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Lets say i have 2 equations in the form ... = x

How would i superimpose them so that when i graph the equations i can see both lines in the same equation.
I thought i could just take each part and multiply them like this:

equation 1: ... = 0
equation 2: ... = 0

superimposed equation: (...) * (...) = 0

However when i tried this it did not work, the equations i used were:

$x\ -\ 3\ -\ \ 0\ \sqrt{-\frac{\left|\left|y\ -\ 2\right|-1\right|}{\left|y\ -\ 2\right|\ -\ 1}}=\ 0$

and

$y-1\cdot\ \sqrt{-\frac{\left|\left|x\ -\ 2\right|\ -\ 1\right|}{\left|x\ -\ 2\right|\ -\ 1}}=0$

The superimposition i tried doing (which didnt work) was:
$\left(y-1\cdot\ \sqrt{-\frac{\left|\left|x\ -\ 2\right|\ -\ 1\right|}{\left|x\ -\ 2\right|\ -\ 1}}\right)\cdot\left(x\ -\ 3\ -\ \ 0\ \sqrt{-\frac{\left|\left|y\ -\ 2\right|-1\right|}{\left|y\ -\ 2\right|\ -\ 1}}\right)=0$
however Desmos just shows nothing

If you plot them separately (i used Desmos for this) you will get a horizontal line and a vertical line,
superimposed they should thus form a 90 degree angle

Luke_
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  • If you plot $x-y=0$ and $x+y=0$ you get two straight lines, which is all the points where the two equations are true. If you multiply you get $x^2+y^2 = 0$, resulting in the single point $(0,0)$, which is where this equation is true. No idea what you are trying to do, but it is not supposition in any sense. – Paul Jul 18 '21 at 16:00
  • Yes, i was expecting to be wrong, but my question was how do i do superposition, the reason i thought this would work is because in the batman equation, they do actually multiply the different parts/lines to combine/superimpose them @Paul – Luke_ Jul 18 '21 at 16:02
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    Given $f(x, y) = 0$ and $g(x,y)=0$ , then $h(x,y) = f(x,y) g(x,y) = 0$ will contain both the graphs of $f(x,y) = 0$ and $g(x,y) = 0$ simply because $h(x,y) = 0$ implies that either $f(x,y) = 0$ or $g(x,y) = 0$. So, if you take $f(x,y) = x - y$ and $g(x,y) = x + y$ then $h(x,y) = x^2 - y^2 $. The graph of $h(x,y) = 0$ contains the two lines $y = x$ and $y=-x$ which are the graphs of $f(x,y) = 0$ and $g(x,y)=0$ respectively. – Hosam Hajeer Jul 18 '21 at 16:42
  • To create a piece of a curve to plot with another piece, you need to restrict the values of x (or y) for which each equation is valid. For example, $f(x) = x$ for $0\le x\le 1$ which give you a line segment from $(0,0)$ to $(1,1)$. – Paul Jul 18 '21 at 16:43
  • Oops, stuck in a +in my comment sorry & fooled myself... – Paul Jul 18 '21 at 16:47
  • @GeometryLover as i said before, i tried doing that (in the example posted in my question) but that did not work, maybe i did it wrong – Luke_ Jul 18 '21 at 17:30
  • The points that satisfy one equation don't just make the other equation untrue, they make it undefined due to division by zero, so when you're plugging in, say, $(2,1)$, you're testing if (undefined)*(0)=0 which Desmos just gives up on as soon as it encounters a division by 0. – TomKern Jul 18 '21 at 17:32
  • I tried to fix this by replacing the function $\frac{|x|}{x}$ by $\text{sign}(x)$, which is 0 when $x = 0$ and Desmos knows about. However this also doesn't work as Desmos doesn't handle complex numbers: it treats square roots of negatives as undefined.

    I tried to fix it again by replacing your square roots by ${1\leq x \leq 3}$, which Desmos knows to be equal to 1 if the inequality is true and undefined otherwise. This still didn't work.

    In fact, the graph of $(y-1)(x-3)=0$ isn't working correctly. Does anyone have any insights as to why?

     – TomKern Jul 18 '21 at 17:33

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