Verifying that all solutions to a system of eq. are found

[Mr Davis 97]

I am given the following problem: Find all ordered pairs (a, b) such that $2a + b = 12$ and $ab = 3$. Given this system of equations, I simply use substitution, and then solve the quadratic $2a^2 - 12a + 3 = 0$. Solving this, I obtain two ordered pairs: $\displaystyle (\frac{6 + \sqrt{30}}{2},~6 - \sqrt{30})$ and $\displaystyle (\frac{6 - \sqrt{30}}{2},~6 + \sqrt{30})$. These are all of the ordered pairs that I found. The problem asks to find all of the ordered pairs that satisfy the system. How can I be absolutely sure (through some kind of informal logic) that only two such ordered pairs exist, and these two are those ordered pairs?

 

[fresh_42]

I am given the following problem: Find all ordered pairs (a, b) such that $2a + b = 12$ and $ab = 3$. Given this system of equations, I simply use substitution, and then solve the quadratic $2a^2 - 12a + 3 = 0$. Solving this, I obtain two ordered pairs: $\displaystyle (\frac{6 + \sqrt{30}}{2},~6 - \sqrt{30})$ and $\displaystyle (\frac{6 - \sqrt{30}}{2},~6 + \sqrt{30})$. These are all of the ordered pairs that I found. The problem asks to find all of the ordered pairs that satisfy the system. How can I be absolutely sure (through some kind of informal logic) that only two such ordered pairs exist, and these two are those ordered pairs?

You can rule out $ab=0$, so you may substitute $b=\frac{3}{a}$ and multiply the equation by $a$ without changing the solutions.
Now you see, that the equation is of order $2$, i.e. there cannot be more than $2$ solutions.
To verify that your solutions are valid, simply insert them into the original equations and see whether they satisfy them.

You could also argue geometrically. A line (eq.1) intersects a hyperbola (eq.2) in at most $2$ points.

 

[Lucas SV]

Because you have deduced it. Assume $(a,b)$ is a par of real numbers satisfying your equations. this implies your quadratic equations. The quadratic equation only has two solutions. so the initial assumption implies $a$ is either of the two values you found. but as fresh pointed out, $a$ uniquely determines $b$. So indeed there can only be two solutions.

 

[fresh_42]

Because you have deduced it.

Ok, in this small example it might be correct, if the calculation steps are carefully described.
It is not true in general, however. A deduction is $x \in A ⇒ x \in B$. It does not guarantee $x \in B ⇒ x \in A$.
Therefore all deduction steps will have to be equivalence relations. Your wording "because you deduced it" often leads to errors.
There is a fundamental difference between a necessary condition and a sufficient condition!

 

[Lucas SV]

Ok, in this small example it might be correct, if the calculation steps are carefully described.
It is not true in general, however. A deduction is $x \in A ⇒ x \in B$. It does not guarantee $x \in B ⇒ x \in A$.
Therefore all deduction steps will have to be equivalence relations. Your wording "because you deduced it" often leads to errors.
There is a fundamental difference between a necessary condition and a sufficient condition!

Yes this is all true, that is why I put an explanation later, although maybe i didn't make myself clear that i was filling the steps. If you want a set theoretic proof, you can take $A'$ be the solutions of the quadratic equation and $A$ to be pairs of real numbers which solve the first equation. Then show $\phi:A'\rightarrow A$, $\phi(a)=(a,3/a)$ is a set isomorphism. So $A$ and $A'$ must have the same number of elements, or cardinality if you like. Since we know $A'$ has two elements, so has $A$.

The details of the proof come in computing $A'$ and in showing that $\phi$ is a well defined function, and is both injective and surjective.