Skip to content

Maths explained

Simultaneous equations explained step by step, with examples

Simultaneous equations explained step by step: the elimination and substitution methods, a word problem, common mistakes, and practice questions with answers.

For students. By the Qlive team. . 5 min read.

This guide is for you if you are in roughly Grades 9 and 10 and need a reliable method for simultaneous equations. In CBSE they are taught in Class 10 as Pair of Linear Equations in Two Variables. In GCSE and IGCSE they are called simultaneous equations, and ICSE covers them in Class 9. Before you start, you should be comfortable solving linear equations with one unknown.

The idea in plain words

One equation with two unknowns, like x + y = 10, has lots of possible answers: 5 and 5, 7 and 3, 2 and 8, and so on.

Add a second equation, like x - y = 4, and usually only one pair of values works for both at the same time. That is what “simultaneous” means. Here, x = 7 and y = 3.

On a graph, each linear equation is a straight line. The solution is the point where the two lines cross.

There are two main methods:

Method Idea Best when
Elimination Add or subtract the equations to remove one unknown Both equations look like ax + by = c
Substitution Replace one unknown with an expression from the other equation One equation already says y = … or x = …

Worked examples

Example 1: elimination by adding

Solve:

  • 2x + y = 11 … (1)
  • x - y = 1 … (2)

The y terms are +y and -y. They have opposite signs, so adding will remove them.

  1. Add (1) and (2): (2x + x) + (y - y) = 11 + 1, so 3x = 12
  2. x = 4
  3. Substitute x = 4 into (1): 2(4) + y = 11, so 8 + y = 11, so y = 3

Answer: x = 4, y = 3

Check in (2): 4 - 3 = 1. Correct.

Example 2: elimination with multiplying first

Solve:

  • 3x + 2y = 16 … (1)
  • 2x + 5y = 18 … (2)

Neither unknown has matching numbers, so make the x terms match.

  1. Multiply (1) by 2: 6x + 4y = 32 … (3)
  2. Multiply (2) by 3: 6x + 15y = 54 … (4)
  3. The x terms are both +6x (same sign), so subtract (3) from (4): 15y - 4y = 54 - 32, so 11y = 22
  4. y = 2
  5. Substitute into (1): 3x + 2(2) = 16, so 3x + 4 = 16, so 3x = 12, so x = 4

Answer: x = 4, y = 2

Check in (2): 2(4) + 5(2) = 8 + 10 = 18. Correct.

Example 3: substitution

Solve:

  • y = 2x - 1 … (1)
  • 3x + 2y = 19 … (2)

Equation (1) already tells you what y is, so substitute it into (2).

  1. Replace y in (2) with (2x - 1): 3x + 2(2x - 1) = 19
  2. Expand: 3x + 4x - 2 = 19
  3. Simplify: 7x - 2 = 19, so 7x = 21, so x = 3
  4. Substitute into (1): y = 2(3) - 1 = 5

Answer: x = 3, y = 5

Check in (2): 3(3) + 2(5) = 9 + 10 = 19. Correct.

Example 4: a word problem

2 pens and 3 notebooks cost ₹120. 4 pens and 1 notebook cost ₹90. Find the cost of one pen and one notebook.

  1. Choose letters: let p be the cost of a pen and n the cost of a notebook.
  2. Write the equations:
    • 2p + 3n = 120 … (1)
    • 4p + n = 90 … (2)
  3. From (2): n = 90 - 4p
  4. Substitute into (1): 2p + 3(90 - 4p) = 120
  5. Expand: 2p + 270 - 12p = 120
  6. Simplify: -10p + 270 = 120, so -10p = -150, so p = 15
  7. n = 90 - 4(15) = 90 - 60 = 30

Answer: a pen costs ₹15 and a notebook costs ₹30.

Check in (1): 2(15) + 3(30) = 30 + 90 = 120. Correct.

Common mistakes and how to avoid them

Subtracting negatives wrongly. When you subtract one equation from another, every sign in the second equation flips. For example, (5x + 2y) - (3x - 4y) = 2x + 6y, not 2x - 2y. Write the subtraction out in full if you are unsure.

Multiplying only some terms. When you multiply an equation by 3, the number on the right side must be multiplied too. 2x + 5y = 18 becomes 6x + 15y = 54, not 6x + 15y = 18.

Stopping after one unknown. Finding x is only half the answer. Always substitute back to find y.

Choosing add when you should subtract. Same signs, subtract. Different signs, add. Look before you combine.

Not checking. Substitute both values into the equation you did not use to find the second value. If it works, you can be confident.

Messy word problems. Write down what each letter stands for, including units, before you write any equation.

Try these

  1. x + y = 10 and x - y = 4
  2. 3x + y = 14 and x + y = 6
  3. 2x + 3y = 13 and 5x - 2y = 4
  4. y = 3x + 1 and 2x + y = 16
  5. 2 adult tickets and 3 child tickets cost 550. 1 adult and 1 child ticket together cost 225. Find the price of each ticket.

Answers

  1. Add: 2x = 14, so x = 7, then y = 3.
  2. Subtract: 2x = 8, so x = 4, then y = 2.
  3. Multiply the first by 2 and the second by 3: 4x + 6y = 26 and 15x - 6y = 12. Add: 19x = 38, so x = 2. Then 4 + 3y = 13, so y = 3.
  4. Substitute: 2x + 3x + 1 = 16, so 5x = 15, x = 3, and y = 10.
  5. 2a + 3c = 550 and a + c = 225. From the second, a = 225 - c. Then 450 - 2c + 3c = 550, so c = 100 and a = 125. Adult 125, child 100.

Still unsure?

If simultaneous equations still feel shaky, the trouble is often in negative numbers or expanding brackets rather than the method itself. A Qlive teacher can work through it with you one-to-one, starting with a free assessment that finds the exact gap. You can read more on our maths page and our CBSE page.

Common questions

Should I use elimination or substitution?

Use substitution when one equation already says y = something or x = something. Use elimination when both equations are in the form ax + by = c. Both methods give the same answer, so pick whichever needs fewer steps.

How do I know whether to add or subtract the equations?

Look at the signs of the terms you want to eliminate. If they have the same sign, subtract. If they have different signs, add. A simple memory aid is "same sign subtract".

What does it mean if both variables disappear?

If you get something like 0 = 5, the lines are parallel and there is no solution. If you get 0 = 0, the two equations describe the same line and there are infinitely many solutions. CBSE Class 10 covers both cases.

Read next

Book a free assessment