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.

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.

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
**Posts:**1628**Joined:**Mon Dec 08, 2008 4:22 pm-
**Contact:**

Non-linear systems tend to be ugly, no matter what. But you might be able to simplify a bit byFind all real solutions of the system:

x - y = 3

x^3 - y^3 = 387

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Plug "3" in for the "x - y", and simplify by dividing through by the 3. Then solve the resulting

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

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:

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This is a radical equation; you begin to

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Finish the simplification, and then verify that each x-value is "allowed" inside the square root in the equation with the "\mbox{[*]}[/tex]" 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.