Find a basis for the subset of $\mathcal{P}_3$ defined as $V = \{f(x) \in \mathcal{P}_3: f(1) + f(-2) = 0, f'(-1) = f'(-2)\}$. Let $\mathcal{P}_3$ be the vector space of real polynomials of variable $x$ with degree 3 or less (with respect to usual addition of polynomials and multiplication of scalars with polynomials). Let $V$ be a subset of $\mathcal{P}_3$ defined as $$V = \{f(x) \in \mathcal{P}_3: f(1) + f(-2) = 0, f'(-1) = f'(-2)\}$$.
So we know elements of $\mathcal{P}_3$ have the form $a_0 + a_1x + a_2x^2 + a_3x^3$ for $a_0, a_1, a_2, a_3 \in \mathbb{R}$. Plugging the function values from above into this form, we get:
$$ 2a_0-a_1+5a_2-7a_3=0 $$
$$ 6a_2-9a_3=0 $$.
First I have to prove that $V$ is a subspace of $\mathcal{P}_3$. Am I allowed to combine the above two conditions based on their common equality to 0, as follows?
$$ 2a_0-a_1+5a_2-7a_3=6a_2-9a_3 \implies 2a_0-a1-a_2+2a_3 = 0$$.
Combining this would yield a parameterized solution with 3 span vectors. However, if I took the augmented matrix (based on the two conditions above)
$$ \begin{bmatrix} 2 & -1 & 5 & -7 & | & 0 \\ 0 & 0 & 6 & -9 & | & 0 \end{bmatrix} $$ and reduced it, I would get \begin{bmatrix} 1 & -0.5 & 0 & 0.25 & | & 0 \\ 0 & 0 & 1 & -1.5 & | & 0 \end{bmatrix} which only gives me 2 free variables and thus just a parameterized solution with just 2 span vectors? Do I lose information when setting the two conditions equal to each other?
I'm not sure how to tackle this dilemma so I'd appreciate any help in advance. Thank you!
 A: Short answer: what you did was fine.
It is certainly valid to say, because both expressions equal $0$, they equal each other. You therefore know that any solutions to this system of equations would have to satisfy all conditions that you derive afterwards.
The question is, once you've rewritten the equation, do you gain any erroneous solutions? If you fail to use all the relevant information available to you, your revised list of conditions may allow for new solutions that were not present in the original problem. This is a pitfall of most solving techniques: the logic only flows the one way, helping you narrow down your possible solutions to a manageable number (finite, or maybe a family involving only a few parameters). The onus is then on you to check your possible solutions, and see which (if any) are erroneous (and this would be a valid strategy here: you can create your two parameter family of polynomials, and verify that both conditions for membership in $V$ are satisfied).
However, Gaussian elimination is an exception. It is provable (without much difficulty) that Gaussian elimination neither gains nor loses solutions. The solution set you have at the end, provided there were no errors, is guaranteed to be the solution set of the system you started with.
Now, in your case, you didn't use Gaussian elimination, but instead performed a substitution. While the method of solving by substitution isn't talked about so much in linear algebra, the particular substitution you did was equivalent to an elementary row operation. You went from
$$\begin{cases}
2a_0 - a_1 + 5a_2 - 7a_3 &= 0 \\
6a_2 - 9a_3 &= 0
\end{cases}$$
to
$$\begin{cases}
2a_0 - a_1 + 5a_2 - 7a_3 &= 6a_2 - 9a_3 \\
6a_2 - 9a_3 &= 0
\end{cases} \iff
\begin{cases}
2a_0 - a_1 + (5 - 6)a_2 + (-7 + 9)a_3 &= 0 \\
6a_2 - 9a_3 &= 0
\end{cases}$$
That is to say, you ended up simply subtracting one equation from the other. This is a perfectly valid elementary row operation, which does not change the solution set, so no erroneous solutions were introduced.
Mind you if you had combined the two assumptions in this way, and rejected both of them, leaving you with just
$$2a_0−a_1−a_2+2a_3=0,$$
then you would be losing information, and there would be an extra parameter's worth of erroneous solutions.
