4 SIMULTANEOUS 틀 Activity 5 Solve for \( x \) and \( y \) in: \( 5.1 \quad 2 y=3+x \) and \( 7=x^{2}+4 y^{2} \)

To solve the simultaneous equations \( 2y = 3 + x \) and \( 7 = x^2 + 4y^2 \), we can start by expressing \( y \) in terms of \( x \) from the first equation. From \( 2y = 3 + x \), we can express \( y \) as: \[ y = \frac{3 + x}{2} \] Next, we substitute this expression for \( y \) into the second equation: \[ 7 = x^2 + 4\left(\frac{3 + x}{2}\right)^2 \] Now, simplifying the second equation: \[ 7 = x^2 + 4\left(\frac{(3 + x)^2}{4}\right) \] \[ 7 = x^2 + (3 + x)^2 \] \[ = x^2 + (9 + 6x + x^2) \] \[ = 2x^2 + 6x + 9 \] Setting this equal to 7 gives: \[ 7 = 2x^2 + 6x + 9 \] \[ 0 = 2x^2 + 6x + 9 - 7 \] \[ 0 = 2x^2 + 6x + 2 \] Dividing the whole equation by 2, we get: \[ 0 = x^2 + 3x + 1 \] Now we can use the quadratic formula to solve for \( x \): \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-3 \pm \sqrt{{3^2 - 4 \cdot 1 \cdot 1}}}}{2 \cdot 1} \] \[ = \frac{{-3 \pm \sqrt{9 - 4}}}{2} \] \[ = \frac{{-3 \pm \sqrt{5}}}{2} \] So, the solutions for \( x \) are: \[ x_1 = \frac{{-3 + \sqrt{5}}}{2}, \quad x_2 = \frac{{-3 - \sqrt{5}}}{2} \] Next, we substitute these values back into the equation for \( y \): For \( x_1 \): \[ y_1 = \frac{3 + \frac{{-3 + \sqrt{5}}}{2}}{2} = \frac{3 - 3 + \sqrt{5}}{4} = \frac{\sqrt{5}}{4} \] For \( x_2 \): \[ y_2 = \frac{3 + \frac{{-3 - \sqrt{5}}}{2}}{2} = \frac{3 - 3 - \sqrt{5}}{4} = \frac{-\sqrt{5}}{4} \] Thus, the solutions are: \[ (x_1, y_1) = \left(\frac{-3 + \sqrt{5}}{2}, \frac{\sqrt{5}}{4}\right), \quad (x_2, y_2) = \left(\frac{-3 - \sqrt{5}}{2}, \frac{-\sqrt{5}}{4}\right) \]