2 equations and 3 variables

$x+y=2\iff y=2-x\implies x(2-x)-z^2=1$ $\iff x^2-2x+(z^2+1)=0$ $\implies D=-4z^2\ge 0\implies z=0.$

Hence $z=0$ . This leaves us with a simple system of equations: \begin{cases}x+y=2\\xy=1\end{cases}\implies x+\frac{1}{x}=2\stackrel{x\neq 0}\iff x^2-2x+1=0\implies x=1\implies y=1.

After checking $(1;1;0)$ , we see it is a solution. Hence $1+1+0=\boxed{2}$ .

Since z has even power(squared), it will take the values of the form k,-k.

If we sum up all the values, it will always be zero. So we just have to add x and y.

So, answer is x + y, which is 2 as per question.

So, answer is 2.

Because ${ z }^{ 2 }$ is at least 0, then $xy$ must be at least 1. But since $x+y=2$ , the sole case where $xy$ is at least 1 is when x=1 and y=1. Then $xy$ can only be exactly 1, and $z$ can only be 0. So the sum is 1+1+0=2.

using equation 1: y=2-x and sub this into equation 2 to give: x(2-x)-z^2=1. This can be rearranged to get x^2 +2x+1 +z^2 = 0, which can factorized into the following equation: (x-1)^2 +z^2 =0. Both parts of the equation must equal zero, therefore x=1 and z=0. Now it follows that y=2-x=1 and hence x+y+z= 2

since z^2 is always nonnegative we know that xy must be at least 1, meaning it must be positive. this means either x and y are both positive or both negative. the sum of any two negative numbers must be negative and thus cannot equal 2, so x and y have to be both positive. this leave three outcomes for (x,y) based on the first equation. they are (0,2) (1,1) and (2,0). the part established earlier about xy being at least 1 is not met by (0,2) or (2,0), which leaves only (1,1) as our solution for (x,y). some quick substitution on the second equation gives 1*1-z^2=1. z=0. the only solution set is (1,1,0) which, when added together, gives our answer of 2.

that's kind of my "makes sense" solution which is not a mathematically sound proof, which is something I need to work on. my "three outcomes" statement up there is a bit rough, so let me see if I can clean that section up here. rearranging the first equation gives y=2-x. knowing both x and y to be positive gives us the solution set (x,2-x) for all 0<=x<=2. the other part we established earlier (the part about xy at least 1) can be rewritten using the new y as x(2-x)>=1 or 2x-x^2>=1. this inequality is only true at x=1. using the first equation gets us the single solution for (x,y) of (1,1).

Using x+y=2 we get xy = y(2-y) = 2y - y².The we replace it in the second equation: 2y - y² -z² = 1 then we adjust it to get some squares: y² - 2y +1 = -z² or (y-1)² = -z² which can only be true when z=0. And since we have x+y=2 then x+y+z=2.

i just thinking about if z = 0, then xy = 1. and to make xy=1, x and y must 1.. and its correct

Now xy-z2=1 hence x=(1+z2)/y playing this in eqn x+y=2 we get z2=-(y-1)2 this is a complex no. Hence neglecting z the summation is 2 from the eqn

From $x + y = 2$ , we can deduce (or differentiate $xy$ ) that the maximum value of $xy$ is $1$ and $z^2$ has to be larger or equal to zero if $z$ is real number.

The maximum of $xy$ minus the minimum of $z^2$ is equal to $1$ , which means the solution to these equations is unique with $x=1, y=1, z=0$ .

Hence $x + y + z = 2$ .

We have $x + y >= 0$ .

Moreover, $xy - z^2 = 1 \Leftrightarrow z^2 = xy - 1 \Leftrightarrow xy \geq 1 \Leftrightarrow x \geq 1/y$

We know that $y + \dfrac {1}{y} = 2$ with the equality catched when $y = 1 \Leftrightarrow x = y = 1$

Finally, we get $z^2 = xy - 1 = 1 - 1 = 0$

So, the only solution of the system is $(x, y, z) = (1, 1, 0)$ .