Lesson 4 of 12 · Algebra Foundations
Linear Equations & Systems
Solving a system is mechanical once set up correctly. The two skills that actually distinguish strong performance here are recognizing how many solutions a system has before fully solving it, and translating a word problem into equations without hesitation.
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Learning objectives
- Solve two-variable linear systems fluently by both substitution and elimination, and choose the faster one on sight.
- Determine whether a system has no solution, exactly one, or infinitely many, from the coefficients alone.
- Solve three-variable linear systems by systematic elimination.
- Translate a word problem into a system of equations directly, without an intermediate "let x be..." false start.
The core idea
Every linear system question is really one of two things: "solve it" or "determine what kind of solution set it has." The second type is often faster than it looks, because two linear equations in two variables represent two lines, and lines can only relate to each other three ways — they cross once, they're identical, or they're parallel and distinct. You can often answer "how many solutions" from the coefficients alone, without solving anything.
Classifying a 2×2 linear system without solving
ax + by = c and dx + ey = f
Compare the ratios a/d and b/e. If they're unequal, the lines cross: exactly one solution. If they're equal, compare c/f too — if all three ratios match, the equations describe the same line: infinitely many solutions. If a/d = b/e but c/f differs, the lines are parallel and distinct: no solution.
Elimination, in general
Multiply equations so a variable's coefficients match (or are opposite), then add or subtract. For three variables, apply elimination twice — first to remove one variable and reduce to a two-variable system, then solve that system normally.
Worked example 1 — Substitution
Problem
Solve the system: 2x + 3y = 12, x − y = 1.
Key insight
The second equation already isolates x in terms of y with almost no work: x = y + 1.
Solution
Substituting: 2(y+1) + 3y = 12 → 5y = 10 → y = 2, and then x = 3. Solution: (x, y) = (3, 2).
Takeaway
Substitution wins when one equation already isolates a variable cheaply — look for that before defaulting to elimination.
Worked example 2 — Classifying without solving
Problem
Without fully solving, determine how many solutions the system 4x + 6y = 10, 2x + 3y = 7 has.
Key insight
Compare the coefficient ratios: 4/2 = 2 and 6/3 = 2, so the left sides are proportional — but check the constants too.
Solution
10/7 ≠ 2, so the constants aren't in the same ratio as the coefficients. The lines are parallel but distinct: no solution.
Takeaway
All three ratios must match for infinitely many solutions; two matching and one not means no solution at all.
Worked example 3 — Three-variable elimination
Problem
Solve: x + y + z = 6, x − y + z = 2, x + y − z = 0.
Key insight
Each pair of equations shares two variables with matching signs — subtracting pairs eliminates one variable at a time cheaply.
Solution
Subtracting the third equation from the first eliminates y directly: 2z=6 → z=3. Subtracting the second from the first eliminates z: 2y=4 → y=2. From the first equation: x + 2 + 3 = 6 → x = 1. Solution: (x, y, z) = (1, 2, 3).
Takeaway
Before mechanically eliminating in a fixed order, look for pairs of equations where subtraction cancels a variable in one step.
Worked example 4 — Word problem setup
Problem
A store sells pens at $2 each and markers at $3 each. A customer buys 10 items total for $27. How many of each did they buy?
Key insight
"10 items total" and "$27 total" are two independent pieces of information — each becomes its own equation.
Solution
Let p = pens, m = markers: p + m = 10 and 2p + 3m = 27. From the first, p = 10 − m; substituting: 2(10−m) + 3m = 27 → m = 7, p = 3. 3 pens and 7 markers.
Takeaway
Every independent quantity mentioned in the problem ("total items," "total cost") is one equation — count them before writing anything down.
Strategy notes
Check for a cheap elimination before committing to a method
Scan both equations for a variable whose coefficients are equal, opposite, or simple multiples of each other. That variable eliminates in one step — use it, rather than mechanically applying substitution every time.
"How many solutions" questions rarely need full solving
If a problem only asks for the number of solutions (not the solutions themselves), compare coefficient ratios directly instead of solving the system all the way through.
Common mistakes
Sign errors when subtracting equations
Subtracting an entire equation means subtracting every term, including the constant on the right side. Distribute the subtraction across the whole equation, not just the variable terms.
Concluding "infinitely many" from matching coefficients alone
Matching coefficient ratios only guarantee the lines are parallel. You must also check the constant term's ratio to distinguish "same line" (infinite solutions) from "parallel and distinct" (no solution).
Writing one equation for two unknowns in a word problem
A system needs as many independent equations as unknowns. If you can only extract one equation from a word problem with two unknowns, you've missed a piece of given information.
Practice
20 questions across four difficulty tiers. Solve-the-system questions want
your answer as comma-separated values in order (e.g. 6, 4 for
x, y — or 2, 3, 4 for x, y, z). Classification questions want
a short phrase like "no solution" or "infinitely many solutions." Up to
three tries per question before the solution is shown.