The hardest part of systems of equations is not solving them — it is setting them up correctly. The mechanics of substitution or elimination are straightforward; the translation from a described situation to a pair of equations is where most students lose marks.
The general approach
Identify how many unknowns you have. For a system of two equations, you need exactly two unknowns and exactly two independent relationships between them.
Name each unknown explicitly, with units. The discipline of writing "let x = the number of adult tickets sold" rather than "let x = adult tickets" prevents a category of error where you confuse the count of something with the cost of something.
Write each constraint as an equation. A "constraint" is anything the problem states that restricts the values of your unknowns.
Example 1: ticket pricing
An event sells adult tickets for $12 and child tickets for $7. Total attendance: 350 people. Total revenue: $3,450.
Let a = number of adult tickets sold. Let c = number of child tickets sold.
Constraint 1 (attendance): a + c = 350 Constraint 2 (revenue): 12a + 7c = 3,450
Solve by substitution: a = 350 − c into constraint 2: 12(350 − c) + 7c = 3,450 4,200 − 12c + 7c = 3,450 −5c = −750 c = 150, a = 200.
Example 2: mixing solutions in chemistry
You need 500 mL of a 30% saline solution. You have a 20% saline solution and a 50% saline solution. How much of each?
Let x = mL of 20% solution. Let y = mL of 50% solution.
Constraint 1 (volume): x + y = 500 Constraint 2 (salt content): 0.20x + 0.50y = 0.30 × 500 = 150
Solve: x = 500 − y, then 0.20(500 − y) + 0.50y = 150 → 100 − 0.20y + 0.50y = 150 → 0.30y = 50 → y ≈ 166.7 mL, x ≈ 333.3 mL.
Example 3: distance problems with two vehicles
Two vehicles leave from the same point at the same time, travelling in opposite directions. Vehicle A travels at 55 mph. Vehicle B travels at 70 mph. After how many hours are they 375 miles apart?
Here we have one unknown only: time t.
Distance of A = 55t. Distance of B = 70t. Total distance = 55t + 70t = 125t = 375. t = 3 hours.
This is technically one equation with one unknown — but it is a useful calibration problem. Students who try to force two variables often make it harder.
What to do when a problem has three unknowns
You need three equations. They must be independent (no one is a multiple of another). Standard problems: three people sharing costs with given pairwise totals; three chemical components mixed to given specifications; three unknowns related by interest-rate constraints. Set up one variable per unknown, identify three constraints, and use elimination or matrices.
The error to avoid: double-counting constraints
A constraint must carry new information. "a + c = 350" and "2a + 2c = 700" are not two independent constraints — one is twice the other. A system where every pair of equations is proportional has infinitely many solutions and represents a badly set-up problem.
Frequently Asked Questions
What if I cannot find two independent equations for two unknowns?
The problem may be underdetermined (not enough information), or you may have found dependent equations. Check whether your two equations are multiples of each other — if they are, you have one equation in disguise, and you need more information.
Can a system of equations have no solution?
Yes — an inconsistent system. This happens when the two lines are parallel (same slope, different intercept). Geometrically: two lines that never intersect. Algebraically: you end up with a contradiction like 0 = 5 during elimination.