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Jobisdone [24]
4 years ago
7

Solve the system of equations x^3 = 3xy - 4 y^3 = 3xy - 5

Mathematics
1 answer:
mrs_skeptik [129]4 years ago
8 0

A graphing calculator shows the one real solution is near (x, y) = (-0.904, -1.202).

Rewriting each equation to give an expression for 3xy lets you write the relationship

... y³ = x³ -1

This lets you write an expression for y that can be substituted into the first equation to get

... x³ = 3x∛(x³-1)-4

Rewriting this as

... f(x) = x³ -3x∛(x³-1) +4

gives an equation that lets you use iteration methods to refine the root value to any desired degree. We get the real solution to be

... (x, y) = (-0.904012352468, -1.20249332242)

_____

Another approach that can get to a solution is to let x³=z. The the above equation for f(x) becomes

... z -3∛(z(z-1)) +4 = 0

... (z +4)³ = 27z(z -1) . . . . subtracting the radical term, then cubing both sides

... z³ -15z² +75z +64 = 0 . . . . rearranging to standard form

The one real root of this cubic can be found by any of a variety of means, including use of a graphing calculator or the cubic formula. Its solution near

... z = -0.738793548317

is the cube of the solution for x, and is 1 more than the cube of the solution for y in the original system of equations. That is,

... (x, y) = (∛z, ∛(z-1)) = <same x, y values as above>

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