Hornoes \( \lambda^{2} x+2 y=\lambda+2 \) \( 2 \lambda x+\lambda y=2 \lambda \)

To solve the system of equations given by: 1. \( \lambda^{2} x + 2y = \lambda + 2 \) (Equation 1) 2. \( 2\lambda x + \lambda y = 2\lambda \) (Equation 2) we can express the equations in a more manageable form. ### Step 1: Rearranging the equations From Equation 1, we can express \( y \) in terms of \( x \): \[ 2y = \lambda + 2 - \lambda^{2} x \] \[ y = \frac{\lambda + 2 - \lambda^{2} x}{2} \tag{3} \] From Equation 2, we can express \( y \) in terms of \( x \): \[ \lambda y = 2\lambda - 2\lambda x \] \[ y = \frac{2\lambda - 2\lambda x}{\lambda} \quad (\lambda \neq 0) \] \[ y = 2 - 2x \tag{4} \] ### Step 2: Setting the expressions for \( y \) equal Now we can set the two expressions for \( y \) from Equations (3) and (4) equal to each other: \[ \frac{\lambda + 2 - \lambda^{2} x}{2} = 2 - 2x \] ### Step 3: Clearing the fraction Multiply both sides by 2 to eliminate the fraction: \[ \lambda + 2 - \lambda^{2} x = 4 - 4x \] ### Step 4: Rearranging the equation Rearranging gives: \[ \lambda^{2} x - 4x = 4 - (\lambda + 2) \] \[ (\lambda^{2} - 4)x = 2 - \lambda \] ### Step 5: Solving for \( x \) Assuming \( \lambda^{2} - 4 \neq 0 \): \[ x = \frac{2 - \lambda}{\lambda^{2} - 4} \tag{5} \] ### Step 6: Finding \( y \) Now substitute \( x \) back into either expression for \( y \). Using Equation (4): \[ y = 2 - 2\left(\frac{2 - \lambda}{\lambda^{2} - 4}\right) \] \[ y = 2 - \frac{4 - 2\lambda}{\lambda^{2} - 4} \] \[ y = \frac{2(\lambda^{2} - 4) - (4 - 2\lambda)}{\lambda^{2} - 4} \] \[ y = \frac{2\lambda^{2} - 8 - 4 + 2\lambda}{\lambda^{2} - 4} \] \[ y = \frac{2\lambda^{2} + 2\lambda - 12}{\lambda^{2} - 4} \] ### Summary of the solution Thus, the solutions for \( x \) and \( y \) in terms of \( \lambda \) are: \[ x = \frac{2 - \lambda}{\lambda^{2} - 4} \] \[ y = \frac{2\lambda^{2} + 2\lambda - 12}{\lambda^{2} - 4} \] ### Special Cases 1. If \( \lambda^{2} - 4 = 0 \) (i.e., \( \lambda = 2 \) or \( \lambda = -2 \)), we need to check the original equations for consistency. 2. If \( \lambda = 2 \): - Both equations become \( 4x + 2y = 4 \) and \( 4x + 2y = 4 \), which are consistent. 3. If \( \lambda = -2 \): - Both equations become \( 4x + 2y = 0 \) and \( -4x - 2y = -4 \), which are also consistent. Thus, the solution is valid for all \( \lambda \) except \( \lambda = 2 \) and \( \lambda = -2 \) where we have consistent solutions.