1. **State the problem:** We are given two equations: - Quadratic: $x^2 + y - 8 = 6$ - Linear: $y + 5x - 20 = 0$ We need to solve for $x$ by finding values of $x$ and $y$ that satisfy both equations simultaneously. 2. **Rewrite the equations:** - Quadratic: $x^2 + y - 8 = 6 \implies x^2 + y = 14$ - Linear: $y + 5x - 20 = 0 \implies y = 20 - 5x$ 3. **Substitute the linear equation into the quadratic:** Replace $y$ in the quadratic equation with $20 - 5x$: $$x^2 + (20 - 5x) = 14$$ 4. **Simplify the equation:** $$x^2 + 20 - 5x = 14$$ $$x^2 - 5x + 20 = 14$$ $$x^2 - 5x + 6 = 0$$ 5. **Solve the quadratic equation:** Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$, $b=-5$, and $c=6$. Calculate the discriminant: $$\Delta = (-5)^2 - 4 \times 1 \times 6 = 25 - 24 = 1$$ Since $\Delta > 0$, there are two real solutions: $$x = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}$$ 6. **Find the two values of $x$:** - For $+$ sign: $x = \frac{5 + 1}{2} = 3$ - For $-$ sign: $x = \frac{5 - 1}{2} = 2$ 7. **Find corresponding $y$ values:** Using $y = 20 - 5x$: - For $x=3$: $y = 20 - 5 \times 3 = 20 - 15 = 5$ - For $x=2$: $y = 20 - 5 \times 2 = 20 - 10 = 10$ **Final solutions:** $$\boxed{(x,y) = (3,5) \text{ or } (2,10)}$$