Phase Portraits of a non linear ODE

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Let $$x'=x(1-x)- \frac{\alpha xy}{\delta +x}$$ $$y'=\beta y(\gamma+y) -\beta xy$$ in the first quadrant of $\mathbb R^2$, with $\alpha, \beta, \delta >0$ and $0<\gamma<1$.

Sketch the phase portrait of the system with its ceroclines.

I could get $x-$cerocline and $y-$cerocline. Nonetheless, I could not find an easy way to obtain the fixed points, so I could sketch de phase portrait. If there is another way for sketching the phase portrait, I would be glad to know it.

Thank you in advance

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The stream plot can be built with the help of the null-clines. Choosing a value to the parameters, as $\left(\alpha=\beta=\gamma=\delta=\frac 12\right)$ we have:

enter image description here

The red dots are the equilibrium points obtained by the crossing of the curves

$$ \left\{ \begin{array}{l} (1-x) x-\frac{x y}{2 \left(x+\frac{1}{2}\right)} =0\\ \frac{1}{2} y \left(y+\frac{1}{2}\right)-\frac{x y}{2} =0\\ \end{array} \right. $$

The equilibrium points can be qualified according to the jacobian eigenvalues.

$$ \left\{ \begin{array}{lccccc} &1 & 2 & 3 & 4 & 5\\ \text{equilibrium points} & \{0.,-0.5\} & \{1.,0.\} & \{-0.866025,-1.36603\} & \{0.866025,0.366025\} & \{0.,0.\} \\ \text{eigenvalues} & \{1.5,-0.25\} & \{-1.,-0.25\} & \{5.14238,-0.544308\} & \{-0.837913,0.239836\} & \{1.,0.25\} \\ \text{kind} & \text{saddle} & \text{sink} & \text{saddle} & \text{saddle} & \text{source} \end{array} \right. $$