Algebra

Systems of Equations: Substitution, Elimination, Graphing

One system, solved three ways, with the coefficient patterns that tell you in advance which method will be shortest.

By the Gauth editorial teamHow this content is produced and checkedUpdated 10 September 2026

The idea

A single linear equation in two unknowns has infinitely many solutions. Every point on the line $2 x + 3 y = 12$ works, so the equation on its own pins down nothing. A second equation cuts that infinite set down: you are now asking which points lie on both lines. Generically two distinct lines cross once, and that intersection is the solution.

Only two things can spoil this. If the lines are parallel they never meet and there is no solution. If the two equations secretly describe the same line — $2 x + 3 y = 12$ and $4 x + 6 y = 24$ are the same line written twice — then every point on it qualifies and there are infinitely many solutions. Those are the only three outcomes for a pair of straight lines, and recognising which one you are in is half of what the topic is about.

All three methods do the same underlying thing: use one equation to remove a variable from the other, leaving a single equation in one unknown that you already know how to solve. Substitution removes it by replacement, elimination removes it by cancellation, and graphing removes it by looking. The choice is about arithmetic cost, not correctness.

When you need it

The structural cue is two unknowns and two independent pieces of information. In word problems that usually reads as two different totals about the same two quantities:

A count and a value. Twelve coins worth £2.10 in total, made of 10p and 20p pieces: one equation counts coins, the other counts money.
A mixture. Two solutions at different concentrations combined to hit a target volume and a target strength.
Two tariffs compared. Two phone plans with different fixed fees and per-minute rates, asked to find the usage where the cost is identical.
Distance problems with a current or a wind, where the same craft has two different effective speeds.
Fitting a line through two known points, where the unknowns are the slope and the intercept rather than $x$ and $y$ .

The reliable signal that you are looking at a system rather than one equation is that the problem gives you two distinct sentences of fact and asks for two numbers. Getting from the English to those two equations is a separate skill, covered in the guide to turning words into equations.

The method

The interesting question is not how to execute each method — that is mechanical — but how to choose. Here is the reasoning, then the rule.

Take a general system and eliminate $x$ by hand to see what governs the outcome:

a_{1} x + b_{1} y a_{2} x + b_{2} y = c_{1} = c_{2}

Multiply the first equation by $a_{2}$ and the second by $a_{1}$ , so both have the same $x$ coefficient:

a_{1} a_{2} x + a_{2} b_{1} y a_{1} a_{2} x + a_{1} b_{2} y = a_{2} c_{1} = a_{1} c_{2}

Subtract the first from the second:

(a_{1} b_{2} - a_{2} b_{1}) y = a_{1} c_{2} - a_{2} c_{1}

Everything hinges on $a_{1} b_{2} - a_{2} b_{1}$ . If it is non-zero you can divide and there is exactly one $y$ , hence exactly one solution. If it is zero, the left side vanishes and you are left with a statement about constants only: false means no solution, true means infinitely many. That quantity is the determinant of the coefficient matrix, and it is the same number that decides whether a matrix can be inverted. Checking it on our system gives $2 (- 1) - 4 (3) = - 14$ , comfortably non-zero, so a unique solution exists before any solving begins.

The working rule:

Normalise both equations. Multiply out fractions and decimals, expand brackets, and put the variables in the same order on the same side. Comparing coefficients is impossible until the equations look alike.
Scan for a coefficient of 1 or −1. If one exists, isolating that variable introduces no fractions, and substitution is the cheapest route. This is the only condition that reliably favours substitution.
Otherwise scan for a variable whose coefficients cancel or divide neatly. Coefficients of $4$ and $- 4$ mean you can add immediately; $3$ and $6$ mean one doubling. Elimination then avoids fractions entirely.
Solve the resulting single-variable equation. When substituting, put the expression into the equation you did not rearrange, or you will produce a true but empty identity.
Back-substitute and write an ordered pair. A system in two unknowns has a two-part answer. Half an answer earns half the marks.
Check in both original equations. One substitution into each takes seconds and is the only check that catches a rearrangement error.

One system, three methods

Below is the same system solved by substitution, elimination, and graphing. Try to predict each step before revealing it — the point is to feel the difference in arithmetic cost between the three routes, not just to reach the answer three times.

2x + 3y = 12 and 4x − y = 10

0 of 11 steps

Try it yourself first — you'll remember it longer.

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First solution free

Three worked examples

The first is built for substitution, the second for elimination, and the third arrives buried in fractions and decimals so that most of the work happens before any method is chosen.

Worked example

y = 2 x - 1 and 3 x + y = 14

1
One equation is already solved for y. That is substitution's ideal case, and there is no reason to consider anything else.
$y = 2 x - 1$
2
Replace y in the second equation. The first equation has now been used, so it cannot give any further information.
$3 x + (2 x - 1) = 14$
3
Collect and solve.
$5 x - 1 = 14 ⟹ 5 x = 15 ⟹ x = 3$
4
Back-substitute into the isolated form.
$y = 2 (3) - 1 = 5$
5
Check both originals: 5 = 2(3) − 1 holds, and 3(3) + 5 = 14 holds.
$3 (3) + 5 = 14 ✓$

Answer

(x, y) = (3, 5)

Worked example

5 x + 4 y = 6 and 3 x - 4 y = 10

1
No coefficient is 1 or −1, so substitution would create fractions. But the y coefficients are already +4 and −4, which is elimination handed to you.
$+ 4 y and - 4 y$
2
Add the equations term by term. No scaling is needed at all.
$(5 x + 3 x) + (4 y - 4 y) = 6 + 10 ⟹ 8 x = 16$
3
Solve for x.
$x = 2$
4
Back-substitute into either original. The first is as easy as the second.
$5 (2) + 4 y = 6 ⟹ 4 y = - 4 ⟹ y = - 1$
5
Check the pair in the equation you did not use for back-substitution: 3(2) − 4(−1) = 6 + 4 = 10.
$3 (2) - 4 (- 1) = 10 ✓$

