Proving subset is a subspace of vector space R2

$$W = \begin{bmatrix}a\\b\end{bmatrix} \in \mathbb{R}^2 : 2a − 3b = 0$$

I know the zero vector exists, but I need some help proving the addition and multiplication.

Thanks

1) As you say, the zero vector $$\underline 0 \in W$$ because $$a=b=0 \implies 2a - 3b = 0$$.
2) Suppose $$\underline u = \begin{bmatrix} u_1 \\ u_2 \end{bmatrix} \in W$$ and $$\underline v = \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}\in W$$. Then $$2 u_1 - 3u_2 = 0 = 2 v_1 - 3 v_2$$. From this it follows that $$2u_1 - 3u_2 + 2v_1 - 3v_2 = 2(u_1 + v_1) -3(u_2 + v_2) = 0$$, so $$\underline u + \underline v \in W$$.
3) Now suppose that we have $$\underline u$$ as above, and $$\alpha \in \mathbb{R}$$. Then $$\alpha \underline u$$ = $$\begin{bmatrix}\alpha u_1 \\ \alpha u_2 \end{bmatrix}$$. Since $$2u_1 - 3u_2 = 0$$, we have $$\alpha(2u_1 - 3u_2) = 2(\alpha u_1) - 3(\alpha u_2) = 0$$ and so $$\alpha \underline u \in W$$.
So $$W$$ is a subspace of $$\mathbb{R}^2$$.
$$2(a_1+a_2)-3(b_1+b_2)= (2a_1-3b_1) + ( 2a_2-3b_2)$$ Also $$2\lambda a -3\lambda b =\lambda (2a-3b)$$