A problem about the dimension of the intersection of two subspaces I'm trying to solve the problem below.

$U$ and $W$ are subspaces of polynomials over $\mathbb{R}$.
$U = Span(t^3 + 4t^2 - t + 3, t^3 + 5t^2 + 5, 3t^3 + 10t^2 -5t + 5)$
  $W = Span(t^3 + 4t^2 + 6, t^3 + 2t^2 - t + 5, 2t^3 + 2t^2 -3t + 9)$ 
What is $dim(U \cap W)$?

I have solved it using the fact that $dim(U) + dim(W) - dim(U \cap W) = dim(U \cup W)$, but was wondering how to solve it without using this fact.
In order to find $dim(U \cap W)$, I first try and find $U \cap W$.
Clearly if $v \in U \cap W$, then $$\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5) = \beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$ for some $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3 \in \mathbb{R}$.
Using this fact, you can reduce a system of linear equations to work out that:
$\alpha_1 + 5\alpha_3 - \beta_2 - 3\beta_3 = 0$
$\alpha_2 -2 \alpha_3 + 2\beta_2 + 6\beta_3 = 0$
$\beta_1 + 2\beta_2 + 5\beta_3 = 0$
But I don't know where to go from here.
Any help would be greatly appreciated.
 A: Consider a polynomial $p(t)=at^{3}+bt^{2}+ct+d$ belonging to $U$ and $V$ at the same time and solve two separated systems to find the conditions for $p(t)$ to be in $U$ and beside find the coditions for $p(t)$ to be in $V$ and once you find those two list of conditions consider a system on the unkowns $a,b,c$. 
$$p(t)=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$ 
$$ p(t)=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$ 
then
$$at^{3}+bt^{2}+ct+d=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$ 
$$at^{3}+bt^{2}+ct+d=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$ 
this implies that
For $U$ you need to find the conditions on $a,b$ and $c$ for the following system to have solution:
$$at^{3}+bt^{2}+ct+d=\alpha_1(t^3 + 4t^2 - t + 3) +\alpha_2(t^3 + 5t^2 + 5) +\alpha_3(3t^3 + 10t^2 -5t + 5)$$ 
$$A_{U}=\left(\begin{array}{ccc|c}
1&1&3&a\\
4&5&10&b\\
-1&0&-5&c\\
3&5&5&d
\end{array}\right)$$
For $V$ you need to find the conditions on $a,b$ and $c$ for the following system to have solution:
$$at^{3}+bt^{2}+ct+d=\beta_1(t^3 + 4t^2 + 6) + \beta_2(t^3 + 2t^2 - t + 5) + \beta_3(2t^3 + 2t^2 -3t + 9)$$ 
$$A_{V}=\left(\begin{array}{ccc|c}
1&1&2&a\\
4&2&2&b\\
0&-1&-3&c\\
6&5&9&d
\end{array}\right)$$
The row reduced echelon form of $A_{U}$ will give you the conditions for a polynomial to be in $U$. The same for $A_{V}$. These conditions are linear equations in $a,b,c$ that you can put in a linear system of equations in the variables $a,b,c$ to finally find the conditions for $p(t)$ to be in the intersection of $U$ and $V$. 
A: Consider the matrix:
\begin{bmatrix}
    1       & 4 & -1 & 3  \\
    1       & 5 & 0 & 5  \\
    3       & 10 & -5 & 5  \\   
    1       & 4 & 0 & 6  \\ 
    2       & 2 & -3 & 9  \\
\end{bmatrix}
Row reduce the matrix to its Row reduced echelon form to get the basis of  $U + W = \text{Span}\{v_1,v_2,v_3,v_4,v_5,v_6 \}$.
Then use the formula:
$$\dim(U + W) = \dim(U) + \dim(W) - \dim(U \cap W)$$
