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Equation games (Posted on 2002-07-03) Difficulty: 2 of 5
Can you design an equation with two unknowns and still find the value of one?

See The Solution Submitted by Dulanjana    
Rating: 3.3750 (8 votes)

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re: Anything better? | Comment 11 of 24 |
(In reply to Anything better? by levik)

Hello everyone. I am new to Flooble, so go easy on me for this one.

equation 1: (x-4)(y^2+3)=0 (assuming x and y are real)
there are two variables, but we know that x=4 is one of them. This equation is the same concept as the original solution. There is a product that equals zero, but one of them (y^2+3) can not equal therefore it has to be the other, thus x=4.

Others that fit the same pattern are (x+3)e^y=0 and we know x must be 3

equation 2: (x-5)(y-5)=0
This one is questionable...see what you think. It does answer the question "...and still find the value of one?" We know that "one" of the variables must = 5, we just don't know which one it is.

  Posted by Greg on 2003-01-07 18:28:41

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