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

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UpStudy AI · Accepted Answer

1. Solve the first equation for $$ y $$:
   $$
   3y = 2x \quad \Longrightarrow \quad y = \frac{2}{3}x.
   $$

2. Substitute $$ y = \frac{2}{3}x $$ into the second equation:
   $$
   x^2 - y^2 + 2x - y = 1 \quad \Longrightarrow \quad x^2 - \left(\frac{2}{3}x\right)^2 + 2x - \frac{2}{3}x = 1.
   $$

3. Simplify the equation:
   $$
   x^2 - \frac{4}{9}x^2 + 2x - \frac{2}{3}x = 1.
   $$
   Combine $$ x^2 $$ terms:
   $$
   x^2 - \frac{4}{9}x^2 = \frac{9}{9}x^2 - \frac{4}{9}x^2 = \frac{5}{9}x^2.
   $$
   Combine $$ x $$ terms:
   $$
   2x - \frac{2}{3}x = \frac{6}{3}x - \frac{2}{3}x = \frac{4}{3}x.
   $$
   So the equation becomes:
   $$
   \frac{5}{9}x^2 + \frac{4}{3}x - 1 = 0.
   $$

4. Multiply the entire equation by $$ 9 $$ to eliminate fractions:
   $$
   5x^2 + 12x - 9 = 0.
   $$

5. Solve the quadratic equation using the quadratic formula:
   $$
   x = \frac{-12 \pm \sqrt{12^2 - 4 \times 5 \times (-9)}}{2 \times 5}.
   $$
   Calculate the discriminant:
   $$
   \Delta = 12^2 - 4 \times 5 \times (-9) = 144 + 180 = 324.
   $$
   Taking the square root:
   $$
   \sqrt{324} = 18.
   $$
   Thus:
   $$
   x = \frac{-12 \pm 18}{10}.
   $$
   This gives two solutions:
   $$
   x = \frac{-12 + 18}{10} = \frac{6}{10} = \frac{3}{5}, \quad
   x = \frac{-12 - 18}{10} = \frac{-30}{10} = -3.
   $$

6. Find $$ y $$ for each value of $$ x $$ using $$ y = \frac{2}{3}x $$:
   - For $$ x = \frac{3}{5} $$:
     $$
     y = \frac{2}{3} \times \frac{3}{5} = \frac{2}{5}.
     $$
   - For $$ x = -3 $$:
     $$
     y = \frac{2}{3} \times (-3) = -2.
     $$

7. The solutions to the system are:
   $$
   \left( x,y \right) = \left(\frac{3}{5}, \frac{2}{5}\right) \quad \text{and} \quad \left( x,y \right) = (-3, -2).
   $$
The solutions are $$ \left( \frac{3}{5}, \frac{2}{5} \right) $$ and $$ (-3, -2) $$.

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