Solving Systems by Substitution

Learning goals

Replace a variable with an equal expression to leave one unknown
Choose a coefficient of $1$ to avoid fractions
Substitute into the other equation, then back-substitute
Read the pair as the crossing point of two lines
Interpret a cancelled variable as parallel lines or one line
Check both coordinates in both original equations

Turning two unknowns into one

A system of two linear equations presents two equations that must hold at the same time, and asks for the pair $(x, y)$ that satisfies both. In the last lesson you saw the picture: each equation draws a line, and the one pair solving both sits exactly where the two lines cross. Reading that crossing point off a graph is quick to picture but hard to do exactly, especially when the coordinates are fractions. Substitution finds it precisely, and it needs nothing beyond the equation solving from the Linear Equations chapter.

The method rests on a single principle you have used since your very first equation: equals may be replaced by equals. If one equation tells you that $y = 7 - 2x$ , then $y$ and $7 - 2x$ are two names for the same number. So anywhere $y$ appears you may write $7 - 2x$ instead without changing what is true.

Watch it clear an entire variable out of the way. Take

y = x + 1, \qquad 2x + y = 7.

The first equation says $y$ is the same as $x + 1$ . Replace $y$ in the second equation with $x + 1$ :

2x + (x + 1) = 7.

The $y$ is gone. What remains has only $x$ in it, and you already know how to finish:

3x + 1 = 7, \qquad 3x = 6, \qquad x = 2.

Then recover $y$ from the first equation: $y = x + 1 = 3$ . The solution is $(2, 3)$ , the very point where these two lines cross.

The substitution method step by step

The example above followed a fixed routine, and every system yields to the same five moves.

Solve one equation for one variable. Pick a variable that is easy to isolate, ideally one whose coefficient is $1$ or $-1$ so no fractions appear. This gives an expression for that variable in terms of the other.
Substitute into the other equation. Replace that variable, in the equation you did not use in step 1, with the expression you found. The result is a single equation in one variable.
Solve the one-variable equation. Use the ordinary methods from the Linear Equations chapter.
Back-substitute. Put that value into the expression from step 1 to get the second variable.
Check. Substitute the pair into both original equations. A genuine solution satisfies each one.

Steps 2 and 5 are the two most often skipped. Substituting into the other equation is what makes progress, and checking in both originals is what catches a slip.

Worked example 1 Solve $x - 2y = 1$ and $3x + y = 17$

Neither equation is solved for a variable yet, so start with step 1. In $x - 2y = 1$ the variable $x$ has coefficient $1$ , so isolating it is clean. Add $2y$ to both sides:

x = 1 + 2y.

Substitute this expression for $x$ into the other equation, $3x + y = 17$ :

3(1 + 2y) + y = 17.

Now there is only $y$ to find. Distribute and collect:

3 + 6y + y = 17, \qquad 3 + 7y = 17, \qquad 7y = 14, \qquad y = 2.

Back-substitute $y = 2$ into $x = 1 + 2y$ :

x = 1 + 2(2) = 5.

So the candidate solution is $(5, 2)$ . Check it in both original equations, not just the one you solved:

x - 2y = 5 - 2(2) = 1 \ \checkmark, \qquad 3x + y = 3(5) + 2 = 17 \ \checkmark.

Both hold, so $(5, 2)$ is the solution.

Check your understanding

In the system $y = x - 2$ and $3x + y = 10$ , what do you get right after substituting for $y$ in the second equation?

3x + (x - 2) = 10

3x + y = 10

x - 2 = 10

3(x - 2) + y = 10

Why substitution works

Substitution can feel almost too easy, so it is worth seeing exactly why the pair it produces is guaranteed to be the solution of the system. The argument below also shows that no solutions are gained and none are lost.

Why replacing a variable preserves the solution set#

A system of two equations asks for every pair $(x, y)$ that makes both equations true at once, and that collection of pairs is the solution set. Substitution swaps the system for a smaller problem, so we must confirm it keeps this set exactly.

Suppose the first equation can be solved for $y$ , giving $y = E$ , where $E$ is some expression built from $x$ (for instance $E = x + 1$ ). Solving an equation does not change which pairs satisfy it. Every step is reversible: adding the same amount to both sides can be undone, and so can dividing both sides by a nonzero number. So the pairs satisfying the first equation are precisely the pairs for which $y = E$ . A pair lies on the first line exactly when its $y$ -coordinate equals $E$ .

