{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: \[ (x, y) = (1, 2), \; (1, -2), \; (-1, 2), \; (-1, -2) \]