Solving Simultaneous Equations by Substitution

Solve a pair of linear equations exactly by making one variable the subject of one equation and substituting it into the other, and check the answer in both equations.

Before you start

A graph gives the solution of simultaneous equations only as an estimate when the lines meet between grid lines (lesson 2.5). Algebra gives it exactly.
Expanding with a negative number in front: $-2(2x - 3) = -4x + 6\text{.}$ Every term in the bracket is multiplied by $-2$ .
Solving a linear equation: $3x - 2(2x - 3) = 1$ gives $3x - 4x + 6 = 1$ , so $-x = -5$ and $x = 5$ .

The Substitution Method

Two equations in $x$ and $y$ are hard to solve at once. If one equation tells you what $y$ is in terms of $x$ , you can put that expression in place of $y$ in the other equation. What is left has only $x$ in it, and you know how to solve that.

Number the equations (1) and (2) so that each step can say which equation it uses.

Solving by substitution

(1) Make one variable the subject of one equation. Choose a variable whose coefficient is $1$ or $-1$ if there is one.

(2) Substitute that expression, in brackets, into the other equation.

(3) Solve the equation in one unknown.

(4) Substitute the value back to find the other unknown.

(5) Check both values in both original equations.

Example 1: One equation already gives y

Solve the simultaneous equations $y = 2x + 1$ and $3x + y = 11$ .

Solution

$y = 2x + 1$ $\quad$ (1)

$3x + y = 11$ $\quad$ (2)

Equation (1) already has $y$ as the subject. Substitute $2x + 1$ for $y$ in (2):

$3x + (2x + 1) = 11$

$5x + 1 = 11$

$5x = 10$ , so $x = 2$ .

Substitute $x = 2$ into (1): $y = 2(2) + 1 = 5$ .

Check in (1): $2(2) + 1 = 5$ , true. Check in (2): $3(2) + 5 = 11$ , true.

On a graph, the lines $y = 2x + 1$ and $3x + y = 11$ meet at $(2, 5)$ , the same answer:

$x = 2$ , $y = 5$

Making a Variable the Subject First

Usually neither equation gives $x$ or $y$ on its own. Look for a variable whose coefficient is $1$ or $-1$ : making it the subject needs no fractions.

Example 2: A coefficient of 1

Solve $x + 2y = 7$ and $3x - y = 7$ .

Solution

$x + 2y = 7$ $\quad$ (1)

$3x - y = 7$ $\quad$ (2)

In (1), $x$ has coefficient $1$ . Subtract $2y$ from both sides: $x = 7 - 2y$ $\quad$ (3)

Substitute (3) into (2): $3(7 - 2y) - y = 7$

$21 - 6y - y = 7$

$21 - 7y = 7$

$-7y = -14$ , so $y = 2$ .

Substitute $y = 2$ into (3): $x = 7 - 2(2) = 3$ .

Check in (1): $3 + 2(2) = 7$ , true. Check in (2): $3(3) - 2 = 7$ , true.

You could instead make $y$ the subject of (2): $y = 3x - 7$ . Then $x + 2(3x - 7) = 7$ , $7x - 14 = 7$ , $x = 3$ , $y = 2$ : the same answer.

$x = 3$ , $y = 2$

Example 3: A coefficient of −1, and a minus sign in front of the bracket

Solve $2x - y = 3$ and $3x - 2y = 1$ .

Solution

$2x - y = 3$ $\quad$ (1)

$3x - 2y = 1$ $\quad$ (2)

In (1), $y$ has coefficient $-1$ . Add $y$ to both sides and subtract $3$ : $y = 2x - 3$ $\quad$ (3)

Substitute (3) into (2), keeping the brackets: $3x - 2(2x - 3) = 1$

The $-2$ multiplies both terms in the bracket: $-2 \times 2x = -4x$ and $-2 \times (-3) = +6$ .

$3x - 4x + 6 = 1$

$-x + 6 = 1$

$-x = -5$ , so $x = 5$ .

Substitute $x = 5$ into (3): $y = 2(5) - 3 = 7$ .

Check in (1): $2(5) - 7 = 3$ , true. Check in (2): $3(5) - 2(7) = 15 - 14 = 1\text{,}$ true.

$x = 5$ , $y = 7$

When Both Equations Give y

If both equations have $y$ as the subject, the two expressions for $y$ must be equal at the solution. Set them equal: this is the same as the meeting point of two graphs in lesson 2.5.

Example 4: Setting the two expressions equal

Solve $y = 3x - 4$ and $y = 6 - 2x$ .

Solution

$y = 3x - 4$ $\quad$ (1)

$y = 6 - 2x$ $\quad$ (2)

Both equal $y$ , so $3x - 4 = 6 - 2x$ .

