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Let's solve each system of equations step by step: **Problem 6:** \[ \begin{cases} 3x^2 + y^2 = 36 \\ 8x^2 + 4y^2 = 56 \end{cases} \] 1. Divide the second equation by 4: \[ 2x^2 + y^2 = 14 \] 2. Subtract the second equation from the first: \[ (3x^2 + y^2) - (2x^2 + y^2) = 36 - 14 \\ x^2 = 22 \] 3. Substitute \( x^2 = 22 \) into \( 2x^2 + y^2 = 14 \): \[ 2(22) + y^2 = 14 \\ 44 + y^2 = 14 \\ y^2 = -30 \] No real solutions since \( y^2 \) cannot be negative. **Problem 7:** \[ \begin{cases} 3x^2 + y^2 = 45 \\ 12x^2 + 8y^2 = 50 \end{cases} \] 1. Divide the second equation by 4: \[ 3x^2 + 2y^2 = 12.5 \] 2. Subtract the second equation from the first: \[ (3x^2 + y^2) - (3x^2 + 2y^2) = 45 - 12.5 \\ -y^2 = 32.5 \] No real solutions since \( y^2 \) cannot be negative. **Problem 8:** \[ \begin{cases} x^2 + y^2 = 9 \\ 4x^2 + 9y^2 = 45 \end{cases} \] 1. Multiply the first equation by 4: \[ 4x^2 + 4y^2 = 36 \] 2. Subtract the first from the second: \[ (4x^2 + 9y^2) - (4x^2 + 4y^2) = 45 - 36 \\ 5y^2 = 9 \\ y^2 = \frac{9}{5}, \quad y = \pm\sqrt{\frac{9}{5}} \] 3. Substitute \( y^2 = \frac{9}{5} \) into \( x^2 + y^2 = 9 \): \[ x^2 + \frac{9}{5} = 9 \\ x^2 = 9 - \frac{9}{5} \\ x^2 = \frac{36}{5}, \quad x = \pm\sqrt{\frac{36}{5}} \] Real solutions are: \[ x = \pm\sqrt{\frac{36}{5}}, \quad y = \pm\sqrt{\frac{9}{5}} \] **Problem 9:** \[ \begin{cases} x^2 + 3y^2 = 16 \\ 4x^2 + 2y^2 = 24 \end{cases} \] 1. Divide the second equation by 2: \[ 2x^2 + y^2 = 12 \] 2. Multiply the first equation by 2: \[ 2x^2 + 6y^2 = 32 \] 3. Subtract the second equation from the first: \[ (2x^2 + 6y^2) - (2x^2 + y^2) = 32 - 12 \\ 5y^2 = 20 \\ y^2 = 4, \quad y = \pm 2 \] 4. Substitute \( y^2 = 4 \) into \( x^2 + 3y^2 = 16 \): \[ x^2 + 3(4) = 16 \\ x^2 = 16 - 12 \\ x^2 = 4, \quad x = \pm 2 \] Real solutions are: \[ x = \pm 2, \quad y = \pm 2 \]