How to find a basis with 2 constraints

If V is a subspace with $(x_1,x_2,x_3,x_4)\in R^4$ such that $x_1 -2x_2+x_3=0, 2x_1-3x_2+x_3 = 0$

How would I find a basis for this? I cant seem to find a way other than inspection because normally I would rewrite the constraints that are given but it isnt working in this case because both equations have the same variables.

$2x_2 - x_3 = x_1, 3x_2 - x_3 = 2x_1\to x_2 = x_1 \to x_3 = x_1 \to (x_1,x_2,x_3,x_4) = (x_1, x_1,x_1,x_4) = x_1(1,1,1,0) + x_4(0,0,0,1) \to \mathcal{B} = \{(1,1,1,0),(0,0,0,1)\}$.

• can you elaborate in your alrgebra please – Dan Mar 11 '15 at 16:08
• you solve for $x_2,x_3$ in terms of $x_1$, and rewrite the quadriple using $x_1,x_4$, and the basis will emerge as shown.... – DeepSea Mar 11 '15 at 16:11