10.2 Plot on the coordinate plane the set of points satisfying the equation $2 x^{2}+3 x y+y^{2}+x=1$.

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Answer. The sought set of points is the union of two lines $\mathrm{y}=-\mathrm{x}-1$ and $\mathrm{y}=-2 \mathrm{x}+1$. Hint Consider the given equation as a quadratic equation in $y$. Its discriminant is $\mathrm{D}=(3 \mathrm{x})^{2}-4\left(2 \mathrm{x}^{2}+\mathrm{x}-1\right)=\mathrm{x}^{2}-4 \mathrm{x}+4=(\mathrm{x}-2)^{2}, \quad$ and the roots are $\mathrm{y}_{1}=(-3 \mathrm{x}+\mathrm{x}-2) / 2=-\mathrm{x}-1$, $\mathrm{y}_{2}=(-3 \mathrm{x}-\mathrm{x}+2) / 2=-2 \mathrm{x}+1$. Their graphs are two lines on the plane.