solving non-linear system: x - y = 3, x^3 - y^3 = 387

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testing
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solving non-linear system: x - y = 3, x^3 - y^3 = 387

Find all real solutions of the system:

x - y = 3
x^3 - y^3 = 387

Either this is going to be really nasty, or I'm missing some "tricK", I think. Thoughts? Thanks in advance.

stapel_eliz
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testing wrote:Find all real solutions of the system:

x - y = 3
x^3 - y^3 = 387

Non-linear systems tend to be ugly, no matter what. But you might be able to simplify a bit by factoring:

. . . . .$x^3\, -\, y^3\, =\, (x\, -\, y)(x^2\, +\, xy\, +\, y^2)\,=\, 387$

Plug "3" in for the "x - y", and simplify by dividing through by the 3. Then solve the resulting literal equation for, say, y in terms of x, using the Quadratic Formula:

. . . . .$y^2\, +\, xy\, +\, x^2\, -\, 129\, =\, 0$

. . . . .$1(y^2)\, +\, x(y)\, +\, (x^2\, -\, 129)\, =\, 0$

. . . . .$y\, =\, \frac{-x\, \pm\, \sqrt{(x)^2\, -\, 4(1)(x^2\, -\, 129)}}{2(1)}$

. . . . .$y\, =\, \frac{-x\, \pm\, \sqrt{516\, -\, 3x^2}}{2}\, \mbox{ [*]}$

This gives you two solutions, presumably one from the "plus" and the other from the "minus" (which you can see on your calculator, if you graph Y1=X-3 and Y2=(X^3-387)^(1/3) in the same window). The solutions will be the places where a "half" above crosses the other line, y = x - 3, assuming there is a crossing. The solution to one "half" might start like this:

. . . . .$x\, -\, 3\, =\, -\frac{x}{2}\, -\, \frac{\sqrt{516\, -\, 3x^2}}{2}$

. . . . .$2x\, -\, 6\, =\, -x\, -\, \sqrt{516\, -\, 3x^2}$

. . . . .$3x\,-\, 6\, =\, -\sqrt{516\, -\, 3x^2}$

This is a radical equation; you begin to solve by squaring both sides:

. . . . .$9x^2\, -\, 36x\, +\, 36\, =\, 516\, -\, 3x^2$

. . . . .$12x^2\, -\, 36x\, -\, 480\, =\, 0$

. . . . .$x^2\, -\, 3x\, -\, 40\, =\, 0$

. . . . .$x\, =\, \frac{-(-3)\, \pm\, \sqrt{(-3)^2\, -\, 4(1)(-40)}}{2(1)}$

. . . . .$x\, =\, \frac{3\, \pm\, \sqrt{169}}{2}\, =\, \frac{3\, \pm\, 13}{2}$

Finish the simplification, and then verify that each x-value is "allowed" inside the square root in the equation with the "$\mbox{[*]}$" above. Once you've found which, if any, of the values is allowable, back-solve (using "y = x - 3") for the corresponding y-values for that "half".

Then note that, due to the squaring, solving the other "half" should look very similar.

Eliz.

testing
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Joined: Sun Dec 07, 2008 12:26 am
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Thanks.