1. **State the problem:** We need to solve the simultaneous equations: $$x^2 + y^2 = 89$$ $$y = x + 3$$ 2. **Substitute the linear equation into the circle equation:** Since $y = x + 3$, substitute $y$ in the circle equation: $$x^2 + (x + 3)^2 = 89$$ 3. **Expand and simplify:** $$x^2 + (x^2 + 6x + 9) = 89$$ $$2x^2 + 6x + 9 = 89$$ 4. **Bring all terms to one side:** $$2x^2 + 6x + 9 - 89 = 0$$ $$2x^2 + 6x - 80 = 0$$ 5. **Divide the entire equation by 2 to simplify:** $$x^2 + 3x - 40 = 0$$ 6. **Solve the quadratic equation using the quadratic formula:** The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=3$, and $c=-40$. Calculate the discriminant: $$\Delta = b^2 - 4ac = 3^2 - 4(1)(-40) = 9 + 160 = 169$$ Calculate the roots: $$x = \frac{-3 \pm \sqrt{169}}{2} = \frac{-3 \pm 13}{2}$$ 7. **Find the two values of $x$:** - For $x = \frac{-3 + 13}{2} = \frac{10}{2} = 5$ - For $x = \frac{-3 - 13}{2} = \frac{-16}{2} = -8$ 8. **Find corresponding $y$ values using $y = x + 3$:** - When $x=5$, $y = 5 + 3 = 8$ - When $x=-8$, $y = -8 + 3 = -5$ **Final solutions:** $$\boxed{(x,y) = (5,8) \text{ or } (-8,-5)}$$