Simultaneous equations — substitution method — Mathematics

Before this

y = mx + c · coming soon
Solving equations with brackets and fractions · coming soon
Substituting values into formulae · coming soon

This leads to

The elimination method
One linear, one quadratic
Solution sets: unique, infinite, none · coming soon

One clue is not enough

Amara thinks of two numbers. She gives you one clue: "The second number is one more than double the first."

Call the first number $x$ and the second number $y$ . The clue says $y = 2 x + 1$ . Can you find her numbers?

You cannot. Try $x = 0$ : then $y = 1$ . Try $x = 1$ : then $y = 3$ . Try $x = 2$ : then $y = 5$ . Every choice of $x$ gives a $y$ that fits. The clue is true for infinitely many pairs, so it cannot tell you which pair Amara chose.

So Amara gives a second clue: "Three times the first number, plus the second number, makes 16." That is $3 x + y = 16$ . On its own, this clue also has infinitely many pairs: $(0, 16)$ , $(1, 13)$ , $(2, 10)$ , and so on.

Now put the two lists side by side.

$x$	$y$ from clue 1: $y = 2 x + 1$	$y$ from clue 2: $3 x + y = 16$
0	1	16
1	3	13
2	5	10
3	7	7
4	9	4
5	11	1

Only one row agrees. When $x = 3$ , both clues give $y = 7$ . Each clue alone allows endless answers. Together they allow exactly one. Two equations that must be true at the same time are called simultaneous equations, and solving them means finding the values that make both true.

But the table only worked because the answer was a small whole number in the first few rows. If the answer were $x = \frac{34}{11}$ , you could search a table forever. We need a method that finds the answer directly. This lecture builds the first one: substitution.

The picture behind the table

Each equation is a straight line. Every point on the line $y = 2 x + 1$ is a pair that fits clue 1. Every point on the line $3 x + y = 16$ is a pair that fits clue 2. A pair that fits both clues must lie on both lines. That can only be the point where the lines cross.

Each line holds every pair that fits one clue. The only pair that fits both is the crossing point, (3, 7).

So solving simultaneous equations is finding where two lines meet. A graph shows the idea well, but it is not exact. If the crossing point were at $x \approx 3.09$ , you could not read it from a sketch. Algebra gives the exact answer.

The key move: equal things can replace each other

Look at clue 1 again: $y = 2 x + 1$ . This equation does more than describe $y$ . It says that $y$ and $2 x + 1$ are the same number, written in two different ways.

Now look at clue 2: $3 x + y = 16$ . This $y$ is the same $y$ , because we want a pair that fits both clues. So wherever we see $y$ in clue 2, we can write $2 x + 1$ instead. That changes nothing about the value. It only changes how the value is written.

3 x + (2 x + 1) = 16

Look at what just happened. The new equation has only one letter, $x$ . You already know how to solve an equation with one unknown. Collect the $x$ terms:

5 x + 1 = 16

Subtract 1 from both sides to get $5 x = 15$ , then divide by 5 to get $x = 3$ .

Now that you know $x$ , clue 1 gives $y$ at once: $y = 2 (3) + 1 = 7$ . That is it. That is the whole idea. Use one equation to write one letter in terms of the other, then put that expression into the second equation. Two unknowns become one.

This is the same thing you did with formulae in earlier years. To use $A = lw$ with $l = 5$ , you replace $l$ by 5. Here you replace $y$ by $2 x + 1$ . The only difference is that the replacement is an expression, not a number.

Stop and think

Why did we write $3 x + (2 x + 1)$ with a bracket, when the bracket does not seem to change anything here?

Show the answer

Because $y$ is one number, and $2 x + 1$ must be treated as one number when it takes $y$ 's place. Here the bracket makes no difference, because the $y$ is only added. But if the equation had $3 y$ or $- y$ , the whole of $2 x + 1$ would be multiplied. For example, $3 y$ becomes $3 (2 x + 1) = 6 x + 3$ , not $6 x + 1$ . Writing the bracket every time is a habit that protects you when it matters.

The substitution method

1. Choose one equation and write one letter on its own, in terms of the other letter (for example $y = 10 - 2 x$ ).

2. Substitute that expression, in a bracket, into the other equation.

3. Solve the new equation, which has one unknown.

4. Put that value back into your rearranged equation to find the second unknown.

5. Check the pair in both original equations.

Worked examples

Worked Example 1 — One letter is already on its own

Solve $y = 2 x + 1$ and $3 x + y = 16$ .

