\( 3 y=2 x \ldots \) \( x^{2}-y^{2}+2 x-y=1 \ldots \) (1)

To solve the given system of equations, let's begin with the first equation \(3y = 2x\). We can express \(y\) in terms of \(x\) as follows: \[ y = \frac{2}{3}x. \] Now, let's substitute this expression for \(y\) into the second equation: \[ x^2 - \left(\frac{2}{3}x\right)^2 + 2x - \left(\frac{2}{3}x\right) = 1. \] Simplifying \(\left(\frac{2}{3}x\right)^2\): \[ \left(\frac{2}{3}x\right)^2 = \frac{4}{9}x^2. \] Substituting this back into the equation, we get: \[ x^2 - \frac{4}{9}x^2 + 2x - \frac{2}{3}x = 1. \] Combining like terms, we can rewrite \(x^2\) as \(\frac{9}{9}x^2\): \[ \frac{9}{9}x^2 - \frac{4}{9}x^2 + 2x - \frac{2}{3}x = 1. \] This simplifies to: \[ \frac{5}{9}x^2 + 2x - \frac{2}{3}x = 1. \] Writing \(-\frac{2}{3}x\) with a common denominator of 3 gives \(-\frac{2}{3}x = -\frac{2}{3}x\): \[ \frac{5}{9}x^2 + \left(2 - \frac{2}{3}\right)x = 1. \] Calculating \(2 - \frac{2}{3} = \frac{6}{3} - \frac{2}{3} = \frac{4}{3}\): \[ \frac{5}{9}x^2 + \frac{4}{3}x = 1. \] To eliminate the fractions, multiply through by 9: \[ 5x^2 + 12x = 9. \] Rearranging gives: \[ 5x^2 + 12x - 9 = 0. \] Now we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a = 5\), \(b = 12\), and \(c = -9\): First, calculate the discriminant: \[ b^2 - 4ac = 12^2 - 4(5)(-9) = 144 + 180 = 324. \] \[ \sqrt{324} = 18. \] Now, apply the quadratic formula: \[ x = \frac{-12 \pm 18}{2 \times 5} = \frac{-12 + 18}{10} = \frac{6}{10} = \frac{3}{5} \] or \[ x = \frac{-12 - 18}{10} = \frac{-30}{10} = -3. \] Now let's find the corresponding \(y\) values: For \(x = \frac{3}{5}\): \[ y = \frac{2}{3} \left(\frac{3}{5}\right) = \frac{2 \times 3}{3 \times 5} = \frac{2}{5}. \] For \(x = -3\): \[ y = \frac{2}{3}(-3) = -2. \] Thus, the solutions to the system of equations are: \[ \left(\frac{3}{5}, \frac{2}{5}\right) \text{ and } (-3, -2). \]