{2х²+3у² =14; х²-у²=-3

To solve the system of equations given by: 1. \( 2x^2 + 3y^2 = 14 \) (Equation 1) 2. \( x^2 - y^2 = -3 \) (Equation 2) we can start by expressing \( x^2 \) and \( y^2 \) in a more manageable form. ### Step 1: Solve for \( x^2 \) from Equation 2 From Equation 2: \[ x^2 = y^2 - 3 \] ### Step 2: Substitute \( x^2 \) in Equation 1 Now, substitute \( x^2 \) into Equation 1: \[ 2(y^2 - 3) + 3y^2 = 14 \] This simplifies to: \[ 2y^2 - 6 + 3y^2 = 14 \] Combine like terms: \[ 5y^2 - 6 = 14 \] Add 6 to both sides: \[ 5y^2 = 20 \] Divide by 5: \[ y^2 = 4 \] ### Step 3: Solve for \( y \) Take the square root of both sides: \[ y = \pm 2 \] ### Step 4: Solve for \( x^2 \) when \( y^2 \) is known Now, plug \( y^2 = 4 \) back into Equation 2 to find \( x^2 \): \[ x^2 - 4 = -3 \] Add 4 to both sides: \[ x^2 = 1 \] ### Step 5: Solve for \( x \) Take the square root: \[ x = \pm 1 \] ### Step 6: List all possible pairs \( (x, y) \) We find the pairs of solutions for \( (x, y) \): 1. \( (1, 2) \) 2. \( (1, -2) \) 3. \( (-1, 2) \) 4. \( (-1, -2) \) These are the solutions to the given system of equations. Final answers are: \[ (x, y) = (1, 2), \; (1, -2), \; (-1, 2), \; (-1, -2) \]