The first equation already gives $y$ . Substitute it into the second: $3 x + (2 x + 1) = 16$ , so $5 x + 1 = 16$ , so $5 x = 15$ , so $x = 3$ .

Then $y = 2 (3) + 1 = 7$ . The solution is $x = 3$ , $y = 7$ .

Check in both: $2 (3) + 1 = 7$ ✓ and $3 (3) + 7 = 16$ ✓.

Worked Example 2 — Rearrange first, and the answer is a fraction

Solve $2 x + y = 10$ and $5 x - 3 y = 4$ .

Neither letter is on its own. In the first equation, $y$ has no number in front of it, so it is easy to isolate: $y = 10 - 2 x$ .

Substitute into the second equation, with a bracket: $5 x - 3 (10 - 2 x) = 4$ .

This is the hard step. Spend time here. The $- 3$ multiplies both terms in the bracket: $- 3 \times 10 = - 30$ , and $- 3 \times (- 2 x) = + 6 x$ . So $5 x - 30 + 6 x = 4$ .

Collect: $11 x - 30 = 4$ , so $11 x = 34$ , so $x = \frac{34}{11}$ .

Then $y = 10 - 2 \times \frac{34}{11} = \frac{110}{11} - \frac{68}{11} = \frac{42}{11}$ .

A fraction is not a sign that something went wrong. Most pairs of lines do not cross at whole numbers. Leave the answer exact: $x = \frac{34}{11} \approx 3.09$ and $y = \frac{42}{11} \approx 3.82$ (3 s.f., three significant figures).

Check in both: $2 \times \frac{34}{11} + \frac{42}{11} = \frac{110}{11} = 10$ ✓ and $5 \times \frac{34}{11} - 3 \times \frac{42}{11} = \frac{170 - 126}{11} = \frac{44}{11} = 4$ ✓.

In Example 2, the sign inside the bracket is where most mistakes happen. Find the mistake in this student's work.

Spot the error — a sign lost in the bracket

A student solves $3 x + y = 8$ and $5 x - 2 y = 6$ by substitution. Exactly one line contains a mistake. Tap it.

A student's solution of 3x + y = 8 and 5x − 2y = 6. Line 3 is wrong.

Line 1. From the first equation: y = 8 − 3x. Correct.
Line 2. Substitute: 5x − 2(8 − 3x) = 6. Correct, with the bracket.
Line 3. Expand: 5x − 16 − 6x = 6. Wrong: −2 × (−3x) = +6x, so it should be 5x − 16 + 6x = 6.
Line 4. Collect: −x − 16 = 6, so x = −22. This follows from line 3, so it carries the mistake.
Line 5. Then y = 8 − 3(−22) = 74. Also carries the mistake.
Line 6. Check: 5(−22) − 2(74) = −258, not 6. The check is correct, and it shows that something went wrong earlier. The right answer is x = 2, y = 2.

Notice the last line. The check did its job: it showed that the answer was wrong before anyone else had to. This is why step 5 is part of the method, not an extra.

Choosing what to isolate

With two equations and two letters, there are four ways to start. You could isolate $x$ or $y$ , from either equation. They all lead to the same answer, but some are much easier than others.

Worked Example 3 — Look at all four options first

Solve $3 x + 4 y = 18$ and $x - 2 y = 0$ .

The four options:

$y$ from the first: $y = \frac{18 - 3 x}{4}$ . A fraction at once.
$x$ from the first: $x = \frac{18 - 4 y}{3}$ . Also a fraction.
$y$ from the second: $y = \frac{x}{2}$ . A small fraction.
$x$ from the second: $x = 2 y$ . No fraction at all.

Choose $x = 2 y$ . The rule for choosing: look for a letter with no number in front of it (a coefficient of 1 or −1). Isolating that letter needs no division, so no fractions appear.

Substitute into the first equation: $3 (2 y) + 4 y = 18$ , so $10 y = 18$ , so $y = \frac{9}{5}$ . Then $x = 2 \times \frac{9}{5} = \frac{18}{5}$ .

Check in both: $3 \times \frac{18}{5} + 4 \times \frac{9}{5} = \frac{54 + 36}{5} = 18$ ✓ and $\frac{18}{5} - 2 \times \frac{9}{5} = 0$ ✓.

Worked Example 4 — A fraction in the equation

Solve $\frac{x}{2} + y = 7$ and $y = x - 2$ .

The second equation gives $y$ . Substitute: $\frac{x}{2} + (x - 2) = 7$ .

