8.3 Solving Systems of Linear Equations: Graphing, Substitution & Elimination
Key Takeaways
- A system's solution is the ordered pair (x, y) where both equations are true simultaneously — graphically, the intersection point of the two lines
- Use substitution when one equation is already solved for a variable; use elimination when both equations are in standard form Ax + By = C
- Parallel lines (same slope, different y-intercept) mean no solution; identical lines (same slope, same y-intercept) mean infinitely many solutions
- When adding equations for elimination, multiply every term by the scaling constant, not just the target variable's term
- Always check a system's solution in both original equations, not just one, before finalizing an answer
Why Systems of Linear Equations Matter on the GED
Assessment target A.2.d requires you to solve a system of linear equations — two equations with two unknowns solved together. This skill appears in its own dedicated GED questions and also underlies two-quantity comparison word problems like the ticket-sales and mixture scenarios common on the test. Because a system question can be presented as a graph, a table, or a pair of equations, the GED expects you to move comfortably between three solving methods — graphing, substitution, and elimination — and to recognize which one is fastest for a given pair of equations. Choosing the right method under time pressure (115 minutes for the whole test) is often the difference between a 90-second system problem and one that eats five minutes of your limited time.
Core Terms
A system of equations is a set of two (or more) equations considered together. A solution to a system is the ordered pair (x, y) that makes both equations true at the same time — graphically, it's the point where the two lines intersect. Depending on how the two lines relate to each other, a system has exactly one of three possible outcomes:
| System behavior | Slopes and intercepts | Number of solutions |
|---|---|---|
| Intersecting lines | Different slopes | Exactly one solution |
| Parallel lines | Same slope, different y-intercept | No solution |
| Coincident (identical) lines | Same slope, same y-intercept | Infinitely many solutions |
The Three Solving Methods
| Method | Best used when | Basic steps |
|---|---|---|
| Graphing | You need a quick visual estimate or the equations are already in slope-intercept form | Graph both lines on the same coordinate plane; the intersection point is the solution |
| Substitution | One equation is already solved for a variable, or easily rearranged into that form | Solve one equation for one variable, substitute that expression into the other equation, solve for the remaining variable, then back-substitute |
| Elimination (addition) | Both equations are in standard form (Ax + By = C) with matching or easily-matched coefficients | Multiply one or both equations by a constant so one variable's coefficients become opposites, add the equations to cancel that variable, solve, then back-substitute |
Worked Example 1 — Substitution
Solve the system: y = 2x + 1 and 3x + y = 11
- The first equation is already solved for y, so substitute (2x + 1) for y in the second equation:
- 3x + (2x + 1) = 11
- Combine like terms: 5x + 1 = 11
- Subtract 1: 5x = 10
- Divide by 5: x = 2
- Back-substitute into y = 2x + 1: y = 2(2) + 1 = 5
- Solution: (2, 5)
Worked Example 2 — Elimination Without Multiplying
Solve the system: 2x + 3y = 12 and 4x − 3y = 6
- The y-terms (+3y and −3y) are already opposites, so add the two equations directly:
- (2x + 4x) + (3y − 3y) = 12 + 6 → 6x = 18
- Divide by 6: x = 3
- Substitute x = 3 into the first equation: 2(3) + 3y = 12 → 6 + 3y = 12 → 3y = 6 → y = 2
- Solution: (3, 2)
Worked Example 3 — Elimination That Requires Multiplying First
Solve the system: x + 2y = 8 and 3x − y = 3
- Neither variable's coefficients are opposites yet. Multiply the second equation by 2 so the y-coefficients become opposites: 2(3x − y) = 2(3) → 6x − 2y = 6
- Add this to the first equation: (x + 6x) + (2y − 2y) = 8 + 6 → 7x = 14
- Divide by 7: x = 2
- Substitute into x + 2y = 8: 2 + 2y = 8 → 2y = 6 → y = 3
- Solution: (2, 3)
Real-World System: Ticket Sales
"A theater sold 100 tickets for a total of $650. Adult tickets cost $8 and child tickets cost $5. How many of each type were sold?"
- Let a = number of adult tickets and c = number of child tickets.
- Total tickets: a + c = 100
- Total revenue: 8a + 5c = 650
- Solve the first equation for c: c = 100 − a
- Substitute into the second equation: 8a + 5(100 − a) = 650 → 8a + 500 − 5a = 650 → 3a = 150 → a = 50
- Then c = 100 − 50 = 50. 50 adult tickets and 50 child tickets.
Common Traps
- Sign errors when adding equations — if one equation must be subtracted (or multiplied by a negative), a lost negative sign is the single most frequent elimination mistake.
- Multiplying only part of an equation rather than every term when scaling an equation for elimination.
- Reporting only one coordinate — a system's solution is always an ordered pair (x, y), not a single number; report both values.
- Not checking the solution in both original equations — a value that satisfies only one equation is not a valid system solution.
Solve the system: y = 3x − 4 and 2x + y = 6. What is the value of x?
A system of two linear equations has lines with the same slope and the same y-intercept. How many solutions does the system have?
A parking garage charges a flat entry fee plus an hourly rate. Two customers' receipts give the system: f + 2r = 14 and f + 5r = 26, where f is the flat fee and r is the hourly rate. What is the hourly rate, r?