\( x^{2}-2 x y-3 y^{2}=0 \) and \( y-x=2 \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x^{2}-2xy-3y^{2}=0\\y-x=2\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x^{2}-2xy-3y^{2}=0\\x=-2+y\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(-2+y\right)^{2}-2\left(-2+y\right)y-3y^{2}=0\) - step3: Simplify: \(4-4y^{2}=0\) - step4: Move the constant to the right side: \(-4y^{2}=0-4\) - step5: Remove 0: \(-4y^{2}=-4\) - step6: Change the signs: \(4y^{2}=4\) - step7: Divide both sides: \(\frac{4y^{2}}{4}=\frac{4}{4}\) - step8: Divide the numbers: \(y^{2}=1\) - step9: Simplify the expression: \(y=\pm \sqrt{1}\) - step10: Simplify: \(y=\pm 1\) - step11: Separate into possible cases: \(y=1\cup y=-1\) - step12: Rearrange the terms: \(\left\{ \begin{array}{l}x=-2+y\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-2+y\\y=-1\end{array}\right.\) - step13: Calculate: \(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=-1\end{array}\right.\) - step14: Check the solution: \(\left\{ \begin{array}{l}x=-1\\y=1\end{array}\right.\cup \left\{ \begin{array}{l}x=-3\\y=-1\end{array}\right.\) - step15: Rewrite: \(\left(x,y\right) = \left(-1,1\right)\cup \left(x,y\right) = \left(-3,-1\right)\) To solve the system of equations given by: 1. \( x^{2} - 2xy - 3y^{2} = 0 \) 2. \( y - x = 2 \) we can find the solutions step by step. ### Step 1: Substitute \( y \) from the second equation into the first equation. From the second equation, we can express \( y \) in terms of \( x \): \[ y = x + 2 \] ### Step 2: Substitute \( y \) into the first equation. Now, substitute \( y = x + 2 \) into the first equation: \[ x^{2} - 2x(x + 2) - 3(x + 2)^{2} = 0 \] ### Step 3: Expand and simplify the equation. Expanding the equation: \[ x^{2} - 2x^{2} - 4x - 3(x^{2} + 4x + 4) = 0 \] This simplifies to: \[ x^{2} - 2x^{2} - 4x - 3x^{2} - 12x - 12 = 0 \] Combining like terms: \[ -4x^{2} - 16x - 12 = 0 \] ### Step 4: Multiply through by -1 to simplify. \[ 4x^{2} + 16x + 12 = 0 \] ### Step 5: Divide the entire equation by 4. \[ x^{2} + 4x + 3 = 0 \] ### Step 6: Factor the quadratic equation. Factoring gives: \[ (x + 1)(x + 3) = 0 \] ### Step 7: Solve for \( x \). Setting each factor to zero gives: \[ x + 1 = 0 \quad \Rightarrow \quad x = -1 \] \[ x + 3 = 0 \quad \Rightarrow \quad x = -3 \] ### Step 8: Find corresponding \( y \) values. Using \( y = x + 2 \): 1. For \( x = -1 \): \[ y = -1 + 2 = 1 \] 2. For \( x = -3 \): \[ y = -3 + 2 = -1 \] ### Final Solutions The solutions to the system of equations are: \[ (x, y) = (-1, 1) \quad \text{and} \quad (x, y) = (-3, -1) \]