Practice Maths

Simultaneous Equations: Elimination Method

Key Ideas

Key Terms

elimination method
Adds or subtracts equations to remove one variable entirely.
equal
Subtract the equations. If they are opposite, add them.
multiply
One or both equations by a suitable number first.
No solution
If elimination leads to a false statement (e.g. 0 = 5), the lines are parallel.
Infinite solutions
If elimination leads to a true statement (e.g. 0 = 0), the equations describe the same line.

Choosing the most efficient method

Use elimination when the coefficients of one variable are the same or easily made the same by multiplying. Use substitution when one variable is already isolated.

Hot Tip Line up the equations neatly with variables in the same columns before adding or subtracting. A small alignment error can cost you the whole solution!

Worked Example

Question: Solve simultaneously:   2x + y = 7   and   x − y = 2

Step 1 — Identify the variable to eliminate.
The y coefficients are +1 and −1. Adding the equations eliminates y.

Step 2 — Add the equations.
   2x + y = 7
+   x − y = 2
 ———————
   3x      = 9

Step 3 — Solve for x.
x = 3

Step 4 — Substitute to find y.
3 − y = 2 → y = 1

Step 5 — Verify.
Eq (1): 2(3) + 1 = 7 ✓    Eq (2): 3 − 1 = 2 ✓
Solution: x = 3, y = 1

The Idea Behind Elimination

The elimination method works by adding or subtracting the two equations so that one variable disappears completely, leaving a single equation with one unknown. This is possible when the coefficients of one variable are equal (or can be made equal by multiplying). Once one variable is eliminated, you solve for the remaining variable and then substitute back to find the other.

Adding or Subtracting Equations

If the coefficients of one variable have opposite signs, add the equations to eliminate that variable. If the coefficients have the same sign, subtract one equation from the other. For example:

Equations: 3x + 2y = 11 and 3x − y = 5. The x-coefficients are both 3 (same sign), so subtract: (3x + 2y) − (3x − y) = 11 − 5, giving 3y = 6, so y = 2. Substitute back: 3x + 2(2) = 11, so 3x = 7, x = 7/3. Check both equations.

Second example: 2x + y = 8 and x − y = 1. The y-coefficients are +1 and −1 (opposite signs), so add: 3x = 9, x = 3. Then y = 8 − 2(3) = 2. Solution: (3, 2).

Multiplying to Match Coefficients

When the coefficients do not already match, multiply one or both equations by a constant to make a pair of coefficients equal. For example: 2x + 3y = 12 and 5x + y = 13. Multiply the second by 3 to make the y-coefficients both 3: 15x + 3y = 39. Now subtract: (15x + 3y) − (2x + 3y) = 39 − 12, giving 13x = 27, x = 27/13. For neater numbers, consider multiplying to eliminate x instead: multiply equation 1 by 5 and equation 2 by 2: 10x + 15y = 60 and 10x + 2y = 26. Subtract: 13y = 34, y = 34/13.

Key tip: Choose to eliminate the variable whose coefficients are easier to match. If one variable has small coefficients (like 1, 2, 3) and the other has large ones (like 7, 9, 11), eliminate the one with small coefficients to keep the arithmetic simple. Also: when you subtract, be especially careful with the signs — expand the subtraction fully before simplifying.

When to Choose Elimination over Substitution

Elimination is generally the better choice when both equations are in the form ax + by = c (sometimes called "standard form") and neither variable has a coefficient of 1. Substitution is better when one equation is already solved for a variable (y = ...). Both methods always give the same answer — choosing wisely just saves time and reduces the chance of sign errors.

Mastery Practice

  1. Solve each pair by adding or subtracting the equations directly. No multiplication needed. Fluency

    1. 3x + y = 11   and   x + y = 5
    2. 4x + 3y = 19   and   2x + 3y = 13
    3. 5x + 2y = 16   and   3x − 2y = 0
    4. x + 4y = 14   and   x − 4y = −2
    5. 6x + y = 20   and   6x − y = 10
    6. 2x + 5y = 24   and   2x + y = 8
    7. 7x − 3y = 25   and   x + 3y = 7
    8. 4x + 7y = 30   and   4x + 3y = 18

  2. Multiply one or both equations to make coefficients equal, then eliminate. Fluency

    1. 2x + y = 9   and   3x + 2y = 14
    2. 3x + 2y = 16   and   x + y = 7
    3. 4x + 3y = 23   and   2x + y = 9
    4. 3x + 4y = 26   and   2x + y = 10
    5. 5x + 2y = 29   and   3x + y = 16
    6. 2x + 3y = 13   and   3x + 2y = 12

  3. Mixed method questions. State which method you will use and why, then solve. Verify each solution. Understanding

    1. y = 3x − 1   and   2x + y = 9
    2. 4x − 3y = 1   and   5x − 3y = −2
    3. 3x + 2y = 12   and   6x + 4y = 24   (What do you notice?)
    4. 2x + y = 5   and   4x + 2y = 15   (What do you notice?)

