The idea: the answer is where the lines cross

A system of equations is two (or more) equations that must be true at the same time, like x + y = 6 and x − y = 2. A solution is a pair (x, y) that makes both true. Each equation ax + by = c on its own has infinitely many solutions, and they form a straight line. A point that solves both equations lies on both lines, so the answer is the point where the lines cross. Here that is (4, 2): 4 + 2 = 6 and 4 − 2 = 2.

Left: the equations x + y = 6 in blue and x − y = 2 in orange. Right: their graphs, a blue line falling and an orange line rising, crossing at the point (4, 2), with the checks 4 + 2 = 6 and 4 − 2 = 2
Each equation is a line; the one point on both lines, (4, 2), is the solution of the system.

There are three classroom methods. They all find the same point; they differ in how much work the numbers take.

Graphing

Find two points on each line (two points fix a line), draw both lines and read off the crossing. Then put the point into both equations to check it. Graphing shows at once whether there is one answer, none or infinitely many, but it is only exact when the crossing lands on grid lines: a crossing at (13/7, 2/7) cannot be read off a grid. (Demo: graphing, read the crossing (4, 2))

Substitution

Solve one equation for one unknown, then put that expression into the other equation. With y = 2x − 3 and x + 2y = 4, replace y by (2x − 3): x + 2(2x − 3) = 4, so 5x − 6 = 4, 5x = 10 and x = 2. Then y = 2(2) − 3 = 1. On the graph, y = 2x − 3 is the same blue line, just written as y = mx + k. The equation x = 2 is a vertical line through the crossing, and y = 1 a horizontal line through it. (Demo: substitution, y = 2x − 3 into x + 2y = 4)

Pick the unknown whose coefficient is 1 or −1, so no fractions appear. In 3x + 2y = 12 and 2x − 3y = −5 no coefficient is ±1, and substitution gives y = −(3/2)x + 6; it still works, but elimination is easier there.

Elimination

Add or subtract the equations so that one unknown cancels. With x + y = 6 and x − y = 2 the y terms are already +y and −y, and adding gives 2x = 8, so x = 4. Put it back into the first equation: 4 + y = 6, y = 2. (Demo: elimination by adding)

When the y terms do not match, multiply first. For 3x + 2y = 12 and 2x − 3y = −5, multiply the first by 3 and the second by 2: 9x + 6y = 36 and 4x − 6y = −10. Adding gives 13x = 26, x = 2, and then y = 3. (Demo: elimination after multiplying)

Why is this allowed? Multiplying both sides of an equation by a number other than 0 keeps exactly the same solutions, so its line does not move. The sum of two equations is a new equation that is true wherever both are true, so its line passes through the same crossing. Choosing the multipliers so y cancels makes that new line vertical: it reads x = 4 directly.

Before: the lines x + y = 6 and x − y = 2 crossing at (4, 2). After adding the equations: the new equation 2x = 8 is the vertical line x = 4, drawn in green through the same crossing point
Adding the equations makes a new line through the same crossing; with y cancelled it is the vertical line x = 4.

One, none or infinitely many

Two lines in a plane either cross once, are parallel, or are the same line. The algebra says which: if both unknowns cancel, the statement left over decides it.

linesslopesthe algebra ends insolutions
crossdifferentx = 4, y = 2exactly one
parallelsame, different y-interceptsa false statement, 0 = −4none (inconsistent)
the same linesame, same y-intercepta true statement, 0 = 0 or 6 = 6infinitely many (dependent)

x − y = 1 and 2x − 2y = 6 both have slope 1, so they are parallel; doubling the first and subtracting gives 0 = −4 (Demo: no solution, parallel lines). x + y = 3 and 2x + 2y = 6 are the same line, and substitution ends in 6 = 6 (Demo: infinitely many, the same line).

Three small graphs. One solution: two lines crossing at one point. No solution: two parallel lines, the algebra ends in 0 = −4. Infinitely many: two equal lines on top of each other, the algebra ends in 0 = 0
Cross once, parallel, or the same line: one solution, none, or infinitely many.

Which method when

methodbest whenweak spot
graphingyou want the picture, or a rough answernot exact off the grid
substitutionone equation is already y = … or has a coefficient ±1fractions otherwise
eliminationboth equations are in ax + by = c formneeds the right multipliers

For three or more unknowns, elimination written as rows of numbers is Gaussian Elimination, which also shows what each row operation does to the lines. A line fitted to data points comes from a 2×2 system too: see Linear Regression. Solving a system is undoing a matrix, as in Linear Transformations. When one equation is a parabola, substitution gives a quadratic: see Quadratic Equations.

What the page leaves out

Only five preset systems with whole-number coefficients, two unknowns and no vertical lines are shown, and elimination always removes y. Cramer's rule (x = det₁/det), matrices, three equations in three unknowns (planes meeting at a point), nonlinear systems such as a line and a circle, and systems of inequalities (shaded regions) are only mentioned or left out.