\( \left. \begin{array} { l } { y = 3 x ^ { 2 } - 2 x - 8 \ldots } \\ { y = 5 x - 7 \ldots } \end{array} \right. \)

Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=3x^{2}-2x-8\\y=5x-7\end{array}\right.\) - step1: Substitute the value of \(y:\) \(3x^{2}-2x-8=5x-7\) - step2: Move the expression to the left side: \(3x^{2}-2x-8-\left(5x-7\right)=0\) - step3: Subtract the terms: \(3x^{2}-7x-1=0\) - step4: Solve using the quadratic formula: \(x=\frac{7\pm \sqrt{\left(-7\right)^{2}-4\times 3\left(-1\right)}}{2\times 3}\) - step5: Simplify the expression: \(x=\frac{7\pm \sqrt{\left(-7\right)^{2}-4\times 3\left(-1\right)}}{6}\) - step6: Simplify the expression: \(x=\frac{7\pm \sqrt{61}}{6}\) - step7: Separate into possible cases: \(\begin{align}&x=\frac{7+\sqrt{61}}{6}\\&x=\frac{7-\sqrt{61}}{6}\end{align}\) - step8: Evaluate the logic: \(x=\frac{7+\sqrt{61}}{6}\cup x=\frac{7-\sqrt{61}}{6}\) - step9: Rearrange the terms: \(\left\{ \begin{array}{l}x=\frac{7+\sqrt{61}}{6}\\y=3x^{2}-2x-8\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{7-\sqrt{61}}{6}\\y=3x^{2}-2x-8\end{array}\right.\) - step10: Calculate: \(\left\{ \begin{array}{l}x=\frac{7+\sqrt{61}}{6}\\y=\frac{-7+5\sqrt{61}}{6}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{7-\sqrt{61}}{6}\\y=-\frac{7+5\sqrt{61}}{6}\end{array}\right.\) - step11: Check the solution: \(\left\{ \begin{array}{l}x=\frac{7+\sqrt{61}}{6}\\y=\frac{-7+5\sqrt{61}}{6}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{7-\sqrt{61}}{6}\\y=-\frac{7+5\sqrt{61}}{6}\end{array}\right.\) - step12: Rewrite: \(\left(x,y\right) = \left(\frac{7+\sqrt{61}}{6},\frac{-7+5\sqrt{61}}{6}\right)\cup \left(x,y\right) = \left(\frac{7-\sqrt{61}}{6},-\frac{7+5\sqrt{61}}{6}\right)\) To solve the system of equations given by: 1. \( y = 3x^2 - 2x - 8 \) 2. \( y = 5x - 7 \) we can set the two equations equal to each other since they both equal \( y \): \[ 3x^2 - 2x - 8 = 5x - 7 \] Now, let's rearrange this equation to bring all terms to one side: \[ 3x^2 - 2x - 5x - 8 + 7 = 0 \] This simplifies to: \[ 3x^2 - 7x - 1 = 0 \] Next, we can use the quadratic formula to find the values of \( x \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 3 \), \( b = -7 \), and \( c = -1 \). Calculating the discriminant: \[ b^2 - 4ac = (-7)^2 - 4 \cdot 3 \cdot (-1) = 49 + 12 = 61 \] Now substituting back into the quadratic formula: \[ x = \frac{7 \pm \sqrt{61}}{6} \] This gives us two possible values for \( x \): 1. \( x_1 = \frac{7 + \sqrt{61}}{6} \) 2. \( x_2 = \frac{7 - \sqrt{61}}{6} \) Next, we substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. Using \( y = 5x - 7 \): For \( x_1 \): \[ y_1 = 5\left(\frac{7 + \sqrt{61}}{6}\right) - 7 = \frac{35 + 5\sqrt{61}}{6} - 7 = \frac{35 + 5\sqrt{61} - 42}{6} = \frac{-7 + 5\sqrt{61}}{6} \] For \( x_2 \): \[ y_2 = 5\left(\frac{7 - \sqrt{61}}{6}\right) - 7 = \frac{35 - 5\sqrt{61}}{6} - 7 = \frac{35 - 5\sqrt{61} - 42}{6} = \frac{-7 - 5\sqrt{61}}{6} \] Thus, the solutions to the system of equations are: 1. \( \left( \frac{7 + \sqrt{61}}{6}, \frac{-7 + 5\sqrt{61}}{6} \right) \) 2. \( \left( \frac{7 - \sqrt{61}}{6}, \frac{-7 - 5\sqrt{61}}{6} \right) \)