System of equations in 3 variables

Algebra Level pending

Given that x , y , x,y, and z z are real numbers that satisfy the following system of equations

x + y + z = 5 x+y+z=5 x 2 + y 2 + z 2 = 15 x^2+y^2+z^2=15 x y = z 2 xy=z^2

find the value of

1 x + 1 y + 1 z . \frac{1}{x}+\frac{1}{y}+\frac{1}{z}.


The answer is 5.

Victor Loh by Victor Loh , 19, Singapore

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1 solution

We have ( x + y + z ) 2 = x 2 + y 2 + z 2 + 2 ( x y + x z + y z ) = 25 (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2(xy + xz + yz) = 25 , and so

x y + x z + y z = 5 xy + xz + yz = 5 .

But since x y = z 2 xy = z^{2} this means that

z ∗ ( z + x + y ) = 5 ⟹ z = 1 ⟹ x + y = 5 − 1 = 4 z*(z + x + y) = 5 \Longrightarrow z = 1 \Longrightarrow x + y = 5 - 1 = 4 .

Now 1 x + 1 y + 1 z = x + y x y + 1 z = 4 1 + 1 1 = 5 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{x+y}{xy} + \frac{1}{z} = \frac{4}{1} + \frac{1}{1} = \boxed{5} .

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1 x + 1 y + 1 z = x y + y z + z x z 3 = 5 1 = 5 \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{z^3}=\frac{5}{1}=\boxed{5} , so you didn't really need to find x + y x+y .

mathh mathh - 6 years, 10 months ago

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Yes, good point.

Also, I just wanted to get a message to you regarding the "Let me get this straight" question you were helping me with. I wasn't able to change the solution value from 25 25 to the one you suggested so I just deleted the question entirely. I still think that it was an interesting question but I had made such a mess of it that it was just best to 'go back to the drawing board'. Thanks again for your help. :)

Brian Charlesworth - 6 years, 10 months ago

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