Solve: \( \begin{array}{l}3 x=y+4 \quad \text { and } \\ y^{2}-x y=9 x+7 \\ \text { Simultaneously }\end{array} \)

To solve the system of equations \[ \begin{aligned} 1) & \quad 3x = y + 4 \\ 2) & \quad y^2 - xy = 9x + 7 \end{aligned} \] we can first express \(y\) from the first equation: \[ y = 3x - 4. \] Now, substitute \(y\) in the second equation: \[ (3x - 4)^2 - x(3x - 4) = 9x + 7. \] Expanding \((3x - 4)^2\): \[ 9x^2 - 24x + 16. \] Now substituting into the second equation, we get: \[ 9x^2 - 24x + 16 - (3x^2 - 4x) = 9x + 7. \] This simplifies to: \[ 9x^2 - 24x + 16 - 3x^2 + 4x = 9x + 7. \] Combining like terms on the left side: \[ 6x^2 - 20x + 16 = 9x + 7. \] Now, move all terms to one side: \[ 6x^2 - 20x - 9x + 16 - 7 = 0, \] which gives: \[ 6x^2 - 29x + 9 = 0. \] Next, we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 6\), \(b = -29\), and \(c = 9\): Calculate the discriminant: \[ b^2 - 4ac = (-29)^2 - 4(6)(9) = 841 - 216 = 625. \] Now apply the quadratic formula: \[ x = \frac{29 \pm \sqrt{625}}{2(6)} = \frac{29 \pm 25}{12}. \] Calculating the two possible values for \(x\): 1. \(x = \frac{54}{12} = \frac{9}{2} = 4.5\) 2. \(x = \frac{4}{12} = \frac{1}{3}\) Now, substitute back to find the corresponding \(y\) values for each \(x\): 1. For \(x = 4.5\): \[ y = 3(4.5) - 4 = 13.5 - 4 = 9.5. \] 2. For \(x = \frac{1}{3}\): \[ y = 3\left(\frac{1}{3}\right) - 4 = 1 - 4 = -3. \] Thus, the solutions are: \((x, y) = (4.5, 9.5)\) and \((x, y) = \left(\frac{1}{3}, -3\right)\).