Find the number of ordered pairs satisfying the following:

Question

What are the number of ordered pairs of (x,y) satisfying the following system of simultaneous equations? \begin{align*} |x^2-2x|+y & =1\\ x^2+|y| & =1 \end{align*}

Answer 1

Observe: From the second equation $|y|=1-x^2$ it is clear that $x \in [-1,1]$. Thus $|x^2-2x|=|x||x-2|=|x|(2-x)$ (because for $x \in [-1,1]$ we will have $-3 \leq x-2 \leq -1$).

Now consider the following:

Case(1): Let $y \geq 0$, then the system is \begin{align*} |x|(2-x)+y & =1\\ x^2+y & =1 \end{align*} which implies $$|x|(2-x)=x^2.$$

Case(1a): if $x \geq 0$, then we have $$x(2-x)=x^2$$ Here we have only TWO solutions namely $\color{red}{x=0,1}$.

Case(1b): if $x < 0$, then we have $$-x(2-x)=x^2$$ Here we have NO solutions.

Case(2): Let $y < 0$, then the system is \begin{align*} |x|(2-x)+y & =1\\ x^2-y & =1 \end{align*} Once again eliminate $y$ and proceed. Can you take it from here?

Answer 2

\begin{align} |x^2-2x| &= 1-y\\ |y| &= 1-x^2 \\ \end{align}

First we note that, if $y=0$, then $x = 1$.

Next we note that we need $1-x^2 \ge 0$; that is $-1 \le x \le 1$.

If $y < 0$, then $y=x^2-1$:

\begin{align} |x^2-2x| &= 1-y \\ |x^2-2x| &= 2-x^2 &\text{(Square both sides.)} \\ x^4-4x^3+4x^2 &= x^4-4x^2+ 4 \\ 4x^3-8x^2+4 &= 0 \\ 4(x-1)(x^2-x-1)&= 0 \\ x &\in \{1,\, \frac 12(1-\sqrt 5),\, \frac 12(1+\sqrt 5)\} \\ (x,y) &\in \left\{ \left(\frac{1-\sqrt 5}{2}, \frac{1-\sqrt 5}{2}\right) \right\} \end{align}

If $y \ge 0$, then $y=1-x^2$:

\begin{align} |x^2-2x| &= 1-y \\ |x^2-2x| &= x^2 &\text{(Square both sides.)} \\ x^4-4x^3+4x^2 &= x^4 \\ -4x^2(x-1) &= 0 \\ (x,y) &\in \{(0,1), (1,0) \} \end{align}

So $(x,y) \in \left\{(0,1),(1,0), \left(\frac{1-\sqrt 5}{2}, \frac{1-\sqrt 5}{2} \right)\right\}$