Solve Simultaneous: \( \begin{array}{l}x y=8 \\ 2 x+y=17\end{array} \)
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To solve the simultaneous equations \(xy = 8\) and \(2x + y = 17\), we can start by expressing \(y\) from the first equation. We have \(y = \frac{8}{x}\). Now, we can substitute this expression into the second equation: \[ 2x + \frac{8}{x} = 17. \] To eliminate the fraction, multiply through by \(x\) (assuming \(x \neq 0\)): \[ 2x^2 + 8 = 17x. \] Rearranging gives us the quadratic equation: \[ 2x^2 - 17x + 8 = 0. \] Now, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -17\), and \(c = 8\): First, calculate the discriminant: \[ b^2 - 4ac = (-17)^2 - 4 \cdot 2 \cdot 8 = 289 - 64 = 225. \] Now apply the formula: \[ x = \frac{17 \pm \sqrt{225}}{4} = \frac{17 \pm 15}{4}. \] Calculating the two possible values for \(x\): 1. \(x = \frac{32}{4} = 8\) 2. \(x = \frac{2}{4} = \frac{1}{2}\) Next, we substitute these values back into \(y = \frac{8}{x}\): 1. For \(x = 8\): \(y = \frac{8}{8} = 1\) 2. For \(x = \frac{1}{2}\): \(y = \frac{8}{\frac{1}{2}} = 16\) So the solutions to the system are \((8, 1)\) and \(\left(\frac{1}{2}, 16\right)\).
