What is a system of equations?

It is two (or more) equations considered at the same time, sharing the same unknowns. The solution of a system in two unknowns is a pair of numbers satisfying every equation separately — one equation alone is not enough to pin down two numbers.

How does the substitution method work?

Take one equation and express one unknown from it, say y, then put that expression in place of y in the other equation. What is left is a single equation in a single unknown, solved like any linear equation; substituting the number back gives the second one.

How does the elimination method work?

Multiply the equations by numbers chosen so that the coefficients of one unknown become opposite, then add the equations side by side. That unknown disappears and a single-unknown equation is left. It is handier than substitution whenever no unknown carries a coefficient of 1.

Which method should I choose?

If some equation carries an unknown with coefficient 1 or −1, substitution is quicker, because expressing it introduces no fractions. Otherwise elimination is usually more comfortable. Both give the same result, so the choice is purely about arithmetic convenience.

How do you check the solution of a system?

Substitute the pair you found into both equations and evaluate both sides. The solution is correct only when both equations hold — a pair fitting just one of them is not a solution of the system.

When does a system have no solutions, and when infinitely many?

If both unknowns vanish during the work and a false statement is left, such as 0 = 6, the system is inconsistent and has no solutions — on a graph, two parallel lines. If a true statement is left, such as 0 = 0, the system is dependent and infinitely many pairs satisfy it — the two lines coincide.

What does the solution of a system mean on a graph?

Every linear equation in two unknowns describes a line. The solution of the system is the point the two lines share, so its coordinates are the pair you are looking for. Crossing lines give one solution, parallel lines none, and coinciding lines infinitely many.

Advanced level

Systems of equations

A system of equations puts two conditions on two unknowns at once. Learn substitution and elimination, see the solution as the point where two lines cross, and learn to recognise inconsistent and dependent systems.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Linear equationsA linear equation is an equality with the unknown in the first power. Learn the operations that keep equations equivalent, solve one step by step, check the result, and learn to spot contradictory and identity equations.

All formulas

A system of two linear equations
$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$
the solution is a pair (x, y) satisfying both equations
Condition for a determinate system
$a_1 b_2 - a_2 b_1 \neq 0$
exactly one solution — the lines cross at a single point
Substitution method
$y = \frac{c_1 - a_1 x}{b_1} \;\longrightarrow\; a_2 x + b_2 \cdot \frac{c_1 - a_1 x}{b_1} = c_2$
express one unknown and put it into the other equation
Elimination method
$m \cdot (a_1 x + b_1 y) + n \cdot (a_2 x + b_2 y) = m c_1 + n c_2$
choose m and n so that one unknown cancels out
Determinant formulas
$W = a_1 b_2 - a_2 b_1, \quad x = \frac{W_x}{W}, \quad y = \frac{W_y}{W}$
W_x and W_y come from W by replacing the matching column with the constant terms
Inconsistent system
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
parallel lines — no solutions; when all three ratios are equal the system is dependent

A single equation in two unknowns has infinitely many solutions: $x + y = 5$ is satisfied by $(1, 4)$ , by $(2, 3)$ and by $(10, -5)$ alike. To pin down two numbers you need two conditions at once. That is a system of equations:

\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}

The solution of a system is a pair $(x, y)$ satisfying both equations simultaneously. A pair fitting only one of them is not a solution — and that is the whole difference between a system and two separate equations.

The substitution method

The idea is simple: express one unknown from one equation and put it into the other. What is left is a single equation in a single unknown — a task we already know how to finish.

Solve the system: x + y = 5 and 2x − y = 4.

Watch the bracket while substituting: the whole expression goes in, so a minus in front of it flips the sign of every term inside. This is where most signs get lost.

The elimination method

When no unknown carries a coefficient of 1, expressing one introduces fractions. It is then easier to multiply the equations so that the coefficients of one unknown become opposite, and add the equations side by side — that unknown then cancels itself out.

Solve the system: 3x + 2y = 12 and 5x − 2y = 4.

