Answer
See answer below
Work Step by Step
Let $S$ be the set of solutions to the linear system:
$-2x_2+5x_3=7 \\
4x_1-6x_2+3x_3=0$
We can guess the solutions are $x_1=-\frac{9}{4}\\x_2=-1\\
x_3=1$
since $-2(-1)+5(1)=7 \\
4(-\frac{9}{4})-6(-1)+3(1)=0$
But $4v=(-9,-4,4) \in S$ since $-2.(-4)+5.4=28 \ne 7$
Since $S$ is not closed under scalar multiplication by vectors, $S$ is not a vector space.
