# Why is $y + x = 3$ not the same as $y^2 + x^2 = 9$

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.

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Why are they meant to be equivalent? – Tunococ Sep 10 '12 at 20:38
Why should the two equations be equivalent? Let $y=1$ and $x=2$. Is $1^2+2^2=9$? – André Nicolas Sep 10 '12 at 20:38
What is this question for exactly? – S4M Sep 10 '12 at 20:45

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 $$y^2 + x^2 = 9.$$

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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 $x+y=\text{what?}$

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+1. It's a shame I can only vote once. – Blue Sep 11 '12 at 0:05
Thank you, Blue. – Michael Hardy Sep 11 '12 at 1:06
@MichaelHardy : Have you a book or article conclude pretty examples like above? – Hoseyn Heydari Dec 17 '13 at 10:20

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.

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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 $x,y\in \mathbb N > 0$.

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This does not answer the question in its current (and intended) format; see my comment to Luis Costa's answer. – M Turgeon Sep 10 '12 at 20:57
@MTurgeon - Better? – KeithS Sep 10 '12 at 21:27

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.

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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.

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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).

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It is not supposed to be interpreted as an intersection problem. If we look at the edit history, it is clear that the OP is asking why the two equations do not define the same set of solutions. – M Turgeon Sep 10 '12 at 20:40

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.

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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.

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