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I wanted graph of this equation:
$$left(left(fracleft(xsqrtfracleft-x-5+5right)5right)^2+left(fracy4right)^2-1right)left(left(fracleft(xsqrtfracleft-x-5+5right)3.4right)^2+left(fracy2.72right)^2-1right) left(x-yright)=0$$

I want a x=y line as well. So I kept (x-y) on the function side since (x-y)=0 should ultimately give x=y as graph along with graph of other functions. But I don't get the x=y line as a whole. It's not complete and I don't understand why.

I think there's something to do with which other equations I have used. But that doesn't seem to click. Isn't it mathematically correct to multiply two equations with 0 on right sides and get both equations' graph?

You're completely right about "the other equations" being the source of the problem.

Let me start with something simple. What is the value of
$$
x cdot 0 ?
$$

It's very tempting to say, "It's $0$, of course, because anything times zero is zero!" But what if $x$ is, for instance, a cheese omelet? Then...well...then multiplication by zero isn't even defined, and you'd have to say "$x cdot 0$ is undefined" rather than "it's zero." You might say that "cheese omelet" isn't really math, but what if we replace $x$ like this:
$$
(1/0) cdot 0
$$
Then our first factor is undefined, and multiplication by undefined things is undefined, so once again the product is "undefined" rather than zero.

Now let's look at a more complicated example. I'm going to stick with the real numbers -- no complex numbers allowed! -- and I want to look at
$$
sqrtx cdot 0
$$
You might say "Well, that's definitely zero!", but what if $x = -3$? Then the square root is undefined, and so the product is undefined as well.

Here's a theorem that your question hints at:

Theorem: Suppose $f, g : Bbb R^2 to Bbb R$, and $G$ is the solution set of the equation $g(x, y) = 0$, and $F$ is the solution
set of the equation $f(x, y) = 0$. Then the solution set of the
equation $$ f(x,y) g(x, y) = 0 $$ is exactly $F cup G$.

In your case, you've got $g(x, y) = x - y$, and the solutions for $x-y = 0$ form a line in the plane; that's the set $G$ in the theorem. You've multiplied it by some complicated mess that I'll temporarily call $f(x, y)$, and looked for solutions to
$$
tag1
f(x, y) cdot g(x, y) = 0
$$
(perhaps using Desmos or some other graphing tool), and you're wondering why the solution set of this doesn't include all of $G$.

The answer is "because $f$ is not defined on all of $Bbb R^2$ --- for certain $(x,y)$ pairs, $f(x, y)$ is undefined, and those pairs cannot be part of the solution to equation (1)."

For instance, let's look at the point $(a, b) = (-1, -1)$. Clearly this point satisfies $g(a, b) = a - b = (-1) - (-1) = 0$. But what is $f(a, b)$? Well, your formula for $f$ is
$$
f(x, y) = left(left(fracleft(xsqrtfracleft-x-5+5right)5right)^2+left(fracy4right)^2-1right)left(left(fracleft(xsqrtfracleft-x-5+5right)3.4right)^2+left(fracy2.72right)^2-1right)
$$
and plugging in $-1$ for each of $x$ and $y$ gives, deep in the middle, this:
$$
ldots left(fracleft((-1)sqrtfracleft-(-1)-5+5right)5right)^2 ldots
$$
and the middle square root in that simplifies to
beginalign
&(-1)sqrtfracleft-(-1)-5\
&= -sqrtfracleft1-5\
&= -sqrtfrac-4\
&= -sqrtfrac4-4\
&= -sqrt-1
endalign
which is undefined. So because $f(a, b)$ is undefined, so is $f(a, b) cdot g(a, b)$, and hence this product cannot be zero...so the point $(a, b) = (-1, -1)$ is not part of the plot of the solutions to equation (1).

The modified theorem that gets to what's going on here is this:

Theorem: Suppose $D, E subset Bbb R^2$, $f:D to Bbb R$, $g:E to Bbb R$, $G$ is the solution set of the equation $g(x, y) = 0$, and $F$ is the solution
set of the equation $f(x, y) = 0$. Then the solution set of the
equation $$ f(x,y) g(x, y) = 0 $$ is exactly $(Fcap E) cup (G cap D)$.

which says that only those solutions of $g = 0$ (i.e., $G$) that lie in the domain of $f$ (i.e., $D$) get included in the result, and only solutions of $f = 0$ that lie in the domain of $E$ get included in the result.

Notice that if $f(x) = left(left(fracleft(xsqrtfracleft-x-5+5right)5right)^2+left(fracy4right)^2-1right)left(left(fracleft(xsqrtfracleft-x-5+5right)3.4right)^2+left(fracy2.72right)^2-1right)$

then $f(x)$ is not defined for $x=-5$ because then $sqrt frac -5-x-5$ is dividing by $0$. And if $x > -5$ then $-x-5 < 0$ so $sqrt frac -5-x-5$ is the square root of a negative number.

So $f(x)$ simply doesn't work and does not exist if $x ge -5$.

And if $f(x)$ does not exist for $x ge -5$ then $f(x)cdot (x-y)$ can't exist for $x ge -5$ either.

It doesn't matter that multiplying by $f(x)(x-y) = 0$ and $0$ exists always. If $f(x)$ doesn't exist then neither does $f(x)times something else$. So we have $f(x)(x-y)= 0$ AND $x < -5$. We can't magic things into existence just because they "cancel out".