Movie Theater Problem

Adult tickets cost 12 dollars and child tickets cost 8 dollars. A group bought 15 tickets for a total of 148 dollars. How many of each type of ticket did they buy?

Define variables: Let $a$ = number of adult tickets Let $c$ = number of child tickets = Two unknowns: $a$ and $c$

Write the first equation (total tickets): The group bought 15 tickets total: $a + c = 15$ = $a + c = 15$

Write the second equation (total cost): Adults: $12a$ dollars Children: $8c$ dollars Total: $12a + 8c = 148$ = $12a + 8c = 148$

Solve using substitution: From equation 1: $a = 15 - c$ Substitute into equation 2: $12(15 - c) + 8c = 148$ = $180 - 12c + 8c = 148$

Simplify and solve for c: $180 - 4c = 148$ $-4c = -32$ $c = 8$ = $c = 8$ child tickets

Find a: $a = 15 - c = 15 - 8 = 7$ = $a = 7$ adult tickets

Check the answer: Tickets: $7 + 8 = 15$ ✓ Cost: $12(7) + 8(8) = 84 + 64 = 148$ ✓ = Both equations satisfied!

Answer: The group bought 7 adult tickets and 8 child tickets.

Sum and Difference Problem

The sum of two numbers is 52. Their difference is 18. Find the two numbers.

Define variables: Let $x$ = the larger number Let $y$ = the smaller number = Two unknowns: $x$ and $y$

Write equations from the words: Sum is 52: $x + y = 52$ Difference is 18: $x - y = 18$ = System: $\begin{cases} x + y = 52 \\ x - y = 18 \end{cases}$

Solve using elimination: Add both equations: $(x + y) + (x - y) = 52 + 18$ $2x = 70$ = $x = 35$

Find y: $x + y = 52$ $35 + y = 52$ $y = 17$ = $y = 17$

Check: Sum: $35 + 17 = 52$ ✓ Difference: $35 - 17 = 18$ ✓ = Answer verified!

Answer: The two numbers are 35 and 17.

Coin Problem

Maria has 23 coins consisting of quarters and dimes. The total value is 4.25 dollars. How many of each coin does she have?

Define variables: Let $q$ = number of quarters (worth 0.25 dollars each) Let $d$ = number of dimes (worth 0.10 dollars each) = Unknowns: $q$ and $d$

Write the equations: Total coins: $q + d = 23$ Total value: $0.25q + 0.10d = 4.25$ = Two equations, two unknowns

Eliminate decimals: Multiply the second equation by 100: $25q + 10d = 425$ = $25q + 10d = 425$

Use elimination: From equation 1: $d = 23 - q$ Substitute: $25q + 10(23 - q) = 425$ $25q + 230 - 10q = 425$ = $15q = 195$

Solve for q: $q = 195 \div 15 = 13$ = $q = 13$ quarters

Find d: $d = 23 - 13 = 10$ = $d = 10$ dimes

Verify: Coins: $13 + 10 = 23$ ✓ Value: $13(0.25) + 10(0.10) = 3.25 + 1.00 = 4.25$ ✓ = Correct!

Answer: Maria has 13 quarters and 10 dimes.

Rate Problem

A boat travels 60 km downstream in 2 hours and 60 km upstream in 3 hours. Find the speed of the boat in still water and the speed of the current.

Define variables: Let $b$ = boat speed in still water (km/h) Let $c$ = current speed (km/h) = Two rates to find

Understand the situation: Downstream: boat and current work together → speed = $b + c$ Upstream: boat works against current → speed = $b - c$ = Key relationships identified

Use distance = rate × time: Downstream: $60 = (b + c) \times 2$, so $b + c = 30$ Upstream: $60 = (b - c) \times 3$, so $b - c = 20$ = $\begin{cases} b + c = 30 \\ b - c = 20 \end{cases}$

Solve using elimination: Add equations: $2b = 50$ $b = 25$ = Boat speed: 25 km/h

Find the current speed: $b + c = 30$ $25 + c = 30$ $c = 5$ = Current: 5 km/h

Verify: Downstream: $(25 + 5) \times 2 = 60$ ✓ Upstream: $(25 - 5) \times 3 = 60$ ✓ = Both check out!

Answer: The boat's speed in still water is 25 km/h, and the current's speed is 5 km/h.

Mistake: Using only one variable when two are needed

Why: If you have two unknown quantities, you need two variables and two equations. Using one variable often leads to contradictions or overly complex expressions.

Correct: Always identify ALL unknown quantities first, then assign a different variable to each.

Mistake: Mixing up which equation represents which condition

Why: Writing 'cost equation' when you meant 'quantity equation' gives wrong answers even with correct algebra.

Correct: Label your equations clearly: 'Equation 1 (total quantity)' and 'Equation 2 (total value)'

Mistake: Forgetting to check the answer in the original problem

Why: Calculation errors happen. The answer might solve your equations but not match the word problem conditions.

Correct: Always substitute back into BOTH original conditions and verify in context (not just in your equations).

Mistake: Setting up rate problems incorrectly

Why: Downstream, the boat and current speeds ADD. Upstream, you SUBTRACT the current from boat speed.

Correct: Think about it: going with the current makes you faster (add), going against makes you slower (subtract).

Running a Business

Business owners use systems of equations to make pricing decisions and manage inventory.

A coffee shop sells small coffees for 3 euros and large coffees for 5 euros. Yesterday they sold 120 coffees totaling 480 euros. How many of each size did they sell?

Travel Planning

Planning trips involves calculating distances, speeds, and times that often require systems of equations.

A cyclist traveled 50 km. Going uphill took 3 hours, and downhill took 2 hours. What were the uphill and downhill speeds?

Chemistry and Mixing

Chemists and pharmacists use systems to calculate how much of each substance to mix.

A chemist needs 100 mL of a 25% acid solution. She has 10% and 40% solutions available. How much of each should she mix?

Q: How do I know what to make my variables?

A: Look for phrases like 'how many', 'find the', or 'what is' - these indicate the unknowns. Each unknown quantity gets its own variable. Usually you'll have two unknowns that you're asked to find.

Q: Should I use substitution or elimination?

A: Use substitution when one equation is already solved for a variable (like $y = 2x + 3$) or can easily be solved. Use elimination when the coefficients match up nicely or when substitution would create messy fractions.

Q: What if my answer is negative or a fraction?

A: Check if it makes sense in context. You can't have -3 tickets or 2.7 people. But you CAN have negative temperatures, fractional hours, or decimal amounts of money. If the answer doesn't fit the context, recheck your setup.

Q: How do I set up rate problems?

A: Remember: distance = rate × time. For moving objects, think about whether they're moving in the same direction (subtract speeds to find relative speed) or opposite directions (add speeds). For current/wind problems, going WITH adds speed, going AGAINST subtracts.