Multiply every term by 2 to clear the fraction: $x + 2 x - 4 = 14$ , so $3 x = 18$ , so $x = 6$ . Then $y = 6 - 2 = 4$ .

Check in both: $\frac{6}{2} + 4 = 7$ ✓ and $6 - 2 = 4$ ✓.

When the lines never meet

Two straight lines usually cross at one point. But there are two other possibilities. The lines can be parallel, so they never meet. Or the two equations can describe the same line, so they meet everywhere. Substitution tells you which case you are in, if you read the result carefully.

Try it on the graph below. The dark line is $y = 2 x + 1$ , the first equation from the start of this lecture. The second equation, $3 x + y = 16$ , is the same as $y = - 3 x + 16$ . Change its gradient and its intercept, and watch what the substitution gives.

Graph builder — where do the lines meet?

Move the gradient m and the intercept c of the teal line y = mx + c. Watch the crossing point and the substitution steps. Then make the gradient 2, the same as the dark line, and see what the algebra says.

Gradient m: −3

Intercept c: 16

The line y = 2x + 1 with four different second lines y = mx + c. Substituting y = 2x + 1 into y = mx + c gives 2x + 1 = mx + c, so (2 − m)x = c − 1.

Second line	After substituting	Result
y = −3x + 16	5x = 15	one point: (3, 7)
y = 0.5x + 7	1.5x = 6	one point: (4, 9)
y = 2x + 5	0x = 4, so 0 = 4	impossible: no solution, parallel lines
y = 2x + 1	0x = 0, so 0 = 0	always true: infinitely many solutions, the same line

The lines y = 2x + 1 and y = 2x + 5 have the same gradient, 2, so they never meet. Substitution gives 0 = 4.

What did you see? While the gradients are different, the lines always cross at exactly one point, and the substitution gives one value of $x$ . When the gradient is 2, the $x$ terms cancel completely. Then one of two things is left. If what is left is false, such as $0 = 4$ , there is no pair that fits both equations, and the lines are parallel. If what is left is true, such as $0 = 0$ , every pair on the line fits both, because the two equations describe the same line.

Worked Example 5 — The letters disappear

Solve $y = 3 x - 2$ and $6 x - 2 y = 5$ .

Substitute: $6 x - 2 (3 x - 2) = 5$ , so $6 x - 6 x + 4 = 5$ , so $4 = 5$ .

This is false, whatever $x$ is. So no pair fits both equations: there is no solution. Check with the picture: the second equation rearranges to $y = 3 x - 2.5$ . Both lines have gradient 3, so they are parallel, half a unit apart.

If the second equation had been $6 x - 2 y = 4$ , the same steps would give $4 = 4$ . That is always true, and $6 x - 2 y = 4$ rearranges to $y = 3 x - 2$ , the same line. Every point on it is a solution.

Stop and think

You solve two equations and reach $0 = 5$ . Your friend says, "You must have made a mistake." Is your friend right?

Show the answer

Not necessarily. A false statement such as $0 = 5$ can be the correct result: it means the equations have no common solution, because the lines are parallel. Before you decide, check two things. First, go back over the algebra for a slip. Second, look at the gradients: write both equations as $y = m x + c$ . If the gradients are equal and the intercepts are different, $0 = 5$ is the right conclusion, and the answer is "no solution".

Three classic mistakes

Trap 1 — Substituting without a bracket

With $y = 8 - 3 x$ , the term $- 2 y$ becomes $- 2 (8 - 3 x) = - 16 + 6 x$ . Without the bracket, students write $- 2 \times 8 - 3 x$ or lose the sign on $6 x$ . Always put the whole expression in a bracket, then expand with care.

Trap 2 — Putting the expression back where it came from

From $2 x + y = 10$ you get $y = 10 - 2 x$ . If you substitute this back into $2 x + y = 10$ , you get $2 x + 10 - 2 x = 10$ , so $10 = 10$ . That is true, but it tells you nothing, because you used the same information twice. The expression must go into the other equation.

Trap 3 — Stopping halfway, or checking only one equation

Finding $x = 3$ is half an answer: the question asks for a pair. And checking only the equation you rearranged proves little, because that equation was used to make $y$ fit $x$ . The pair must be checked in both original equations. The check takes a minute and catches most slips.

Try it yourself

Work each one on paper first. Open a hint only when you are stuck.

Solve $y = x + 4$ and $2 x + y = 16$ .

Hint 1
The first equation already gives $y$ . Put $x + 4$ in place of $y$ in the second.

