1. **State the problem:** Solve the simultaneous equations: $$2x - y = 7$$ $$xy = 15$$ 2. **Express one variable in terms of the other:** From the first equation, solve for $y$: $$2x - y = 7 \implies y = 2x - 7$$ 3. **Substitute into the second equation:** Replace $y$ in $xy = 15$: $$x(2x - 7) = 15$$ 4. **Expand and simplify:** $$2x^2 - 7x = 15$$ 5. **Bring all terms to one side:** $$2x^2 - 7x - 15 = 0$$ 6. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=2$, $b=-7$, $c=-15$. Calculate the discriminant: $$\Delta = (-7)^2 - 4 \times 2 \times (-15) = 49 + 120 = 169$$ 7. **Find the roots:** $$x = \frac{7 \pm \sqrt{169}}{2 \times 2} = \frac{7 \pm 13}{4}$$ Two solutions: - $$x = \frac{7 + 13}{4} = \frac{20}{4} = 5$$ - $$x = \frac{7 - 13}{4} = \frac{-6}{4} = -\frac{3}{2}$$ 8. **Find corresponding $y$ values:** Using $y = 2x - 7$: - For $x=5$: $$y = 2(5) - 7 = 10 - 7 = 3$$ - For $x = -\frac{3}{2}$: $$y = 2 \times \left(-\frac{3}{2}\right) - 7 = -3 - 7 = -10$$ 9. **Final solutions:** $$(x, y) = (5, 3) \quad \text{or} \quad \left(-\frac{3}{2}, -10\right)$$ These satisfy both equations.