solution to equations $x(1-y)=0$ and $y(1-x)=0$

Question

I need to solve two equations:

$$ x(1-y)=0 \\ y(1-x)=0 $$

If I had one equation, ex $x(1-y)$, the solution would be $x=0$ or $y = 1$. But how do I go about solving two such equations? Are the solutions as below? $$ x = 0 \lor y=0 \\ x = 0 \lor y=1 \\ x = 1 \lor y=0 \\ x = 1 \lor y=1 $$

Answer 1

Both equations in the system need to be satisfied.

• Equation 1 says: either $x=0$ or $y=1$
• Equation 2 says: either $y=0$ or $x=1$

If you choose $x=0$ from equation 1, then you can't choose $x=1$ for equation 2. So $y=0$.

Likewise, if you choose $y=1$ from equation 1, then you can't choose $y=0$ for equation 2. So $x=1$.

If you want to do it with boolean algebra, you have: $$\begin{aligned} ((x=0) \lor (y=1)) \land ((y=0) \lor (x=1)) &= ((x=0) \land (y=0)) \lor ((x=0) \land (x=1)) \lor ((y=1) \land (y=0)) \lor ((y=1) \land (x=1)) \\&= ((x=0) \land (y=0)) \lor \text{FALSE} \lor \text{FALSE} \lor ((y=1) \land (x=1)) \\&= ((x=0) \land (y=0)) \lor ((y=1) \land (x=1)) \end{aligned}$$

Answer 2

There are many solutions, and you seem to have noticed how to find them. Just plug in numbers for $x$ (or $y$) that satisfy one of the equations, and solve the other :

• If $x = 0$, then your first equation yields $0=0$, so you're left with the equation $y=0$
• If $x = 1$, then your first equation yields $0=0$, so you're left with the equation $y=1$

Of course, if $(x,y)$ is neither $(0,0)$ nor $(1,1)$, the system doesn't have a solution.

So your set of solutions is $\lbrace (0,0), (1,1) \rbrace$.

Answer 3

Subtracting the two equations gives $y=x$.

You have already noticed that $x=0$ or $y=1$ and therefore the solutions are $x=y=0$ or $x=y=1$.