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Find the number of ordered triples (x,y,z) of real numbers that satisfy

 \(\begin{align*} x + y - z &= 0, \\ xz - xy + yz &= 27, \\ xyz &= 54. \end{align*}\)

 Jun 27, 2019
 #1
avatar+2417 
+1

I don't really have a good method, but this is what I did:
 

notice x*y*z = 54

 

So prime factorize 54

 

3 * 3 * 3 * 2

 

Ok so with the prime factorization, it isn't tedious at all to guess and check with those numbers into the other equations.

 

x + y - z = 0 (the first equation) is also:

 

x + y = z

and

xyz = 54

 

It is much easier now to guess and check with only the prime factorization.

 

The second equation is very complicated we can use it to check our three x, y, and z values later.

 

So with our prime factorization 3 * 3 * 3 * 2

 

I guessed that (3, 3, 6) is a solution to the equations we have above. Then we check with the second equation to make sure it is true, which it is.

 

Ok, notice how our answer also could be negative. However, that can't be true because xyz = 54 has to have two negative numbers but the only solution to x + y = z has three.

 

So the answer is 1 I believe...

 

This problem was more of a logic puzzle with guessing in my opinion.

 Jun 27, 2019
edited by CalculatorUser  Jun 27, 2019
 #4
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With guesswork you might just get lucky, as with this example.

The problem is that you are not told that the numbers are integers, you're simply told that they are real numbers.

It could happen that none of your guesses work.

Guest Jun 28, 2019
 #2
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Use substitutions to get these values:
x = -6 and y = 3 and z = -3
x = 3 and y = -6 and z = -3
x = 3 and y = 3 and z = 6

 Jun 27, 2019
 #3
avatar+28205 
+2

Here's a systematic way of approaching this:

 

 Jun 28, 2019
edited by Alan  Jun 28, 2019

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