What it is
A system of linear equations is two or more linear equations in the same unknowns, all of which must hold at the same time. With the unknowns x and y it can look like this:
- (1) y = 2x + 1
- (2) 3x + y = 11
A solution is a pair of values (x, y) that makes every equation true at once. x = 2 and y = 5 is a solution: (1) gives 2 · 2 + 1 = 5 and (2) gives 3 · 2 + 5 = 11. The pair x = 0 and y = 1 satisfies (1) but not (2), because 3 · 0 + 1 is 1, not 11, so it is not a solution of the system.
On a graph each equation is a straight line in a coordinate plane, and the points on the line are the pairs that solve that equation. A solution of the system is a point that lies on every line. For two equations in two unknowns there are three cases:
| The lines | Solutions | Example |
|---|---|---|
| cross at one point | one | y = 2x + 1 and 3x + y = 11 |
| are parallel, with the same gradient and different intercepts | none | y = 2x + 1 and y = 2x + 3 |
| are the same line | infinitely many | y = 2x + 1 and 2y = 4x + 2 |
- No solution. In y = 2x + 1 and y = 2x + 3, substituting gives 2x + 1 = 2x + 3, that is 1 = 3, which holds for no x. The lines have the same gradient and different intercepts, so they are parallel and never meet.
- Infinitely many solutions. In y = 2x + 1 and 2y = 4x + 2, the second equation is the first times 2. Both describe the same line, and every point on it is a solution, for example (0, 1), (1, 3) and (2, 5).
With three unknowns each equation is a plane in space, and a solution is a point (x, y, z) that lies in every plane. The cases are the same: one solution, none or infinitely many. The system x + y + z = 6, 2x − y + z = 3 and x + 2y − z = 2 has the solution x = 1, y = 2 and z = 3.
A system of equations is not linear if an unknown appears squared, in a product like xy, or in a denominator. x² + y = 5 and x + y = 3 have two solutions, (2, 1) and (−1, 4), which a linear system cannot have.
Why it matters
A system of equations is needed when two conditions must hold at once. Plan A costs 99 a month plus 0.50 a minute, and plan B costs 49 plus 1 a minute. With cost y and x minutes, that is y = 0.5x + 99 and y = x + 49. The plans cost the same where the lines meet: 0.5x + 99 = x + 49 gives x = 100, and the cost is then 149. Below 100 minutes B is cheaper, above 100 minutes A is.
You find the solution by drawing the lines or by calculating. The substitution method, the elimination method and Cramer’s rule give the same answer. The system above is worked out in the system of equations calculator, and the graphing calculator draws y = 2x + 1 and y = 11 − 3x and shows the point (2, 5) where they cross.
With inequalities instead of equalities, the subject is linear programming.
Common misconceptions
- “The solution is a number.” The solution is a pair: x = 2 is one half of the answer and y = 5 the other. With three unknowns the solution is three numbers.
- “Two equations in two unknowns have one solution.” Only when the lines have different gradients. With the same gradient there is no solution or infinitely many.
- “If I get 1 = 3, I have made a mistake.” Not necessarily. 1 = 3 means the lines are parallel and the system has no solution, and 0 = 0 means the equations describe the same line.
- “A pair that satisfies one of the equations is enough.” The pair must satisfy all of them. (0, 1) lies on y = 2x + 1 but not on 3x + y = 11.
Common questions
How many solutions can a system of linear equations have?
None, one or infinitely many. Two is impossible: two points fix a single straight line, so two lines with two
points in common are the same line and share all their points.
Which method should I choose?
Substitution suits a system where one variable stands alone or has coefficient 1. Elimination suits one where the
coefficients are easily made equal with opposite signs. The graph gives the picture but only an approximate answer
if the crossing point does not fall on the grid. The methods give the same solution.
Can a system have more equations than unknowns?
Yes. x + y = 3, x − y = 1 and 2x + y = 5 have the solution x = 2 and y = 1, and the third equation adds nothing new.
Replace it with 2x + y = 6 and there is no solution, because the first two allow only x = 2 and y = 1.
How do I write the answer when there are infinitely many solutions?
As the line the solutions lie on. For y = 2x + 1 and 2y = 4x + 2, the solutions are all pairs (x, y) with
y = 2x + 1: every x gives a solution.