\( \left. \begin{array} { c } { 3 y = 2 x \ldots } \\ { x ^ { 2 } - y ^ { 2 } + 2 x - y = 1 } \end{array} \right. \)

To solve the system of equations, start with the first equation: \(3y = 2x\). From this, you can express \(y\) in terms of \(x\): \(y = \frac{2}{3}x\). Next, substitute \(y\) into the second equation: \[ x^2 - \left(\frac{2}{3}x\right)^2 + 2x - \left(\frac{2}{3}x\right) = 1 \] Now, simplify the equation: \[ x^2 - \frac{4}{9}x^2 + 2x - \frac{2}{3}x = 1 \] Combining terms yields: \[ \left(1 - \frac{4}{9}\right)x^2 + \left(2 - \frac{2}{3}\right)x - 1 = 0 \] This simplifies to: \[ \frac{5}{9}x^2 + \frac{4}{3}x - 1 = 0 \] To eliminate the fraction, multiply through by 9: \[ 5x^2 + 12x - 9 = 0 \] Now, apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): \[ x = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 5 \cdot (-9)}}{2 \cdot 5} \] \[ x = \frac{-12 \pm \sqrt{144 + 180}}{10} \] \[ x = \frac{-12 \pm \sqrt{324}}{10} \] \[ x = \frac{-12 \pm 18}{10} \] Calculating the two potential values for \(x\): 1. \(x = \frac{6}{10} = 0.6\) 2. \(x = \frac{-30}{10} = -3\) Now substitute back to find \(y\) using \(y = \frac{2}{3}x\): 1. If \(x = 0.6\), \(y = \frac{2}{3}(0.6) = 0.4\). 2. If \(x = -3\), \(y = \frac{2}{3}(-3) = -2\). Thus, the solutions to the system of equations are \((0.6, 0.4)\) and \((-3, -2)\).