Now look at one such pair, a pair with $y = E$ . For it, the symbol $y$ and the expression $E$ stand for the same number. Substituting $E$ in place of $y$ in the second equation therefore cannot change whether that pair satisfies it, because we are replacing a quantity by an equal quantity. The second equation written with $y$ and the second equation with $E$ in place of $y$ are true for exactly the same pairs of this kind.

Put the two observations together. A pair solves the system exactly when it satisfies the first equation and the second. The first condition says $y = E$ . Granting that, the second condition becomes the substituted equation, which contains only $x$ . So the pairs solving the system are exactly the pairs whose $x$ solves the substituted one-variable equation and whose $y$ equals $E$ . Substitution loses nothing and invents nothing: it trades the system for an equivalent single equation. Every solution of that equation extends to one solution of the system. And every solution of the system arises this way, from exactly one solution of that equation.

This also explains what the graph does. When the substituted equation fixes one value of $x$ , the two lines meet at one point. When it collapses to a false statement such as $1 = 4$ , no $x$ can satisfy it, so the lines share no point and run parallel. When it collapses to a statement that is always true such as $4 = 4$ , every $x$ on the first line works, so the two equations describe the same line.

Choosing what to solve for

Every system yields to substitution, but a smart first choice saves you from fractions. Scan the four coefficients and look for a variable whose coefficient is $1$ or $-1$ . Solving for that variable never introduces a denominator, because isolating it takes only additions and subtractions, never a division. If no coefficient is $1$ or $-1$ , any choice works; you will just carry a fraction through the arithmetic.

Worked example 2 Solve $3x + y = 6$ and $5x + 2y = 11$

Look for the easiest variable to isolate. In $3x + y = 6$ the variable $y$ has coefficient $1$ , so solve that equation for $y$ :

y = 6 - 3x.

Substitute into the other equation, $5x + 2y = 11$ :

5x + 2(6 - 3x) = 11.

Distribute and collect the $x$ terms:

5x + 12 - 6x = 11, \qquad -x + 12 = 11, \qquad -x = -1, \qquad x = 1.

Back-substitute $x = 1$ into $y = 6 - 3x$ :

y = 6 - 3(1) = 3.

The solution is $(1, 3)$ . A quick check confirms it: $3(1) + 3 = 6$ and $5(1) + 2(3) = 11$ , both true. Had we instead solved $5x + 2y = 11$ for $x$ , we would have divided by $5$ and dragged a fraction through every line. Choosing the coefficient of $1$ kept the numbers whole.

Check your understanding

Which first move avoids fractions when solving $3x + 2y = 12$ and $x - 4y = 5$ by substitution?

A Solve

3x + 2y = 12

for

y

B Solve

x - 4y = 5

for

x

C Solve

3x + 2y = 12

for

x

D Solve

x - 4y = 5

for

y

When the variable disappears: no solution and infinitely many

Usually the substituted equation pins $x$ to a single value. Once in a while the variable terms cancel completely and you are left with a bare numerical statement instead. That is not a dead end; it is the system telling you how the two lines lie.

Worked example 3 A system with no solution

Solve

y = 2x + 1, \qquad y = 2x + 4.

Both equations are already solved for $y$ , so set the two expressions for $y$ equal (each equals $y$ , so they equal each other):

2x + 1 = 2x + 4.

Subtract $2x$ from both sides, and every $x$ vanishes:

1 = 4.

This is false: no value of $x$ can make $1$ equal $4$ . So there is no pair $(x, y)$ satisfying both equations, and the system has no solution. Geometrically, both lines climb $2$ units for every step right but start at different heights, so they are parallel and never cross.

Worked example 4 A system with infinitely many solutions

Solve

y = 3x - 2, \qquad 6x - 2y = 4.

Substitute $y = 3x - 2$ into the second equation:

6x - 2(3x - 2) = 4.

Distribute carefully, watching the sign on the $-2$ :

6x - 6x + 4 = 4, \qquad 4 = 4.

Every $x$ has cancelled and the statement left over is always true. That means every pair on the first line already satisfies the second, so the system has infinitely many solutions. In fact the second equation is the first one in disguise: rearranging $y = 3x - 2$ gives $3x - y = 2$ , and doubling that gives $6x - 2y = 4$ . The two equations describe the same line, so its every point, written $(x,\, 3x - 2)$ , is a solution.

