How to Solve a System of Two Linear Equations
A system of linear equations is two equations that share the same two unknowns, usually x and y. Solving the system means finding the values that make both equations true at the same time. The two main algebra methods are substitution and elimination, and both are worked below with checks, followed by the special cases that confuse many students.
1. What a solution actually is
The Common Core grade 8 standards give the key picture: solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
Each linear equation draws a straight line. Where the two lines cross, the point (x, y) sits on both lines, so it works in both equations. That gives three possibilities:
- The lines cross once: one solution.
- The lines are parallel: no solution.
- The lines are the same line: infinitely many solutions.
The same standard asks students to solve systems algebraically and to estimate solutions by graphing. Graphing gives a picture, but algebra gives exact answers, so that is the focus here.
2. Method 1: substitution
Substitution works best when one equation already has a variable by itself.
Solve:
- y = 2x + 1
- 3x + y = 16
Step 1: The first equation says y equals 2x + 1, so replace y in the second equation with 2x + 1. 3x + (2x + 1) = 16
Step 2: Combine like terms and solve. 5x + 1 = 16, so 5x = 15, and x = 3.
Step 3: Put x = 3 back into the simpler equation. y = 2(3) + 1 = 7.
Check in both equations:
- First: 2(3) + 1 = 7. Correct.
- Second: 3(3) + 7 = 9 + 7 = 16. Correct.
The solution is (3, 7).
3. Method 2: elimination
Elimination works best when both equations are in the form ax + by = c. The idea is to add or subtract the equations so that one variable cancels.
Solve:
- 2x + 3y = 12
- 4x - y = 10
Step 1: Make the y terms opposites. The first equation has +3y and the second has -y, so multiply every term of the second equation by 3: 12x - 3y = 30
Step 2: Add this to the first equation. The y terms cancel. (2x + 3y) + (12x - 3y) = 12 + 30, so 14x = 42, and x = 3.
Step 3: Substitute x = 3 into either original equation. Using the second: 4(3) - y = 10, so 12 - y = 10, and y = 2.
Check in both original equations:
- First: 2(3) + 3(2) = 6 + 6 = 12. Correct.
- Second: 4(3) - 2 = 12 - 2 = 10. Correct.
The solution is (3, 2).
Why is it allowed to add equations together like this? The Common Core high school algebra standards ask students to prove exactly this: replacing one equation in a system by the sum of that equation and a multiple of the other produces a system with the same solutions. So scaling and adding never changes the answer, as long as it is done correctly.
The most common slip in elimination is multiplying only some terms of an equation. When you scale an equation, every term, including the number on the right side, gets multiplied.
4. Choosing a method
| Situation | Faster method |
|---|---|
| One variable is already alone (y = ... or x = ...) | Substitution |
| Both equations are in ax + by = c form | Elimination |
| A variable has coefficient 1 or -1 | Either works; substitution is often quick |
| Coefficients line up as opposites already | Elimination with no scaling needed |
Both methods always give the same answer for the same system. Pick the one that needs fewer steps.
5. Special case: no solution
The grade 8 standard includes a short example worth knowing: 3x + 2y = 5 and 3x + 2y = 6 have no solution, because 3x + 2y cannot be 5 and 6 at the same time.
If you try elimination here, subtracting the equations leaves 0 = -1, a false statement with no variables left. That false statement is the signal. On a graph, the two lines are parallel and never meet.
6. Special case: infinitely many solutions
Solve:
- x + y = 4
- 2x + 2y = 8
Multiply the first equation by 2 and you get exactly the second equation. Subtracting them leaves 0 = 0, which is always true. The two equations describe the same line, so every point on that line is a solution. You would write the answer as "infinitely many solutions," not as a single point.
7. A word problem
The standards also ask students to solve real-world problems leading to two linear equations.
Problem: A school play sold 20 tickets for a total of $130. Adult tickets cost $8 and student tickets cost $5. How many of each were sold?
Let a be adult tickets and s be student tickets.
- Number of tickets: a + s = 20
- Money collected: 8a + 5s = 130
From the first equation, s = 20 - a. Substitute: 8a + 5(20 - a) = 130, so 8a + 100 - 5a = 130, then 3a = 30, and a = 10. Then s = 20 - 10 = 10.
Check: 10 + 10 = 20 tickets, and 8(10) + 5(10) = 80 + 50 = 130 dollars. Both conditions hold.
8. Key takeaways
- The solution to a system is the point where the two lines intersect, so it must satisfy both equations.
- Use substitution when a variable is already alone, and elimination when both equations are in ax + by = c form.
- When scaling an equation for elimination, multiply every term, including the right side.
- A false result like 0 = -1 means no solution; a true result like 0 = 0 means infinitely many solutions.
- Always check your answer in both original equations, not just one.