Add $2x$ to both sides: $5x - 4 = 6$ . Add $4$ : $5x = 10$ , so $x = 2$ .

Substitute into (1): $y = 3(2) - 4 = 2$ .

Check in (1): $3(2) - 4 = 2$ , true. Check in (2): $6 - 2(2) = 2$ , true.

$x = 2$ , $y = 2$

Negative and Fractional Answers

The method does not change when the answers are negative or fractions. A graph can only estimate a value that falls between grid lines, such as $\frac{1}{3}$ ; algebra gives every value exactly.

Example 5: x = −1 and y = ½

Solve $2x + 4y = 0$ and $3x - 2y = -4$ .

Solution

$2x + 4y = 0$ $\quad$ (1)

$3x - 2y = -4$ $\quad$ (2)

From (1): $2x = -4y$ , so $x = -2y$ $\quad$ (3)

Substitute (3) into (2): $3(-2y) - 2y = -4$

$-6y - 2y = -4$

$-8y = -4$ , so $y = \frac{-4}{-8} = \frac{1}{2}\text{.}$

Substitute into (3): $x = -2 \times \frac{1}{2} = -1\text{.}$

Check in (1): $2(-1) + 4(\frac{1}{2}) = -2 + 2 = 0\text{,}$ true. Check in (2): $3(-1) - 2(\frac{1}{2}) = -3 - 1 = -4\text{,}$ true.

$x = -1$ , $y = \frac{1}{2}$

Why Check in Both Original Equations?

A check must use the original equations

The value found in step (3) always fits the equation you substituted into, and step (4) always fits the rearranged equation, even if the rearranging was wrong. Only the equation you rearranged, in the form it was first given, can catch a mistake made while rearranging. So check in both equations as they were given.

Example 6: A check that catches a mistake

Solve $x + 3y = 10$ and $2x - y = 6$ . Then see what happens if $y$ is made the subject of the second equation wrongly, as $y = 6 - 2x$ .

Solution

$x + 3y = 10$ $\quad$ (1)

$2x - y = 6$ $\quad$ (2)

Correct working. From (2), add $y$ and subtract $6$ : $y = 2x - 6$ . Substitute into (1): $x + 3(2x - 6) = 10$ , so $7x - 18 = 10$ , $7x = 28$ , $x = 4$ . Then $y = 2(4) - 6 = 2$ .

Check in (1): $4 + 3(2) = 10$ , true. Check in (2): $2(4) - 2 = 6$ , true.

With the mistake. $y = 6 - 2x$ into (1): $x + 3(6 - 2x) = 10$ , so $x + 18 - 6x = 10$ , $-5x = -8$ , $x = 1.6$ . Then $y = 6 - 2(1.6) = 2.8\text{.}$

This pair fits the wrong equation $y = 6 - 2x$ (since $6 - 3.2 = 2.8$ ), and it fits (1) (since $1.6 + 8.4 = 10$ ). But in (2), $2(1.6) - 2.8 = 0.4$ , not $6$ . Only the check in (2) shows that something is wrong.

$x = 4$ , $y = 2$ . The wrong pair $(1.6, 2.8)$ is caught only by checking it in the original equation (2).

When to Use Substitution

Substitution is a good choice when one equation already gives $x$ or $y$ , or when a variable has coefficient $1$ or $-1$ , or becomes so after dividing every term by the same number.

When every coefficient is $2$ or more and no division helps, substitution leads to fractions. It still works, but the elimination method of the next lesson is usually easier.

Example 7: Choosing well

(a) Solve $5x + 3y = 7$ and $4x - 2y = 10$ . (b) Solve $5x + 3y = 7$ and $3x + 2y = 5$ by substitution, and say why it is awkward.

Solution

(a) $5x + 3y = 7$ $\quad$ (1), $\quad$ $4x - 2y = 10$ $\quad$ (2)

Divide every term of (2) by $2$ : $2x - y = 5$ , so $y = 2x - 5$ $\quad$ (3)

Substitute into (1): $5x + 3(2x - 5) = 7$ , so $11x - 15 = 7$ , $11x = 22$ , $x = 2$ . Then $y = 2(2) - 5 = -1$ .

Check in (1): $10 - 3 = 7$ , true. Check in (2): $8 - 2(-1) = 8 + 2 = 10\text{,}$ true.

