9.1Solving Two-Variable Linear Systems
Solve 2×2 linear systems by substitution, elimination, and graphing. Classify systems as consistent (one solution), inconsistent (no solution), or dependent (infinitely many solutions).
Systems of equations model real-world situations with multiple constraints — supply and demand, mixture problems, and circuit analysis. Solving them efficiently is a foundational skill for linear algebra and applied mathematics.
Essential Question
What does it mean for an ordered pair to be a solution to a system of equations, and how do substitution, elimination, and graphing each reveal the solution in a different way?
Lesson Overview
- ▸A system of two linear equations consists of two equations sharing the same two variables.
- ▸A solution is an ordered pair (x, y) that satisfies both equations simultaneously.
- Three types of systems
- ✓Consistent & independent: one solution — lines intersect at exactly one point.
- ∞Consistent & dependent: infinitely many solutions — the equations describe the same line.
- ∅Inconsistent: no solution — the lines are parallel and never meet.
- ▸Method 1 — Substitution: solve one equation for one variable, then substitute into the other.
- ▸Method 2 — Elimination: multiply equations to make coefficients opposite, then add to eliminate one variable.
- ▸Method 3 — Graphing: graph both lines and find the intersection (best for estimating, not exact answers).
- ▸Applications: mixture problems, cost/revenue break-even, distance/rate/time.
Worked Examples
Solve by substitution: 2x + y = 7 and x − y = 2.
Step 1 — Solve the second equation for y: y = x − 2.
Step 2 — Substitute into the first equation: 2x + (x − 2) = 7.
Step 3 — Simplify: 3x − 2 = 7 → 3x = 9 → x = 3.
Step 4 — Back-substitute: y = 3 − 2 = 1.
Step 5 — Check in both equations: 2(3)+1 = 7 ✓ and 3−1 = 2 ✓.
Solve by elimination: 3x + 2y = 12 and 5x − 2y = 4.
Step 1 — The y-coefficients are already opposites (+2 and −2).
Step 2 — Add the equations: (3x+5x) + (2y−2y) = 12+4 → 8x = 16.
Step 3 — Solve: x = 2.
Step 4 — Substitute into the first equation: 3(2) + 2y = 12 → 2y = 6 → y = 3.
Step 5 — Check: 3(2)+2(3)=12 ✓ and 5(2)−2(3)=4 ✓.
Solve by elimination (multiply first): 3x + 2y = 7 and x + 4y = 9.
Step 1 — Multiply the first equation by 2: 6x + 4y = 14.
Step 2 — Subtract the second equation: (6x−x) + (4y−4y) = 14−9 → 5x = 5.
Step 3 — Solve: x = 1.
Step 4 — Substitute into x + 4y = 9: 1 + 4y = 9 → 4y = 8 → y = 2.
Step 5 — Check: 3(1)+2(2)=7 ✓ and 1+4(2)=9 ✓.
Identify the system type: 2x − 4y = 6 and x − 2y = 3.
Step 1 — Divide the first equation by 2: x − 2y = 3.
Step 2 — This is identical to the second equation.
Step 3 — Both equations represent the same line → dependent system.
Step 4 — There are infinitely many solutions.
Break-even problem: A company has fixed costs of $500 and variable costs of $8/unit. Revenue is $13/unit. Find the break-even quantity.
Step 1 — Write the cost equation: C = 500 + 8x.
Step 2 — Write the revenue equation: R = 13x.
Step 3 — Set equal for break-even: 500 + 8x = 13x.
Step 4 — Solve: 500 = 5x → x = 100.
Step 5 — Verify: C = 500+800 = $1,300; R = 13(100) = $1,300 ✓.
Guided Practice
Solve by substitution: y = 3x − 1 and 2x + y = 9.
Hint: y is already isolated — substitute directly into the second equation.
Solve by elimination: 4x + 3y = 10 and 2x − 3y = 8.
Hint: The y-coefficients are already opposites — add the equations directly.
Solve: 5x + 2y = 16 and 3x − 4y = 6.
Hint: Multiply the first equation by 2 to make the y-coefficients opposites.
Determine the system type: 6x − 9y = 12 and 2x − 3y = 5.
Hint: Try to make the coefficients match — what happens?
A store sells two types of tickets: adult ($12) and child ($8). If 200 tickets were sold for $2,000, how many of each type were sold?
Hint: Set up two equations — one for total tickets, one for total revenue.
Key Vocabulary
System of equations
Two or more equations with the same variables; the solution satisfies all equations simultaneously.
Consistent system
A system with at least one solution.
Inconsistent system
A system with no solution; the lines are parallel.
Dependent system
A system with infinitely many solutions; the equations represent the same line.
Substitution method
Solving one equation for a variable and substituting into the other equation.
Elimination method
Adding multiples of equations together to cancel (eliminate) one variable.
Break-even point
The quantity at which total cost equals total revenue.
Check Your Understanding
Interactive Practice — 5 Questions
Which ordered pair is a solution to the system x + y = 5 and 2x − y = 1?
Two lines are parallel. The system is:
Solve by elimination: 3x + y = 7 and x − y = 1.
The system 2x − 4y = 8 and x − 2y = 4 is:
A system has exactly one solution. The lines:
Common Mistakes
Common Mistakes
Substituting back into the same equation you solved for the variable.
After finding x, substitute into the OTHER original equation to find y. Using the same equation gives a tautology.
Forgetting to multiply BOTH sides when scaling an equation for elimination.
If you multiply an equation by a constant, every term on both sides must be multiplied.
Declaring a system inconsistent when it is actually dependent.
If you get 0 = 0, the system is dependent (infinitely many solutions). If you get 0 = nonzero, it is inconsistent (no solution).
Stopping after finding x without solving for y.
A solution to a 2-variable system is an ordered pair (x, y). Always find both values.
Math Tips
Math Tips
Choose substitution when one variable is already isolated (e.g., y = 3x + 1). Choose elimination when coefficients are easy to match.
Always check your solution by substituting back into BOTH original equations.
For word problems: define variables clearly, write two equations, then solve the system.
If elimination gives 0 = 0, the system is dependent. If it gives 0 = k (k ≠ 0), it is inconsistent.
Graphing is great for visualizing but gives only approximate answers. Use algebra for exact solutions.