If the coefficients are not opposite to begin with, multiply the equations by suitable numbers — in the system $2x + 3y = 8$ and $3x - 4y = -5$ , multiplying the first by $3$ and the second by $-2$ turns the coefficients of $x$ into $6$ and $-6$ .

Two lines on one graph

Every linear equation in two unknowns describes a line. So a system asks for the point lying on both lines at once:

The solution of the system is the coordinates of the crossing point: x = 3, y = 2.

That picture explains all three possibilities at once. The lines may cross at one point (one solution), be parallel (none), or coincide (infinitely many).

Determinate, inconsistent and dependent systems

Which case you are in is decided by the coefficients alone:

a_1 b_2 - a_2 b_1 \neq 0 \quad \text{— a determinate system, exactly one solution}

Solve the systems: x + y = 4 and 2x + 2y = 10, then x + y = 4 and 2x + 2y = 8.

A practical hint: if both unknowns have vanished during the work, that is not a mistake — that is the answer. All that is left is to read the statement you are holding.

A system as a model

The power of systems comes from problems where two numbers are wanted at once. Two full tickets and three reduced ones cost $96$ zł, while one full and two reduced ones cost $54$ zł. What does each ticket cost?

\begin{cases} 2n + 3u = 96 \\ n + 2u = 54 \end{cases}

From the second equation $n = 54 - 2u$ , so:

2(54 - 2u) + 3u = 96

108 - u = 96 \quad \big| -108

-u = -12 \quad \big| \cdot (-1)

u = 12

Hence $n = 54 - 2 \cdot 12 = 30$ . A full ticket costs $30$ zł and a reduced one $12$ zł. The hardest step is again not the arithmetic — it is writing both conditions down.

In the exercises below, enter both numbers separated by a comma, as $x = 2, y = 3$ (a bare $2, 3$ is accepted too).

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

x + 3y = 14; x + y = 8

Common mistakes

Substituting without a bracket — in $2x - (5 - x)$ the minus applies to both terms.
Checking in one equation only — the pair must fit both, otherwise it does not solve the system.
Adding equations whose coefficients are not opposite — multiply first, add afterwards.
Giving $x$ alone — the solution of a system is a pair of numbers, not one of them.
Reading vanished unknowns as an error — it is the signal of an inconsistent or dependent system.

Formula card

Topic: Systems of equations

A system of two linear equations
$\begin{cases} a_1x + b_1y = c_1 \\ a_2x + b_2y = c_2 \end{cases}$
the solution is a pair (x, y) satisfying both equations
Condition for a determinate system
$a_1 b_2 - a_2 b_1 \neq 0$
exactly one solution — the lines cross at a single point
Substitution method
$y = \frac{c_1 - a_1 x}{b_1} \;\longrightarrow\; a_2 x + b_2 \cdot \frac{c_1 - a_1 x}{b_1} = c_2$
express one unknown and put it into the other equation
Elimination method
$m \cdot (a_1 x + b_1 y) + n \cdot (a_2 x + b_2 y) = m c_1 + n c_2$
choose m and n so that one unknown cancels out
Determinant formulas
$W = a_1 b_2 - a_2 b_1, \quad x = \frac{W_x}{W}, \quad y = \frac{W_y}{W}$
W_x and W_y come from W by replacing the matching column with the constant terms
Inconsistent system
$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
parallel lines — no solutions; when all three ratios are equal the system is dependent

The solution of x + y = 5 and 2x − y = 4 is the point where the two lines cross: (3, 2).

Frequently asked questions

A balance scale in perfect equilibrium rendered as a geometric form, with a straight line crossing an axis like an orbital path, against a cosmic nebula in violet and magenta

Algebra

What x really means — unknown, variable and parameter

The letter x does not always mean the same thing. Sometimes it is a number you are hunting for, sometimes a quantity that varies continuously, sometimes a blank waiting for any number at all. Here is where algebraic notation came from, why it took two thousand years to arrive, what exactly separates an unknown from a variable and a parameter, and why an inequality flips when you multiply by a negative number.

11 min