<s>[INST] Finding a linear combination of a linear combination of vectors Problem
Assume $\vec x, \ \vec y$ are linearly independent.
Let $\vec a = 3\vec x + 6\vec y$ and $\vec b = 9 \vec x + 21\vec y$.
Find an expression for $\vec x + \vec y$ as a linear combination of $\vec a, \vec b$.
I.e., find $c_1, \ c_2 \ \in \ \mathbb R$ such that $\vec x + \vec y = c_1\vec a + c_2\vec b$.
Progress
I have two different answers from two different methods, and I'm not sure which one is right (if any).
Method 1
(Assume all variables are vectors so I don't have to type \vec a bajillion times.)
We have the equations
$$\begin{align}
3x + 6y &= a \\
9x + 21y &= b
\end{align}$$
This gives us the coefficient matrix
$$\begin{bmatrix}
3 & 6 & a \\
9 & 21 & b
\end{bmatrix}$$
whose reduction I've found to be
$$\begin{bmatrix}
1 & 0 & -\frac53a + \frac23b \\
0 & 1 & -a + \frac13 b
\end{bmatrix}$$
Adding the two rows, we get the equation
$$x+y = -\frac83a + b$$
which is my first attempt.
Method 2
This time I used the initial equations
$$\begin{align}
\vec a &= 3x + 6y \\
\vec b &= 9x + 21y\end{align}$$
and found $3a = 9x + 18y$ so that $b-3a = 3y \ \Rightarrow \ y = -a + \frac13b$ by eliminating the $x$ term
and then doing the same with the $y$ term by adding $-\frac72$ of the first equation to the second, giving $x = \frac73 a - \frac32 b$.
Now
$$x+y = \frac43a - \frac76 b$$
which is different from the result of method 1.
Question
One (or both) of these must be wrong.

*

*Is there a way to verify a potential solution here like we do with regular systems of equations?


*Which of my methods went awry? I smell careless mistakes in both of them to be honest.
 [/INST] Eliminate the $x$ term in row two by replacing row two with (row two minus three times row one).
$$
\boxed{\begin{align*}
3x + 6y &= a \\
(9x + 21y) - 3(3x + 6y) &= (b) - 3(a)
\end{align*}}
\implies
\boxed{\begin{align*}
3x + 6y &= a \\
3y &= -3a + b
\end{align*}}
$$
Conveniently, the coefficients of $x$ and $y$ will match if we just replace row one with (row one minus row two).
$$
\boxed{\begin{align*}
(3x + 6y) - (3y) &= (a) - (-3a + b) \\
3y &= a - 3b
\end{align*}}
\implies
\boxed{\begin{align*}
3x + 3y &= 4a - b \\
3y &= a - 3b
\end{align*}}
$$
Finally, scale row one by a factor of a third to obtain:
$$
\boxed{\begin{align*}
\frac{1}{3}(3x + 3y) &= \frac{1}{3}(4a - b) \\
3y &= a - 3b
\end{align*}}
\implies
\boxed{\begin{align*}
x + y &= \frac{4}{3}a - \frac{1}{3} b \\
3y &= a - 3b
\end{align*}}
$$
We can check by substitution:
\begin{align*}
\frac{4}{3}a - \frac{1}{3} b
&= \frac{4}{3}(3x + 6y) - \frac{1}{3} (9x + 21y) \\
&= (4x + 8y) + (-3x + -7y) \\
&= x + y
\end{align*}
</s>