Learning Objectives
Solve a two-variable linear system by substitution or elimination.
Interpret a solution as a pair satisfying both equations.
Build a system from two independent constraints in a word problem.
Prerequisites: solving linear equations and using the distributive property. A system asks for values that make all its equations true at the same time. A pair that satisfies only one equation is not a solution of the system.
1. Choose Substitution When a Variable Is Isolated
If one equation already expresses a variable in terms of the other, substitute that expression into the second equation. This produces a one-variable equation. After solving it, substitute back to find the remaining variable.
Worked Example: Substitution
Solve y = 2x + 1 and x + y = 10. Replace y in the second equation with 2x + 1.
Then x + 2x + 1 = 10, so 3x = 9 and x = 3. Substitute into y = 2x + 1 to obtain y = 7.
Check both equations: 7 = 2(3) + 1 and 3 + 7 = 10. The solution is (3, 7).
2. Choose Elimination When Coefficients Align
Adding or subtracting equations can remove one variable when its coefficients are opposites or equal. If needed, multiply an entire equation by a nonzero constant first. Every term, including the constant on the right, must be multiplied.
Worked Example: Elimination
Solve 2x + 3y = 13 and 4x − 3y = 5. Add the equations to eliminate y.
This gives 6x = 18, so x = 3. Substitute into the first equation: 6 + 3y = 13, giving y = 7/3.
Check the second: 4(3) − 3(7/3) = 12 − 7 = 5. A fractional answer is valid; there is no reason to force an integer.
3. Interpret Special Cases
What elimination gives | System meaning | Graph meaning for two lines |
|---|---|---|
A variable equation | One ordered-pair solution | Intersecting lines |
A false statement | No solution | Distinct parallel lines |
An identity, with equivalent equations | Infinitely many solutions | The same line |
For example, x + y = 4 and 2x + 2y = 8 describe the same line. Doubling the first equation produces the second. But x + y = 4 and 2x + 2y = 9 cannot both hold, because doubling the first would require the total to be 8 rather than 9.
4. Translate Two Constraints
Worked Example: Tickets
A group buys 12 tickets. Adult tickets cost $8 and student tickets cost $5; the total is $78. Let a and s be the counts.
The count equation is a + s = 12. The cost equation is 8a + 5s = 78.
Multiply the first equation by 5 and subtract it from the second: 3a = 18, so a = 6. Then s = 6.
Check both constraints: 6 + 6 = 12 tickets, and 6(8) + 6(5) = $78. Nonnegative integer counts fit the context.
Common Mistakes
When subtracting equations, distribute the subtraction to every term of the equation being subtracted.
Return both variables if an ordered pair is requested.
In a word problem, distinguish a count equation from a cost equation and keep their units consistent.
5. Practice
1. Solve y = 3x − 2 and x + y = 14.
2. Solve 3x + 2y = 16 and 3x − 2y = 8.
3. Classify the system x − 2y = 3 and 2x − 4y = 7.
4. A box has 10 coins, all worth 5 or 10 cents, totaling 70 cents. Find the number of each type.
Worked Solutions
1. Substitute to get x + 3x − 2 = 14, so x = 4 and y = 10. The solution is (4, 10).
2. Adding gives 6x = 24, so x = 4. Then 12 + 2y = 16 gives y = 2.
3. Twice the first equation requires 2x − 4y = 6, which contradicts the second equation. There is no solution.
4. Let f and t count the coins. Then f + t = 10 and 5f + 10t = 70. Subtracting five times the count equation gives 5t = 20, so t = 4 and f = 6.
6. Summary
- A system solution satisfies every equation simultaneously.
- Choose substitution or elimination based on the given structure.
- Use identities and contradictions to recognize special cases.
- Check a word-problem solution against both the equations and the context.