Solving Systems of Two Equations: Tickets, Mixtures, and Break-Even Problems
October 21, 2026
When a problem has two unknown quantities, two independent equations are enough to find them. Systems of equations show up in pricing, chemistry mixtures, budgeting, and comparing plans. The two standard methods are substitution (solve one equation for a variable and plug it into the other) and elimination (add or subtract the equations to cancel a variable).
The hard part is usually translating the words into equations. This article walks through three typical problems.
Skip the math and get your answer instantly:
Open the Scientific Calculator →Example 1: Ticket Sales
A theater sold 100 tickets for $920. Adult tickets cost $12 and child tickets $7. How many of each? Let a be adult tickets and c be child tickets. Equation 1: a + c = 100. Equation 2: 12a + 7c = 920.
Substitution: from equation 1, c = 100 − a. Plug into equation 2: 12a + 7(100 − a) = 920 → 12a + 700 − 7a = 920 → 5a = 220 → a = 44. Then c = 56. Check: 44 × 12 = 528 and 56 × 7 = 392, and 528 + 392 = 920.
Example 2: Mixture Problem
A chemist needs 10 liters of a 30% acid solution by mixing a 20% solution and a 50% solution. Let x be liters of the 20% solution and y liters of the 50% one. Volume: x + y = 10. Acid: 0.20x + 0.50y = 0.30 × 10 = 3.
Elimination: multiply the first equation by 0.20 to get 0.20x + 0.20y = 2. Subtract from the second: 0.30y = 1, so y = 3.33 liters and x = 6.67 liters. Check the acid: 0.2 × 6.67 + 0.5 × 3.33 = 1.33 + 1.67 = 3. The mixture needs twice as much of the weaker solution as the stronger, which makes sense since 30% is twice as close to 20% as to 50%.
Example 3: Break-Even Between Two Options
Gym A charges a $40 monthly fee plus $4 per visit. Gym B charges $10 per visit with no monthly fee. When does A cost less? Let v be visits: Cost A = 40 + 4v, Cost B = 10v. Set equal: 40 + 4v = 10v → v = 6.67. For 7 or more visits a month, Gym A is cheaper. At 12 visits, A costs $88 and B costs $120.
Graphically, the solution is the point where the two lines cross. Before that, B is below A, and after it, A is below B.
| Visits per month | Gym A ($40 + $4/visit) | Gym B ($10/visit) | Cheaper |
|---|---|---|---|
| 4 | $56 | $40 | B |
| 6 | $64 | $60 | B |
| 7 | $68 | $70 | A |
| 12 | $88 | $120 | A |
Choosing a Method
Substitution is convenient when one equation is already solved for a variable or has a coefficient of 1. Elimination is faster when the coefficients line up or can be easily matched. For three or more unknowns, use matrices or a spreadsheet. If the two equations describe the same line, there are infinitely many solutions, and if they describe parallel lines, there is none.
Checking Your Answer
Always substitute the solution back into both original equations. Check that the answer makes sense in context: negative counts or fractional tickets signal an error or an impossible problem. Also check units, such as liters or dollars.
Frequently Asked Questions
How many equations do I need for two unknowns?
Two independent equations.
What is the difference between substitution and elimination?
Substitution replaces a variable using one equation. Elimination adds or subtracts equations to cancel a variable.
What if the system has no solution?
The equations describe parallel lines that never meet, for example x + y = 2 and x + y = 5.
How do I check a solution?
Plug the values into both original equations and verify they are true.
Can I solve a system graphically?
Yes, by graphing both lines and finding their intersection.