A system of equations is two or more equations that share the same variables. The solution is the pair of values that makes every equation true at once. On a graph it is the point where the lines cross.
Three ways to solve
| Method | Use it when |
|---|---|
| Substitution | One equation is already solved for a variable, like y = 2x + 1 |
| Elimination | Both equations are in standard form, like 2x + 3y = 12 |
| Graphing | You have a graphing calculator, or you only need to estimate |
All three give the same answer. Pick the one that means less work for the system in front of you.
Substitution
Solve one equation for one variable, then put that expression into the other equation. You are left with one equation and one unknown.
Example Solve y = 2x + 1 and 3x + y = 16
- The first equation already gives y. Substitute 2x + 1 for y in the second: 3x + (2x + 1) = 16.
- Combine: 5x + 1 = 16, so 5x = 15 and x = 3.
- Put x = 3 back into y = 2x + 1: y = 7.
- Check in the second equation: 3(3) + 7 = 16. ✓
- The solution is (3, 7).
Elimination
Add or subtract the equations so that one variable cancels. If the coefficients don’t match, multiply one or both equations first.
Example Solve 2x + 3y = 12 and 4x − 3y = 6
- The y terms are +3y and −3y. Add the equations: 6x = 18.
- So x = 3.
- Substitute into the first equation: 2(3) + 3y = 12, so 3y = 6 and y = 2.
- Check in the second: 4(3) − 3(2) = 6. ✓
Example Solve 3x + 2y = 16 and 2x − 5y = −2
- Make the y coefficients opposites. Multiply the first equation by 5 and the second by 2: 15x + 10y = 80 and 4x − 10y = −4.
- Add: 19x = 76, so x = 4.
- Substitute: 3(4) + 2y = 16, so y = 2.
- Check: 2(4) − 5(2) = 8 − 10 = −2. ✓
Multiply every term
When you multiply an equation to line up coefficients, multiply the number on the right side too. Forgetting it is the most common elimination error.
Graphing
Graph both lines and read off the intersection. This is quick in a graphing calculator, and on the digital SAT you can type both equations into the built-in Desmos calculator and click the crossing point. By hand, graphing only gives an estimate unless the answer lands on grid lines.
When there is no solution, or infinitely many
Two lines can relate in three ways:
| Lines | Number of solutions | What algebra shows |
|---|---|---|
| Cross once | One | You find a value for each variable |
| Parallel | None | You get a false statement like 0 = 4 |
| Same line | Infinitely many | You get a true statement like 0 = 0 |
Example y = 2x + 3 and y = 2x − 1
Both lines have slope 2 and different intercepts, so they are parallel. Setting them equal gives 2x + 3 = 2x − 1, which simplifies to 3 = −1. No solution.
Example x + y = 4 and 2x + 2y = 8
The second equation is the first one multiplied by 2. They describe the same line, so every point on it is a solution.
A shortcut for standard form: if the x and y coefficients are in the same ratio but the constants are not, there is no solution. If all three are in the same ratio, there are infinitely many.
Word problems
Systems show up whenever a problem has two unknowns and gives you two facts. Name both unknowns, write one equation per fact, then solve.
Example A school play sold 50 tickets for $480. Adult tickets cost $12 and student tickets cost $8. How many of each were sold?
- Let a be adult tickets and s be student tickets.
- Count of tickets: a + s = 50. Money: 12a + 8s = 480.
- From the first, s = 50 − a. Substitute: 12a + 8(50 − a) = 480.
- Simplify: 4a + 400 = 480, so a = 20 and s = 30.
- Check: 12(20) + 8(30) = 240 + 240 = 480. ✓
Example How many liters of a 10% salt solution and a 30% salt solution make 20 liters of a 25% solution?
- Let x be liters of 10% and y be liters of 30%.
- Total volume: x + y = 20. Total salt: 0.10x + 0.30y = 0.25(20) = 5.
- Substitute y = 20 − x: 0.10x + 6 − 0.30x = 5, so −0.20x = −1 and x = 5.
- Then y = 15. Use 5 liters of the 10% solution and 15 liters of the 30% solution.
Try a few
Solve y = x − 4 and 2x + y = 11.
Show answer
(5, 1). Substitute: 2x + x − 4 = 11, so 3x = 15.
Solve x + 2y = 7 and x − 2y = −1.
Show answer
(3, 2). Adding the equations gives 2x = 6. Then 3 + 2y = 7 gives y = 2.
Solve 5x + 2y = 1 and 3x + 4y = 9.
Show answer
(−1, 3). Multiply the first by 2 to get 10x + 4y = 2, subtract the second to get 7x = −7, then 5(−1) + 2y = 1 gives y = 3.
For what value of k does the system 2x + 3y = 6 and 4x + ky = 10 have no solution?
Show answer
k = 6. Then the second equation is 4x + 6y = 10, which has the same left side as twice the first (4x + 6y = 12) but a different constant, so the lines are parallel.
A jar holds 30 coins, all dimes and quarters, worth $4.35. How many of each?
Show answer
21 dimes and 9 quarters. Use d + q = 30 and 10d + 25q = 435. Substituting gives 300 + 15q = 435, so q = 9.
Practice
Questions students ask
Which method is fastest?
If a variable already has a coefficient of 1 or −1, substitution is usually faster. If both equations are in standard form, elimination is. On a test with a graphing calculator, graphing is often fastest of all.
Do I need to find both variables?
Read the question. Many test problems ask only for x, or for a combination like x + y. Find what is asked and stop.
What about three equations?
Use elimination twice to cut the system down to two equations in two unknowns, then solve as usual. The ideas are the same, with more bookkeeping.
New to equations? Start with How to Solve Linear Equations.