(b) $5x + 3y = 7$ $\quad$ (1), $\quad$ $3x + 2y = 5$ $\quad$ (2)

No coefficient is $1$ or $-1$ , and no division helps. From (1): $y = \frac{7 - 5x}{3}$ . Substitute into (2): $3x + \frac{2(7 - 5x)}{3} = 5\text{.}$

Multiply every term by $3$ : $9x + 2(7 - 5x) = 15$ , so $9x + 14 - 10x = 15$ , $-x = 1$ , $x = -1$ . Then $y = \frac{7 - 5(-1)}{3} = \frac{12}{3} = 4\text{.}$

Check in (1): $-5 + 12 = 7$ , true. Check in (2): $-3 + 8 = 5$ , true.

It works, but it needed a fraction and an extra step to clear it. Lesson 2.7 solves pairs like this more easily.

(a) $x = 2$ , $y = -1$ (b) $x = -1$ , $y = 4$ ; substitution was awkward because no variable had coefficient $1$ or $-1$ .

Common Mistakes

Leaving out the brackets. $3x - 2(2x - 3)$ , not $3x - 2 \times 2x - 3$ . Without brackets the $-2$ never reaches the $-3$ .
Losing a sign when making the subject. From $2x - y = 6$ , $y = 2x - 6$ , not $6 - 2x$ .
Substituting into the same equation you rearranged. That only gives a statement that is always true, such as $7 = 7$ , and no value. Substitute into the other equation. (In lesson 2.9, a statement like $0 = 0$ that comes from the two different equations means something else.)
Stopping after one value. The solution is a pair: find $y$ as well as $x$ .
Checking only in the rearranged equation, which always fits (Example 6).

Practice Questions

Solve each pair by substitution on paper before you open its solution, and check in both equations. The questions get harder as you go.

Question 1

Solve (a) $y = x + 4$ and $2x + y = 13$ , (b) $y = 3x$ and $2x + y = 10$ .

ShowHide solution

(a) $2x + (x + 4) = 13$ , so $3x + 4 = 13$ , $3x = 9$ , $x = 3$ . Then $y = 3 + 4 = 7$ . Check: $7 = 3 + 4$ ; $6 + 7 = 13$ .

(b) $2x + 3x = 10$ , so $5x = 10$ , $x = 2$ . Then $y = 3(2) = 6$ . Check: $6 = 3 \times 2$ ; $4 + 6 = 10$ .

(a) $x = 3$ , $y = 7$ (b) $x = 2$ , $y = 6$

Question 2

Solve $x = y + 3$ and $2x + 3y = 16$ .

ShowHide solution

Substitute $y + 3$ for $x$ : $2(y + 3) + 3y = 16$ , so $2y + 6 + 3y = 16$ , $5y = 10$ , $y = 2$ .

Then $x = 2 + 3 = 5$ .

Check: $5 = 2 + 3$ ; $2(5) + 3(2) = 10 + 6 = 16\text{.}$

$x = 5$ , $y = 2$

Question 3

Solve $x + y = 7$ and $3x + 2y = 18$ .

ShowHide solution

From the first: $x = 7 - y$ . Substitute: $3(7 - y) + 2y = 18$ , so $21 - 3y + 2y = 18$ , $21 - y = 18$ , $y = 3$ .

Then $x = 7 - 3 = 4$ .

Check: $4 + 3 = 7$ ; $12 + 6 = 18$ .

$x = 4$ , $y = 3$

Question 4

Solve $4x - y = 14$ and $3x + 2y = 5$ .

ShowHide solution

From the first: $y = 4x - 14$ . Substitute: $3x + 2(4x - 14) = 5$ , so $11x - 28 = 5$ , $11x = 33$ , $x = 3$ .

Then $y = 12 - 14 = -2$ .

Check: $12 - (-2) = 14$ ; $9 + 2(-2) = 5$ .

$x = 3$ , $y = -2$

Question 5

Solve $3x - y = 5$ and $5x - 3y = 3$ .

ShowHide solution

From the first: $y = 3x - 5$ . Substitute: $5x - 3(3x - 5) = 3$ .

The $-3$ multiplies both terms: $5x - 9x + 15 = 3$ , so $-4x = -12$ , $x = 3$ .

Then $y = 9 - 5 = 4$ .

Check: $9 - 4 = 5$ ; $15 - 12 = 3$ .

$x = 3$ , $y = 4$

Question 6

Solve $y = 4x - 3$ and $y = x + 6$ .

ShowHide solution

Set the two expressions for $y$ equal: $4x - 3 = x + 6$ , so $3x = 9$ , $x = 3$ .

Then $y = 3 + 6 = 9$ .

Check: $4(3) - 3 = 9$ ; $3 + 6 = 9$ .

$x = 3$ , $y = 9$

Question 7

Solve $x + 2y = 1$ and $4x - 2y = -11$ .