  4. Set up and solve using the elimination method. Problem Solving

    1. The sum of two numbers is 23 and their difference is 7. Find both numbers.
    2. Two mobile plans: Plan A costs a $20 connection fee plus $0.30 per minute. Plan B costs a $10 connection fee plus $0.50 per minute. For how many minutes do both plans cost the same? What is that cost?
    3. A cinema sells adult tickets for $15 and student tickets for $10. In one session, 80 tickets were sold for a total of $1 050. How many adult and how many student tickets were sold?
    4. A sprinter runs a 400 m race in a tailwind, completing it in 44 seconds. Running in the same conditions with a headwind, it takes 56 seconds. If s is the sprinter’s still-air speed (m/s) and w is the wind speed, write two equations and solve them. (Hint: speed = distance ÷ time)

  5. Each system below requires you to multiply both equations before eliminating. Solve using elimination and verify your solution. Problem Solving

    Strategy. When no single multiplication makes a coefficient match, multiply each equation by the coefficient of the other. For example, to eliminate x from 3x + 2y = 11 and 5x + 3y = 17, multiply equation 1 by 5 and equation 2 by 3 to get 15x in both.
    1. 3x + 5y = 23   and   4x + 3y = 20
    2. 5x + 4y = 33   and   3x + 2y = 19
    3. 2x + 7y = 39   and   5x + 3y = 27

  6. Solve these systems that involve fractions or decimals. First multiply every term to clear fractions/decimals, then use elimination. Problem Solving

    1. 1⁄2x + 1⁄3y = 4   and   1⁄4x + 1⁄2y = 3
    2. 0.3x + 0.2y = 2.6   and   0.5x − 0.4y = 0.2

  7. Number and digit problems. Define two variables, write two equations and solve using elimination. Problem Solving

    1. A two-digit number has a digit sum of 9. When the digits are reversed, the new number is 27 more than the original. Find the original number. (Let t = tens digit, u = units digit.)
    2. The difference between two numbers is 14. Three times the smaller number added to twice the larger number gives 93. Find both numbers.
    3. When twice a number is added to three times another number the result is 38. When the first number is subtracted from the second number the result is 4. Find both numbers.

  8. Finance and investment problems. Write and solve two simultaneous equations. Problem Solving

    1. Zara invests a total of $8 000 in two accounts. Account A earns 4% simple interest per year and Account B earns 6% simple interest per year. After one year, the total interest earned is $400. How much was invested in each account?
    2. A school canteen sells sausage rolls for $2.50 and meat pies for $3.80. On Tuesday, 120 items were sold for a total of $366. How many sausage rolls and how many meat pies were sold?

  9. A rectangle has a perimeter of 64 cm. If the length were increased by 5 cm and the width decreased by 2 cm, the perimeter would still be 64 cm. Problem Solving

    Think carefully. Write two equations using the original dimensions and the modified dimensions. Then interpret whether the system has a unique solution, no solution, or infinitely many solutions, and explain what that means about the rectangle.
    1. Write the equation for the original perimeter in terms of length l and width w.
    2. Write the equation for the new perimeter after the changes.
    3. Solve the system. What do you notice? Explain your finding geometrically.

  10. A tank is being filled by two pipes. Pipe A and Pipe B together fill the tank in 4 hours. Pipe A alone fills the tank in 6 hours less than Pipe B alone. Let a = hours for Pipe A alone and b = hours for Pipe B alone. Problem Solving

    Rate approach. In one hour, Pipe A fills 1⁄a of the tank and Pipe B fills 1⁄b of the tank. Together they fill 1⁄4 of the tank per hour.
    1. Write the two equations relating a and b.
    2. Use substitution or elimination to find a and b. (Hint: let A = 1⁄a and B = 1⁄b to create a linear system.)
    3. State how long each pipe takes to fill the tank alone.
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