Solve for \( x \) and \( y \) \( x-2 y+3=0 \ldots(1) \) \( y=\frac{16}{x} \ldots \) (2)
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To solve the system of equations, we can use equation (2) in equation (1). From equation (2), we have: \[ y = \frac{16}{x} \] Now, substitute \( y \) in equation (1): \[ x - 2\left(\frac{16}{x}\right) + 3 = 0 \] This simplifies to: \[ x - \frac{32}{x} + 3 = 0 \] To eliminate the fraction, multiply through by \( x \) (assuming \( x \neq 0 \)): \[ x^2 + 3x - 32 = 0 \] Now, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 3, c = -32 \). First, calculate the discriminant: \[ b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (-32) = 9 + 128 = 137 \] Now apply the quadratic formula: \[ x = \frac{-3 \pm \sqrt{137}}{2} \] Calculating the two potential values for \( x \): 1. \( x_1 = \frac{-3 + \sqrt{137}}{2} \) 2. \( x_2 = \frac{-3 - \sqrt{137}}{2} \) (which will be negative). Now, substitute \( x_1 \) back into equation (2) to find \( y \): \[ y_1 = \frac{16}{x_1} = \frac{16}{\frac{-3 + \sqrt{137}}{2}} = \frac{32}{-3 + \sqrt{137}} \] Thus, the solution pair is: \[ (x, y) = \left( \frac{-3 + \sqrt{137}}{2}, \frac{32}{-3 + \sqrt{137}} \right) \] This gives you one set of valid values for \( (x, y) \). The second value for \( x \) will lead to a negative \( y \), if needed. You can calculate both to see their exact numerical values.