Check your understanding

Solving a system by substitution, you arrive at $0 = 7$ . What does this tell you?

A The system has no solution. B The system has exactly one solution,

x = 0

. C The system has infinitely many solutions. D You must have erred, since every system has a solution.

Seeing the answer as a crossing point

Substitution never draws a picture, but it is worth remembering that the pair it returns is exactly the point the two lines share. Below are the lines from the opening example, $y = x + 1$ and $2x + y = 7$ . Substitution replaced $y$ in the second equation and returned $(2, 3)$ , and that is precisely where the two lines cross.

The system y = x + 1 and 2x + y = 7 drawn as two lines. Substitution replaces y in the second equation, and the pair it returns, (2, 3), is exactly the point where the two lines cross.

The algebra and the picture agree, which is the point. Substitution is just a way to compute the crossing point without the guesswork of reading it off a grid. The method also stays exact even when the answer is a fraction that no drawing could pin down.

The figure below opens on that same system, $y = x + 1$ as the solid line and $y = -2x + 7$ as the dashed one. The figure reports the crossing point, when there is one, as you move either line. Substitution and the picture are answering the same question, so the coordinates under the figure should always be the pair substitution would return.

Use it to make the two awkward cases from earlier in this lesson concrete, because they are the ones that look like nothing on paper. Give the dashed line the same slope as the solid one, a rise of $1$ over a run of $1$ . Now the crossing point disappears, and that vanishing is what a contradiction like $7 = 1$ is telling you. Then slide the dashed line’s crossing point down onto the solid one’s. Nothing is left but a single visible line, and the sentence changes to say the two equations describe the same line, which is the identity case. The two states differ in one control and nothing else, and they look almost identical, which is exactly why the algebra distinguishes them and your eyes cannot.

The crossing point substitution computes

Rise A Run A Crosses A Rise B Run B Crosses B

y = x + 1. y = -2x + 7. The slopes 1 and -2 are different, so they cross exactly once. They cross at (2, 3).

Two lines on one coordinate plane, one solid and one dashed. When they share a point a dot marks it and the sentence below the figure names its coordinates exactly, as a fraction when the crossing does not land on a grid point. When they are parallel, or are two ways of writing one line, there is no dot and the sentence says which of those it is.

Common mistakes

Stopping after one variable. Finding $x = 2$ is only half the answer. Back-substitute to get $y$ , then state the full pair $(x, y)$ and check it. A system’s solution is a pair, not a single number.
Substituting back into the same equation. If you solve the first equation for $y$ and then plug that expression into the first equation again, every term cancels. You are left with something like $6 = 6$ , true but useless. Always substitute into the other equation.
Dropping the parentheses. When you replace $y$ with $2x + 1$ in $3x - y = 5$ , wrap it up: $3x - (2x + 1) = 5$ . The minus sign must reach both terms, giving $3x - 2x - 1 = 5$ . Without the parentheses you would wrongly write $3x - 2x + 1$ and flip a sign.
Sign slips when isolating a variable. Solving $2x + y = 6$ for $y$ gives $y = 6 - 2x$ , not $y = 6 + 2x$ . A term that crosses the equals sign changes sign.
Misreading a vanished variable. If substitution ends in $0 = 5$ , the answer is “no solution”, and if it ends in $0 = 0$ , the answer is “infinitely many solutions”. Neither one means $x = 0$ .

Key takeaways

Substitution turns a two-variable system into a one-variable equation by replacing a variable with an equal expression, resting on the rule that equals may replace equals.
The steps: solve one equation for a variable (choose a coefficient of $1$ or $-1$ to avoid fractions). Then substitute into the other equation, solve, back-substitute, and check the pair in both originals.
The pair you find is the exact point where the two lines cross, obtained without any graphing.
If the variable cancels and leaves a false statement like $1 = 4$ , the lines are parallel and there is no solution. If it leaves an always-true statement like $4 = 4$ , the equations are the same line and there are infinitely many solutions.
Never call a system done until you have both coordinates and have verified them in each equation.

Solving Systems by Substitution

Learning goals

Turning two unknowns into one

The substitution method step by step

Why substitution works

Choosing what to solve for

When the variable disappears: no solution and infinitely many

Seeing the answer as a crossing point

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)