Discrete Math ProofsAre these steps for finding the solutions of x + 3 = 3...

Step 1
Those steps are correct, but there's one final step required:
6. When $x=1$, the left-hand side is 2 and the right-hand side is 2; when $x=6$, the left-hand side is 3 and the right-hand side is -3. So only the first "solution" is a solution. (Er. When I did this the first time, I wrongly thought that the RHS in the $x=6$ case was 3.)
To explain step 5: we want to find x, given that $(x-1)(x-6)=0$. Let $a=x-1$ and $b=x-6$; then since $ab=0$, we have $a=0$ or $b=0$. So $x-1=0$ or $x-6=0$; so $x=1$ or $x=6$.
Step 2
Why is it the case that if $ab=0$ then $a=0$ or $b=0$? This is precisely the statement that the product of nonzero quantities is nonzero (by taking the contrapositive), which you might find more intuitive; but I see there is another answer which gives algebraic manipulations to prove it.

Step 1
Yes, all the steps are correct. You only need to check those value of x if it satisfies the original equation or not.
Assume that $ab=0$. We want to show that either $a=0$ or $b=0$.
Step 2
If $a=0$ then we are done. Suppose that $a\ne 0$. Then $a^{-1}\in \mathbb{R}$. Thus, using field properties of R we get $b=1b=(a^{-1}a)b=a^{-1}(ab)=a^{-1}0=0.$