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Solve for y. (3xy^2z + 2x^2y^3)/(3x^2yz) = 100
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Solve for y.

(3xy^2z + 2x^2y^3)/(3x^2yz) = 100

User Ladislav Indra
by
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1 Answer

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(3xy^2z + 2x^2y^3)/(3x^2yz) = 100 \\ (xy^2(3z + 2xy))/(3x^2yz) = 100 \\ (y(3z + 2xy))/(3xz) = 100 \\ y(3z + 2xy)=100 * 3xz \\ y(3z + 2xy)=300xz \\ 3zy + 2xy^2-300xz=0 \\ 2xy^2+3zy-300xz=0 \\ \Delta=(3z)^2-4 * 2x *(-300xz) =9z^2+2400x^2z \\ y= (-3z \pm √(9z^2+2400x^2z ) )/(2 * 2x) = (-3z \pm √(3z(3z+800x^2) ) )/(4x)
User Dholbert
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