ShowHide solution

From the first: $x = 1 - 2y$ . Substitute: $4(1 - 2y) - 2y = -11\text{,}$ so $4 - 8y - 2y = -11$ , $-10y = -15$ , $y = 1.5$ .

Then $x = 1 - 2(1.5) = -2\text{.}$

Check: $-2 + 3 = 1$ ; $4(-2) - 2(1.5) = -8 - 3 = -11\text{.}$

$x = -2$ , $y = 1\frac{1}{2}$

Question 8

Solve $x - 2y = -6$ and $3x + y = 3$ .

ShowHide solution

From the first: $x = 2y - 6$ . Substitute: $3(2y - 6) + y = 3$ , so $7y - 18 = 3$ , $7y = 21$ , $y = 3$ .

Then $x = 2(3) - 6 = 0$ .

Check: $0 - 6 = -6$ ; $0 + 3 = 3$ .

$x = 0$ , $y = 3$

Question 9

Wei Jie solved $x + 2y = 8$ and $3x - y = 3$ . He wrote $y = 3 - 3x$ from the second equation and got $x = -0.4$ , $y = 4.2$ . He checked in $x + 2y = 8$ and it worked. Find his mistake and solve the equations correctly.

ShowHide solution

His check: $-0.4 + 2(4.2) = -0.4 + 8.4 = 8\text{.}$ It fits, because he substituted into that equation. In $3x - y = 3$ : $3(-0.4) - 4.2 = -1.2 - 4.2 = -5.4\text{,}$ not $3$ .

His mistake: from $3x - y = 3$ , adding $y$ and subtracting $3$ gives $y = 3x - 3$ , not $3 - 3x$ .

Correctly: $x + 2(3x - 3) = 8$ , so $7x - 6 = 8$ , $7x = 14$ , $x = 2$ . Then $y = 6 - 3 = 3$ .

Check: $2 + 6 = 8$ ; $6 - 3 = 3$ .

He should have written $y = 3x - 3$ . The solution is $x = 2$ , $y = 3$ .

Question 10

Solve $6x + 2y = 8$ and $2x - 3y = 10$ .

ShowHide solution

Divide every term of the first by $2$ : $3x + y = 4$ , so $y = 4 - 3x$ .

Substitute into the second: $2x - 3(4 - 3x) = 10$ , so $2x - 12 + 9x = 10$ , $11x = 22$ , $x = 2$ .

Then $y = 4 - 6 = -2$ .

Check: $12 + 2(-2) = 8$ ; $4 - 3(-2) = 4 + 6 = 10\text{.}$

$x = 2$ , $y = -2$

Question 11

The difference between two numbers is $14$ . Three times the smaller number is $2$ more than the larger. Find the two numbers.

ShowHide solution

Let the larger number be $x$ and the smaller $y$ .

$x - y = 14$ $\quad$ (1), $\quad$ $3y = x + 2$ $\quad$ (2)

From (1), $x = y + 14$ $\quad$ (3).

Substitute (3) into (2): $3y = y + 14 + 2$ , so $3y = y + 16$ , $2y = 16$ , $y = 8$ . Then from (3), $x = 8 + 14 = 22$ .

Check: $22 - 8 = 14$ ; three times $8$ is $24$ , and $22 + 2 = 24$ .

The numbers are $22$ and $8$ .

Question 12

A pen costs $ $1.20$ more than a pencil. Three pens and two pencils cost $ $8.60$ . Find the cost of each.

ShowHide solution

Let a pen cost $ $x$ and a pencil $ $y$ .

$x = y + 1.2$ $\quad$ (1), $\quad$ $3x + 2y = 8.6$ $\quad$ (2)

Substitute (1) into (2): $3(y + 1.2) + 2y = 8.6\text{,}$ so $3y + 3.6 + 2y = 8.6$ , $5y = 5$ , $y = 1$ . Then $x = 2.2$ .

Check: $2.20$ is $1.20$ more than $1$ ; $3(2.2) + 2(1) = 6.6 + 2 = 8.6\text{.}$

A pen costs $ $2.20$ and a pencil costs $ $1$ .

Key Points

Substitution: make one variable the subject of one equation, substitute it in brackets into the other, solve, then find the other value.
Choose a variable with coefficient $1$ or $-1$ (after dividing, if that helps) to avoid fractions.
Keep the brackets: a minus sign in front multiplies every term inside.
If both equations give $y$ , set the two expressions equal.
Answers may be negative, zero or fractions; algebra gives them exactly.
Check in both original equations: the rearranged one always fits, so it cannot catch a mistake.
When no coefficient is $1$ or $-1$ , substitution brings fractions; elimination (lesson 2.7) is usually easier.