1. **Problem statement:** Solve the system of equations $$2x + y = 10$$ $$x^2 + y^2 = 25$$ 2. **Step 1: Express $y$ from the linear equation:** $$y = 10 - 2x$$ 3. **Step 2: Substitute $y$ into the circle equation:** $$x^2 + (10 - 2x)^2 = 25$$ 4. **Step 3: Expand and simplify:** $$x^2 + (100 - 40x + 4x^2) = 25$$ $$x^2 + 100 - 40x + 4x^2 = 25$$ $$5x^2 - 40x + 100 = 25$$ 5. **Step 4: Bring all terms to one side:** $$5x^2 - 40x + 100 - 25 = 0$$ $$5x^2 - 40x + 75 = 0$$ 6. **Step 5: Divide entire equation by 5:** $$\cancel{5}x^2 - \cancel{5}8x + \cancel{5}15 = 0$$ $$x^2 - 8x + 15 = 0$$ 7. **Step 6: Factor quadratic:** $$(x - 3)(x - 5) = 0$$ 8. **Step 7: Solve for $x$:** $$x = 3 \quad \text{or} \quad x = 5$$ 9. **Step 8: Find corresponding $y$ values:** For $x=3$: $$y = 10 - 2(3) = 10 - 6 = 4$$ For $x=5$: $$y = 10 - 2(5) = 10 - 10 = 0$$ 10. **Final solutions:** $$(x,y) = (3,4) \quad \text{or} \quad (5,0)$$ --- 1. **Problem statement:** Solve the inequality $$x^2 - 5x + 6 > 0$$ 2. **Step 1: Factor the quadratic:** $$(x - 2)(x - 3) > 0$$ 3. **Step 2: Find critical points:** $$x = 2, 3$$ 4. **Step 3: Use sign analysis (sign line):** - For $x < 2$, both $(x-2)$ and $(x-3)$ are negative, product is positive. - For $2 < x < 3$, $(x-2)$ positive, $(x-3)$ negative, product negative. - For $x > 3$, both positive, product positive. 5. **Step 4: Write solution:** $$x < 2 \quad \text{or} \quad x > 3$$ 6. **Step 5: Check graphically:** The parabola opens upwards and is above the $x$-axis outside the roots, confirming the solution. **Final answer:** $$x < 2 \quad \text{or} \quad x > 3$$