To solve two linear equations by elimination, make the coefficient of one letter the same size in both equations, then add or subtract the equations so that letter disappears. You are left with one equation in one unknown, which you already know how to solve.
Simultaneous equations appear in most Extended-style algebra questions and again in word problems and graph questions. This lesson is the first method in simultaneous relationships, and it pairs with substitution.
Why does elimination work?
Each equation is a true statement about the same pair of numbers. If you add two true statements, or subtract one from another, the result is still true. So we are free to combine the equations in a way that cancels one letter.
The only skill needed is choosing the combination. That means making one pair of coefficients equal in size, and then deciding between adding and subtracting.
The method, step by step
- Label the equations (1) and (2) and line up the letters in columns.
- Pick a letter to remove. Choose the one whose coefficients are easiest to match.
- Multiply each equation by a whole number so the chosen coefficients have the same size. Multiply every term, including the right-hand side.
- Add if the terms you are removing have opposite signs, subtract if they have the same sign.
- Solve the one-letter equation.
- Substitute back into an original equation to find the other letter.
- Check in the other original equation.
Worked example
Solve: 2x + 3y = 13 (1) and 5x − 2y = 4 (2)
Step 1, choose y. The coefficients are 3 and −2. The lowest common multiple of 3 and 2 is 6, so aim for +6y and −6y.
Step 2, multiply. (1) × 2 gives 4x + 6y = 26. (2) × 3 gives 15x − 6y = 12.
Step 3, add, because +6y and −6y have opposite signs: 19x = 38, so x = 2.
Step 4, substitute x = 2 into (1): 4 + 3y = 13, so 3y = 9 and y = 3.
Step 5, check in (2): 5(2) − 2(3) = 10 − 6 = 4. This matches the right-hand side.
Answer: x = 2, y = 3.
The mistake to watch for
The most common slip is multiplying the left-hand side but leaving the right-hand side alone.
Mistaken working: (1) × 2 gives 4x + 6y = 13, and (2) × 3 gives 15x − 6y = 12. Adding gives 19x = 25.
The student doubled 2x and 3y but kept 13 unchanged, so the new equation is no longer true.
The correction is to treat the equation as a balance. Whatever you do to one side, you do to the other, so 13 becomes 26. A useful habit is to draw an arrow from the multiplier to every term, including the number on the right.
Check yourself
Try these on paper, then open each answer.
1. Solve x + y = 11 and x − y = 3.
Show answer
The y terms have opposite signs, so add: 2x = 14, so x = 7. Then 7 + y = 11 gives y = 4. Check: 7 − 4 = 3, which matches.
x = 7, y = 4
2. Solve 3x + 4y = 18 and 3x + y = 9.
Show answer
The x terms have the same sign, so subtract (2) from (1): 3y = 9, so y = 3. Then 3x + 3 = 9 gives x = 2. Check in (1): 6 + 12 = 18, which matches.
x = 2, y = 3
3. Solve 2x + 5y = 1 and 3x + 2y = −4.
Show answer
Remove x. (1) × 3 gives 6x + 15y = 3. (2) × 2 gives 6x + 4y = −8. Subtract: 11y = 11, so y = 1. Then 2x + 5 = 1 gives x = −2. Check in (2): 3(−2) + 2 = −4, which matches.
x = −2, y = 1
Where this leads next
When one equation already has a letter on its own, substitution is often quicker. After that you can see what the answer means in interpreting an intersection in context, and then test everything in the simultaneous relationships practice set. The non-calculator working trainer lets you check arithmetic steps without a calculator.
Some students can follow every line of a solved example but freeze when the signs change in a new question. Our teachers look for exactly that kind of pattern in online one-to-one Mathematics tuition.