Solve for \( x \) and \( y \) : \[ \begin{array}{l}x=2 y+2 \\ x^{2}-2 x y+3 y^{2}=4\end{array} \]

To solve the given system of equations, we start by substituting the first equation into the second one. 1. The first equation is: \[ x = 2y + 2 \] 2. We can substitute \( x \) into the second equation: \[ x^2 - 2xy + 3y^2 = 4 \] becomes \[ (2y + 2)^2 - 2(2y + 2)y + 3y^2 = 4 \] 3. Expanding the terms: \[ (2y + 2)^2 = 4y^2 + 8y + 4 \] \[ -2(2y + 2)y = -4y^2 - 4y \] 4. Now substitute these back into the equation: \[ 4y^2 + 8y + 4 - 4y^2 - 4y + 3y^2 = 4 \] 5. Combine like terms: \[ (4y^2 - 4y^2 + 3y^2) + (8y - 4y) + 4 = 4 \] \[ 3y^2 + 4y + 4 = 4 \] 6. Subtract 4 from both sides: \[ 3y^2 + 4y = 0 \] 7. Factor out \( y \): \[ y(3y + 4) = 0 \] 8. Setting each factor to zero gives us: \[ y = 0 \quad \text{or} \quad 3y + 4 = 0 \] Therefore, \[ y = 0 \quad \text{or} \quad y = -\frac{4}{3} \] 9. Now we substitute \( y \) back into the first equation to find \( x \): - For \( y = 0 \): \[ x = 2(0) + 2 = 2 \] Thus, one solution is \( (x, y) = (2, 0) \). - For \( y = -\frac{4}{3} \): \[ x = 2\left(-\frac{4}{3}\right) + 2 = -\frac{8}{3} + 2 = -\frac{8}{3} + \frac{6}{3} = -\frac{2}{3} \] Thus, the second solution is \( \left(x, y\right) = \left(-\frac{2}{3}, -\frac{4}{3}\right) \). Finally, the complete set of solutions is: \[ (x, y) = (2, 0) \quad \text{and} \quad \left(-\frac{2}{3}, -\frac{4}{3}\right) \]