: What is a solution for the system of equations x-y2 and y2x-4?

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No.

Whooop!

(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)

-5xy2 + 12xy2 = (-5 + 12)xy2 = 7xy2

x3+xy-x2y2=x(x2+y-xy2)

Related questions

xy2

No.

15x2y2-9xy3 As x2y2 = x xy2 and xy3 = xy2 y then xy2 is in both term you can first factorize 15x2y2-9xy3 = xy2(15x-9y) as 15=3x5 and 9=3x3 15x2y2-9xy3 = 3xy2(5x-3y) and that it !

xy2

xy

XeF2

xy

xy

The GCF is xy

48

Whooop!

(x4 + y4)/(x + y) = Quotient = x3 - x2y + xy2 - y3 Remainder = - 2y4/(x+y) So, x3 - x2y + xy2 - y3 - 2y4/(x+y)