Answer

(x, y) = (2, - 1)

Worked example

\frac{x}{3} + \frac{y}{2} = 4 and 0.4 x - 0.3 y = 1.2

1
Neither equation is usable as written. Clear the first by multiplying every term by 6, the lowest common denominator of 3 and 2.
$6 \cdot \frac{x}{3} + 6 \cdot \frac{y}{2} = 6 \cdot 4 ⟹ 2 x + 3 y = 24$
2
Clear the second by multiplying every term by 10. Multiplying the right-hand side as well is essential — this is where most of the errors in this example live.
$4 x - 3 y = 12$
3
Now compare coefficients. The y terms are +3 and −3, so elimination by addition needs no scaling.
$(2 x + 3 y) + (4 x - 3 y) = 24 + 12 ⟹ 6 x = 36$
4
Solve for x, then back-substitute into the cleared version of the first equation.
$x = 6 ⟹ 2 (6) + 3 y = 24 ⟹ 3 y = 12 ⟹ y = 4$
5
Check against the original fractional and decimal forms, not the cleared ones. A third of six is 2, half of four is 2, and 2 + 2 = 4.
$\frac{6}{3} + \frac{4}{2} = 4 ✓ 0.4 (6) - 0.3 (4) = 2.4 - 1.2 = 1.2 ✓$

Answer

(x, y) = (6, 4)

Where people go wrong

Substituting back into the equation you rearranged

Rearranging $4 x - y = 10$ to $y = 4 x - 10$ and then substituting that into $4 x - y = 10$ gives $4 x - (4 x - 10) = 10$ , which simplifies to $10 = 10$ . True, and completely uninformative. The expression must go into the other equation, because that is the only place new information lives.

Scaling only part of an equation

To turn $4 x - y = 10$ into something with $- 3 y$ , you must multiply all three terms: $12 x - 3 y = 30$ . Multiplying the left side and forgetting the 30 changes what the equation says, and the resulting solution will fail the final check. If your answer does not verify, look here first.

Reading a vanished variable as an error

Solving $3 x - 2 y = 4$ with $6 x - 4 y = 9$ gives $8 = 9$ after doubling the first equation and subtracting. That is not a mistake to hunt down — it is the answer. The lines have identical slopes and different intercepts, so they are parallel and no solution exists. The mirror case, $0 = 0$ , means the equations coincide and every point on the line is a solution.

Trusting an intersection read off a sketch

Graphing is excellent for classification and unreliable for values. A hand sketch cannot distinguish an intersection at $(3, 2)$ from one at $(3.1, 1.95)$ , and a graphing calculator will happily return a rounded decimal where the exact answer was a fraction. Use the picture to decide how many solutions there are, then get the numbers algebraically.

Practice

Decide which method you would use before you start solving. Write answers as $x = \dots, y = \dots$ unless the system has no unique solution.

Four systems, four different best methods

Answers are checked here — nothing is sent anywhere.

1
$x + y = 7, x - y = 1$
2
$2 x + y = 11, y = x + 2$
3
$3 x - 2 y = 4, 6 x - 4 y = 9$
4
$4 x + 3 y = 10, 2 x - y = 0$

Where this goes next

Elimination does not stop at two equations. Applied repeatedly to a three- or four-variable system it becomes Gaussian elimination, and the bookkeeping is exactly what a matrix is for — the connection is made explicit in the two ways to invert a matrix, where the determinant that appeared in the derivation above returns as the test for whether a solution exists. When one of your equations is a curve rather than a line, substitution produces a quadratic and the quadratic formula finishes the job, usually with two intersection points instead of one. For a system you want checked line by line, the system of equations solver runs substitution and elimination on the same input so you can see which was shorter.

Frequently asked questions

Which method is fastest?

Substitution wins when one variable is already isolated or has a coefficient of 1 or -1, because isolating it costs nothing. Elimination wins when both equations are in standard form, and especially when a variable's coefficients are already opposites or easy multiples. Graphing is almost never the fastest route to an exact answer, but it is the fastest way to see whether a system has one solution, none, or infinitely many.

What does it mean when both variables disappear?

You have hit a degenerate case, and the surviving statement tells you which. If you are left with something false like 0 = 5, the lines are parallel and there is no solution. If you are left with something true like 0 = 0, the two equations describe the same line and every point on it is a solution. Neither is a mistake; both are legitimate answers that need to be stated as such.

Can I mix methods within one problem?

Yes, and for three-variable systems you usually should. A common route is to eliminate one variable from two different pairs of equations, which leaves a two-variable system, then finish that by substitution. Nothing requires you to commit to a single method for the whole problem.

How do I check a solution properly?

Substitute the pair into both original equations, not the rearranged ones. If you made an algebra slip while rearranging, checking against your own rearrangement will confirm the error rather than catch it. A correct solution satisfies every equation you were given.

What changes when one of the equations is not a line?

Substitution becomes the default, because eliminating a variable between a line and a curve rarely cancels anything. Substituting the linear equation into the quadratic one produces a single quadratic in one variable, which you then solve as usual. Expect up to two solutions rather than one, and check both against the original equations since squaring or clearing denominators can introduce extraneous roots.

Keep going

System of equations solverSolves by substitution and elimination side by side, with every step shown.Turning words into equationsGetting from two sentences of English to the two equations you need.Inverting a matrix two waysWhere elimination goes once the systems get bigger than two by two.Algebra solverLinear equations, quadratics, systems, and inequalities.