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

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

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

Postby testing on Fri Feb 06, 2009 3:25 pm

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.
testing
 
Posts: 21
Joined: Sun Dec 07, 2008 12:26 am

Sponsor

Sponsor
 

  TOPIC_SOLVED

Postby stapel_eliz on Fri Feb 06, 2009 4:17 pm

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:

. . . . .

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:

. . . . .

. . . . .

. . . . .

. . . . .

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:

. . . . .

. . . . .

. . . . .

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

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

Finish the simplification, and then verify that each x-value is "allowed" inside the square root in the equation with the "" 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. :wink:

Eliz.
User avatar
stapel_eliz
 
Posts: 1784
Joined: Mon Dec 08, 2008 4:22 pm

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

Postby testing on Mon Feb 09, 2009 9:14 pm

Thanks.
testing
 
Posts: 21
Joined: Sun Dec 07, 2008 12:26 am


Return to Advanced Algebra ("pre-calculus")