Does $\{(1, -1),(2, 1)\}$ spans $\mathbb{R}^2$? please correct me Can anyone please correct me? my problem is in the proof part below
Q: Does $\{(1, -1),(2, 1)\}$ spans $\mathbb{R}^2$?
A:
$$c_1(1, -1) + c_2(2, 1) = (x, y)$$
$$c_1 + 2c_2 = x$$
$$-c_1 + c_2 = y$$

$$c_1 = x - 2c_2$$
$$-(x - 2c_2) + c_2 = y$$
$$-x + 2c_2 + c_2 = y$$
$$c_2 = \frac{x + y}{3}$$

$$c_1 + 2\frac{x + y}{3} = x$$
$$c_1 = x - 2\frac{x + y}{3}$$
$$c_1 = x - \frac{2x-2y}{3}$$
Conclusion: it can reach any point $(x, y)$ in $\mathbb{R}^2$
Proof: Get to $(4, 7)$:
so, $c_1 = 6$, $c_2 =\frac{11}{3}$.
$$6(1, -1)+\frac{11}{3}(2, 1) = \left(\frac{40}{3}, -\frac{7}{3}\right)$$
!!! it should've been equal to $(4, 7)$


Perhaps easier:
$$\begin{cases}&\;\;\;c_1+2c_2=x\\{}\\&-c_1+c_2\;\;=y\end{cases}\stackrel{\text{sum both eq's}}\implies3c_2=x+y\implies c_2=\frac{x+y}3$$
and substituting back, say in equation two:
$$c_1=\frac{x-2y}3$$
