1. **State the problem:** Jose bought some tickets for 180 total dollars. He kept 2 tickets and sold the rest at 9 dollars more than the original price each. He made a profit of 36 dollars. We need to find the original price of each ticket. 2. **Define variables:** Let $n$ be the total number of tickets Jose bought. Let $p$ be the original price of each ticket. 3. **Write equations:** Total cost is $$np = 180$$ He kept 2 tickets, so he sold $n-2$ tickets. He sold each ticket for $$p + 9$$ dollars. 4. **Calculate total revenue:** Revenue from selling tickets is $$(n-2)(p+9)$$ 5. **Profit formula:** Profit = Revenue - Cost Given profit is 36, so $$(n-2)(p+9) - 180 = 36$$ 6. **Simplify profit equation:** $$(n-2)(p+9) = 216$$ 7. **Use cost equation to express $n$:** $$n = \frac{180}{p}$$ 8. **Substitute $n$ into profit equation:** $$\left(\frac{180}{p} - 2\right)(p+9) = 216$$ 9. **Expand and simplify:** $$\left(\frac{180 - 2p}{p}\right)(p+9) = 216$$ $$\frac{(180 - 2p)(p+9)}{p} = 216$$ Multiply both sides by $p$: $$(180 - 2p)(p+9) = 216p$$ 10. **Expand left side:** $$180p + 1620 - 2p^2 - 18p = 216p$$ Simplify terms: $$180p - 18p = 162p$$ So: $$162p + 1620 - 2p^2 = 216p$$ 11. **Bring all terms to one side:** $$-2p^2 + 1620 + 162p - 216p = 0$$ $$-2p^2 + 1620 - 54p = 0$$ Multiply entire equation by -1: $$2p^2 - 1620 + 54p = 0$$ Rearranged: $$2p^2 + 54p - 1620 = 0$$ 12. **Divide entire equation by 2:** $$p^2 + 27p - 810 = 0$$ 13. **Solve quadratic equation:** Use quadratic formula: $$p = \frac{-27 \pm \sqrt{27^2 - 4 \times 1 \times (-810)}}{2}$$ Calculate discriminant: $$27^2 = 729$$ $$4 \times 810 = 3240$$ $$\sqrt{729 + 3240} = \sqrt{3969} = 63$$ 14. **Calculate roots:** $$p = \frac{-27 \pm 63}{2}$$ Two solutions: $$p = \frac{-27 + 63}{2} = \frac{36}{2} = 18$$ $$p = \frac{-27 - 63}{2} = \frac{-90}{2} = -45$$ 15. **Interpret solution:** Price cannot be negative, so original price per ticket is $18$. **Final answer:** The original price of each ticket was $18$.