I know this is impossible, but why is the following not possible:
$y + x = 3$ is the same as $y^2 + x^2 = 9$
They're meant to be equivalent.
I know this is impossible, but why is the following not possible:
$y + x = 3$ is the same as $y^2 + x^2 = 9$
They're meant to be equivalent.
Squaring does not "distribute" in general. That is, $ (y + x)^2 \neq y^2 + x^2 $ (except when one or both of the variables happen to be $0$).
You can think of $( \cdots )^n$ as saying "write $(\cdots)$ down $n$ times". In our case, this means $ (y + x)^2 = (y + x)(y + x) = y^2 + 2xy + x^2. $
So, when you square both sides of $y + x = 3$, you get $ y^2 + 2xy + x^2 = 9, $ not $ y^2 + x^2 = 9. $
It's a little bit simpler with (believe it or not) $325$ instead of $9$. $ \begin{align} 1^2 + 18^2 = 325, & & 1 + 18 = 19 \\ 6^2 + 17^2 = 325, & & 6 + 17 = 23 \\ 10^2 + 15^2 = 325, & & 5 + 15 = 20 \end{align} $ So $x^2+y^2 = 325$ is the same as $x+y=\text{what?}$
They are not meant to be equivalent: the set of solutions to $y+x=3$ is a line, whereas the set of solutions to $y^2+x^2=9$ is a circle of radius 3.
The only solution that satisfies the equality is $x=0, y=3$ (or the reverse $x=3,y=0$). There are no $x,y\geq1$ that satisfy this equation.
Simply put:
$x^2 + y^2 = 9, x + y = 3$
$\sqrt{x^2 + y^2} = 3, x+y=3$
$\sqrt{x^2 + y^2} = x + y$
$x^2 + y^2 = (x+y)^2$
$x^2 + y^2 = (x+y)(x+y)$ <-THAT's how squaring a sum works
$x^2 + y^2 = x^2 + 2xy + y^2$
$0 = 2xy$
$0=xy$
We thus see that any $x,y$ that would satisfy both equations must include either $x=0$ or $y=0$. The error you apparently made is in thinking that $(x+y)^2 = x^2 + y^2$, which is incorrect; $(x+y)^2 = (x+y)(x+y) = x^2 + 2xy + y^2 \neq x^2 + y^2$ for any $x,y\in \mathbb N > 0$.
You are mixing up the concepts of equations and equivalences. The system of equations you wrote is possible, but only for specific values of $x$ and $y$ (because it is a system of equations). Assuming both are true:
$x+y=3$
$(x+y)^2=9$
$x^2+2xy+y^2=9$
$2xy=0$
Therefore either $x$ or $y$ is $0$. If one is $0$ then the other is $3$. So both $(3,0)$ and $(0,3)$ work. However, these are the only real values for which both equations are satisfied. So it is not correct to say both are true for all $(x,y)$.
i.e. $x^2+2x+1=(x+1)^2$ is an equality (both sides are equivalent), $x^2+2x-1=(x+1)^2$ is an equation.
Are you trying to say that $\sqrt{x^2 + y^2} = x + y$? Do you think $\sqrt{2} = 2$, e.g., $\sqrt{2} = \sqrt{1 + 1} = 1 + 1 = 2$ You seem to be taking the square root of both sides, using a rule that is false. We can take the square root of both sides, using a real rule, by doing the following: $\sqrt{x^2 + y^2} = 3$ This is equivalent to $x^2 + y^2 = 9$ and we know this because $x^2 + y^2 \geq 0$ for all $x$ and $y$. If it were true that $x + y = 3$ were also equivalent, it would imply that $\sqrt{x^2 + y^2} = x+y$ which is clearly not true.
The question is a little vague. See Grapth's answer for one possible answer or...
That system is possible. It is the equation of a line intersecting a circle. It has two possible solutions (0,3) and (3,0).
Start with your first equation $x+y=3.$ Since the numbers on both sides are equal, then the squares of those numbers must be equal. That is, $(x+y)^2 = 3^2 =9.$ Your error seems to be that you've taken $ (x+y)^2=x^2+y^2$ which is not true in general. The correct formula is $ (x+y)^2 = x^2 + y^2 + 2xy.$ This has been pointed out in a few of the answers, but to understand why it is true, you could look at a nice geometrical explanation of this fact here.
The locus for points of $\{(x,y) \;|\; x+y=3\}$ is a line & for $\{(x,y) \;|\; x^2+y^2=3^2\}$ is a circle. So they're solutions can't be the same.