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Q: How do you use elimination for x-y1 and x y3?

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Y3

12x2 - 24x + 9y

x^2/y^3 = x^2*y^(-3)

no

I think you mustve left out the = sign, but if it's supposed to be x=-y3 I think that would be... y=-3x

Related questions

y^3

y3 x y3 - y (3)3 x 3(3) - 3 9 x 9 - 3 = ? 9 x 9= 81 81 - 3 = 78 I hope that solves your problem

Y3

x6 - y6 = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y)(x2 - xy + y2)(x - y)(x2 + xy + y2)

12x2 - 24x + 9y

x^2/y^3 = x^2*y^(-3)

no

10xy3 * 8x5y3 = (10*8)*(x*x5)*(y3*y3) = 80x6y6

I think you mustve left out the = sign, but if it's supposed to be x=-y3 I think that would be... y=-3x

x-9+y3

The expression is x/y3 - f.

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