Determine numbers $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$ such that Matrix equation holds

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I need to determine a set of numbers $x_{1}$, $x_{2}$, $x_{3}$, $x_{4}$, such that the following holds true:

$(x_1\cdot y_1) + (x_2\cdot y_2) + (x_3\cdot y_3) + (x_4\cdot y_4) = (e_1) + (e_1) + (e_3) -(x_4\cdot e_4)$

Where $y_{1}-y_{4}$ is column vectors of a Matrix $Y$:

\begin{bmatrix}2&-3&8&1\\-4&-3&-2&-2\\2&4&1&3\\2&4&16&1\end{bmatrix}

and $e_{1}-e_{4}$ are column vectors of a Matrix $E$

\begin{bmatrix}\frac{1}{5}&-\frac{4}{5}&\frac{2}{5}&\frac{2}{5}\\-\frac{4}{5}&\frac{1}{5}&\frac{2}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&-\frac{1}{5}&\frac{4}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{4}{5}&-\frac{1}{5}\end{bmatrix}

I am not looking for someone to solve this equation for me, but rather a helping hand as to how I would engage solving this.

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The left-hand side is $YX$ where $X$ is the column vector $[x_1,x_2,x_3,x_4]^T$. The only problem is that you have an unknown $x_4$ on the right. Move the $x_4e_4$ term over to the other side, and you can adjust the $Y$ matrix so that you have a typical system of $4$ linear equations in $4$ unknowns.