show that {$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$:$x_3=5x_4+b$} is a subspace

show that {$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$:$x_3=5x_4+b$} is a subspace

Let $b\in\mathbb{F}$ then
{$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$:$x_3=5x_4+b$} is a subspace iff $b=0$.
This is from axler's Linear algebra done right. $\mathbb{F}$ stands for a
field either the real numbers or complex numbers. I proved the reverse
direction but now I have to show the foward direction. This is what I have
so far.
($\rightarrow$) Assume that $b\in\mathbb{F}$ and let
T={$(x_1,x_2,x_3,x_4)\in\mathbb{F}^4$: $x_3=5x_4+b$} where $T$ is a
subspace of $\mathbb{F}^4$. Suppose not that $b\neq 0$. Since $T$ is a
subspace by assumption that means that $0\in T$ That is
$(x_1,x_2,x_3,x_4)=(0,0,0,0)$ This means that $x_3=0$ and $x_4=0$. If
$x_3=0$ then we have $0=5x_3+b \implies \frac{5}{b}=x_4$ but since $b\neq
0$ we have that $x_4\neq 0$. Suppose that $x_4=0$. Again we have
$x_3=5(0)+b=b$ Since $b\neq 0$ this means $x_3\neq 0$. Thus a
contradiction. Would this be right?