What is a system of equations?
A system of equations is two or more equations that share the same variables. The solution is the set of values that works in all of them at once.
With two variables, each linear equation is a straight line. The solution is the point where the lines cross. For 2x + 3y = 7 and x − y = 1, the lines meet at (2, 1), and the calculator draws both lines so you can see it.
How does the system of equations calculator use elimination?
Multiply one or both equations so a variable has opposite coefficients, then add the equations to cancel it. Solve for the variable that is left and substitute back to find the other.
An elimination method calculator does the same thing in a more organized way. This one uses Gauss-Jordan elimination, scaling and combining rows until each equation has one variable left. Its steps look slightly different from the hand method, but the answer is the same.
How do you solve by substitution?
Solve one equation for one variable, then substitute that expression into the other equation. Substitution is easiest when a variable already has a coefficient of 1.
The calculator itself always uses elimination, so it is not a substitution method calculator. You can still check your substitution work against its answer, since both methods give the same solution.
Here is another one where substitution fits well: y = 2x − 1 and 3x + 2y = 12. Replace y to get 3x + 2(2x − 1) = 12, so 7x = 14 and x = 2. Then y = 2(2) − 1 = 3.
Should you use elimination or substitution?
Pick substitution when one variable is already alone or has a coefficient of 1. Pick elimination when the coefficients line up or can be matched with one multiplication.
| System looks like | Good first choice | Why |
|---|---|---|
| y = 2x − 1, 3x + 2y = 12 | Substitution | y is already isolated |
| x − y = 1, x + y = 5 | Elimination | Adding cancels y right away |
| 2x + 3y = 7, x − y = 1 | Either | One multiplication sets up elimination |
Graphing works too, but it only gives an exact answer when the lines cross at a clean point. Algebra gives exact values every time.
How do you set up a system from a word problem?
Give each unknown its own letter and write one equation for each fact in the problem. Then solve the system the usual way.
Say a school sold 12 tickets for $29. Adult tickets cost $3 and child tickets cost $2. Let a be adult tickets and c be child tickets. The count gives a + c = 12, and the money gives 3a + 2c = 29. Solving gives a = 5 and c = 7. Check it: 5 + 7 = 12 and 15 + 14 = 29.
How do you solve a system with three variables?
Use elimination twice to reduce three equations to two, solve that pair, then substitute back. The calculator handles systems of 3 or 4 equations the same way it handles 2.
Take x + y + z = 6, 2x − y + z = 3, and x + 2y − z = 2. Adding the first and third equations cancels z and gives 2x + 3y = 8. Adding the second and third gives 3x + y = 5. Solving that pair gives x = 1 and y = 2, and then z = 6 − 1 − 2 = 3.
What if there is no solution or infinitely many?
If elimination leaves a false statement like 0 = 2, the system has no solution. If it leaves a true statement like 0 = 0, the system has infinitely many solutions.
For x + y = 3 and 2x + 2y = 8, doubling the first equation gives 2x + 2y = 6, which can never equal 8. The lines are parallel and never meet. For x + y = 3 and 2x + 2y = 6, the second equation is just the first one doubled, so every point on the line works. The calculator writes this as x = 3 − y, with y any real number.
| Result | What elimination shows | Graph |
|---|---|---|
| One solution | A value for each variable | Lines cross at one point |
| No solution | A false statement like 0 = 2 | Parallel lines |
| Infinitely many | A true statement like 0 = 0 | Same line twice |
What are common mistakes when solving systems of equations?
Multiplying only one side of an equation is the most frequent error. If you multiply x − y = 1 by 3, the right side becomes 3 too. Another is stopping after finding x and forgetting to find y.
Always check the pair in both original equations. A solution that works in one equation but not the other means a slip somewhere.