GRE test 9367 #45

Forum for the GRE subject test in mathematics.
deckoff8
Posts: 23
Joined: Tue Nov 08, 2011 11:12 am

GRE test 9367 #45

Postby deckoff8 » Fri Nov 11, 2011 8:34 pm

Consider the system of equations
ax^2 + by^3 = c
dx^2 + ey^3 = f

where a, b, c, d, e, and f are real constants and ae != bd. The maximum number of real solutions (x,y) of the system is:

The answer is two. Though I'm not sure how they got that. If you subtract one from the other you get (a-d)x^2 + (b-e)y^3 = c-f, which has at least 3 solutions, 2 for positive and negatives values of x when y is 0, and 1 for y when x is 0. What am I missing?

owlpride
Posts: 204
Joined: Fri Jan 29, 2010 2:01 am

Re: GRE test 9367 #45

Postby owlpride » Fri Nov 11, 2011 10:26 pm

Careful. Not every solution to "(a-d)x^2 + (b-e)y^3 = c-f" also solves both equations individually.

Let's do a substitutions of variables. Let s = x^2, t = y^3. Now you get a system of linear equations in s and t:

a s + b t = c
d s + e t = f

Since the coefficient matrix is non-singular, the system has precisely one solution (s,t). Now let's undo the change of variables. There are at most two real solutions to x^2 = s, and precisely one (not counting multiplicities) for y^3 = t.

That's at most two real solutions.

deckoff8
Posts: 23
Joined: Tue Nov 08, 2011 11:12 am

Re: GRE test 9367 #45

Postby deckoff8 » Fri Nov 11, 2011 10:37 pm

Damn. That is an awesome explanation. High five and thanks for the response. wish us luck tomorrow ~_~




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