Hint 2
$2 x + (x + 4) = 16$ , so $3 x + 4 = 16$ .

Full answer
$3 x = 12$ , so $x = 4$ . Then $y = 4 + 4 = 8$ . Check: $4 + 4 = 8$ ✓ and $2 (4) + 8 = 16$ ✓.
Solve $x + 3 y = 11$ and $4 x - y = 5$ .

Hint 1
Look for a letter with no number in front of it. There are two: $x$ in the first equation and $y$ in the second. Either works.

Hint 2
Using $x = 11 - 3 y$ : substitute to get $4 (11 - 3 y) - y = 5$ .

Full answer
$44 - 12 y - y = 5$ , so $- 13 y = - 39$ , so $y = 3$ . Then $x = 11 - 3 (3) = 2$ . Check: $2 + 3 (3) = 11$ ✓ and $4 (2) - 3 = 5$ ✓. (Starting from $y = 4 x - 5$ gives $x + 3 (4 x - 5) = 11$ , so $13 x = 26$ , the same answer.)
Solve $2 x + 5 y = 1$ and $3 x - y = 10$ .

Hint 1
In the second equation, $y$ has coefficient −1. Isolate it: $y = 3 x - 10$ .

Hint 2
$2 x + 5 (3 x - 10) = 1$ . The 5 multiplies both terms.

Full answer
$2 x + 15 x - 50 = 1$ , so $17 x = 51$ , so $x = 3$ . Then $y = 3 (3) - 10 = - 1$ . Check: $2 (3) + 5 (- 1) = 6 - 5 = 1$ ✓ and $3 (3) - (- 1) = 10$ ✓. A negative answer is as good as a positive one.
The equations $y = 2 x + k$ and $4 x - 2 y = 6$ contain a number $k$ . For which value of $k$ are there infinitely many solutions? What happens for every other value of $k$ ? Can there ever be exactly one solution?

Hint 1
Substitute anyway, keeping $k$ as a letter. See what happens to the $x$ terms.

Hint 2
$4 x - 2 (2 x + k) = 6$ gives $4 x - 4 x - 2 k = 6$ . The $x$ terms cancel, so you are left with $- 2 k = 6$ .

Full answer
After substituting, the $x$ terms always cancel, leaving $- 2 k = 6$ . If $k = - 3$ , this is true, so every point fits: infinitely many solutions. For any other $k$ , it is false, so there is no solution. There is never exactly one solution. The picture explains why: $4 x - 2 y = 6$ rearranges to $y = 2 x - 3$ , with gradient 2. The first line also has gradient 2. Lines with equal gradients are either the same line ( $k = - 3$ ) or parallel (every other $k$ ). Check with $k = - 3$ : the point $(0, - 3)$ fits both, since $- 3 = 2 (0) - 3$ and $4 (0) - 2 (- 3) = 6$ .

The bigger idea

Substitution rests on one simple rule: if two expressions are equal, either one can stand in the other's place. That rule is used across mathematics and science. In Economics, the equilibrium of a market is where buyers' plans meet sellers' plans. With demand $Q = 500 - 50 P$ and supply $Q = 100 P - 100$ , substituting gives $500 - 50 P = 100 P - 100$ , so $150 P = 600$ , so $P = 4$ and $Q = 300$ . That is a pair of simultaneous equations, and the equilibrium is where the two lines cross. In Physics, you can remove time from two equations of motion in the same way, by writing $t$ in terms of the other quantities and substituting.

Substitution works best when one letter is easy to isolate. When neither letter is easy, as in $2 x + 3 y = 12$ and $5 x - 4 y = 7$ , there is a better tool. The next lecture builds it: instead of replacing a letter, you add or subtract whole equations to make a letter disappear. And because substitution only needs one letter to be isolated, it also works when the second equation is a curve. That comes two lectures from now, when a straight line meets a parabola.

Key concepts in this lesson

Simultaneous equations are two equations that must be true at the same time, and their solution is the pair of values that fits both.
Each linear equation is a straight line, so the solution is the point where the two lines cross.
In the substitution method, one equation gives one letter in terms of the other, and that expression goes, in a bracket, into the other equation to leave one unknown.
Isolating a letter whose coefficient is 1 or −1 avoids fractions, and every answer is checked in both original equations.
If the letters cancel and leave a false statement such as 0 = 4, the lines are parallel and there is no solution.
If the letters cancel and leave a true statement such as 0 = 0, the equations describe the same line and there are infinitely many solutions.