Not even sure where to start, but the equations are:
$\frac{dy}{dx} + 2\frac{dz}{dx}+4y +10z -2 =0 $
$\frac{dy}{dx} +\frac{dz}{dx} +y -z+3 = 0$
where $y=0$ and $z=-2$ when $x=0$.
I tried to eliminate one of the dz/dx's but the z's didnt cancel so it didnt work?
On
First subtract the equations to arrive at $$\frac{dz}{dx}=5-3y-11z$$ Substitute this into the second equation to arrive at $$\frac{dy}{dx}=-8+2y+12z$$ Differentiating this, $$\frac{d^2y}{dx^2}=2\frac{dy}{dx}+12\frac{dz}{dx}$$$$=2\frac{dy}{dx}+12(5-3y-11z)$$ Finally substitute for $z$ from the second line above and get, after simplification, $$\frac{d^2y}{dx^2}+9\frac{dy}{dx}+14y=-28$$ Now you are good to go.
Note that this is a linear system hence it can be solved with a little help of Linear Algebra. The system can be written as $$ \begin{pmatrix}1 & 2 \\ 1 & 1 \end{pmatrix}\cdot \begin{pmatrix}\frac{dy}{dx} \\ \frac{dz}{dx} \end{pmatrix} = \begin{pmatrix}-4y-10z+2 \\ -y+z-3 \end{pmatrix} = \begin{pmatrix}-4 & -10 \\ -1 & 1 \end{pmatrix}\begin{pmatrix}y \\ z\end{pmatrix} + \begin{pmatrix}2\\ -3 \end{pmatrix} $$ Hence we can use the inverse $\begin{pmatrix}-1 & 2 \\ 1 & -1 \end{pmatrix}$ of $\begin{pmatrix}1 & 2 \\ 1 & 1 \end{pmatrix}$ to get
$$ \begin{pmatrix}\frac{dy}{dx} \\ \frac{dz}{dx} \end{pmatrix} = \begin{pmatrix}2 & 12 \\ -3 & -11 \end{pmatrix}\begin{pmatrix}y \\ z\end{pmatrix} + \begin{pmatrix}-8\\ 5 \end{pmatrix} $$ Now I hope you can proceed by solving the associated homogeneous system and constructing the solution to the IVP.
In any case, here are notes on this procedure: https://math.berkeley.edu/~conway/Teaching/old/summer2016-2552bc/exercises/NonhomoSys.pdf