I have to show that the sets $$ H_1=\{(x, y, z, w)\in\mathbb{R}^4:2x+3y+5z-w=0\} $$ and $$ H_2=\{(x, y, z, w)\in\mathbb{R}^4:x+3y-w=2x-4z+6w=0\} $$ are subspaces and I have to determine a basis and the dimension.
My attempt. I easily verify that $H_1$ and $H_2$ are subspaces of $\mathbb{R}^4$. Moreover, $H_1$ is given by the solutions of the linear system of the only equation $$ 2x+3y+5z-3w=0. $$ This system has $\infty^3$ solutions. For any choice of $$ y = s,\quad z = t, \quad w = r $$ there is a solution and it is $$ \left(-\frac{3}{2}s-\frac{5}{2}t+\frac{1}{2}r, s, t, r\right)=s\left(-\frac{3}{2}, 1, 0, 0\right)+t\left(-\frac{5}{2}, 0, 1, 0\right)+r\left(\frac{1}{2}, 0, 0, 1\right). $$ So a basis can be $\left\{\left(-\frac{3}{2}, 1, 0, 0\right), \left(-\frac{5}{2}, 0, 1, 0\right), \left(\frac{1}{2}, 0, 0, 1\right)\right\}$ and the dimension of $H_1$ is $3$.
The same for $H_2$. The associated linear system has $\infty^2$ solutions. For any choice of $$ z = s, \quad w = t $$ there is a solution and it is $$ \left(2s-3t, -\frac{2}{3}s+\frac{4}{3}t, s, t\right)=s\left(2, -\frac{2}{3}, 1, 0\right)+t\left(-3, \frac{4}{3}, 0, 1\right). $$ So a basis can be $\left\{\left(2, -\frac{2}{3}, 1, 0\right), \left(-3, \frac{4}{3}, 0, 1\right)\right\}$ and the dimension of $H_2$ is $2$.
Is my attempt right?
Thank You
It should be $2s-3t$ and not $2s+3t$ for the $x$ coordinate of $H_2$. Adjust the rest in consequence, apart from that, your approach is perfectly adapted.
You also swapped the $x$ coordinates depending on $s$ and $t$ for $H_2$.