# Elimination Method and Sample Problems Contributed by: This pdf describes solving the equation by elimination method step by step with examples for better understanding.
1. Overview
Algebra: Solving Simultaneous Equations -
Elimination Method
This presentation will cover the use of the elimination method to
solve a system of two equations with two unknown variables.
Simultaneous equations Elimination method
I A set of simultaneous equations is a set of equations for In this method we eliminate one variable and from one equation.
which common solutions are sought for a number of variables. The steps involved are:
I You need as least the same number of equations as variables 1. Multiply one or both equations by constants so that one of
to be able to find a solution. the variables has the same coefficient.
I This presentation will only focus on two equations with two 2. Add or subtract one equation from the other so that the
unknown variables. variable with the same coefficient is eliminated.
I There are a number of different ways that you can solve a set 3. Solve this equation to find the value of the variable.
of simultaneous equations. All methods are equally valid. It is 4. Substitute the value of this variable into one of the equations
up to you to choose the method that is easiest for you to use. to find the value of the other variable.
I This presentation will cover the elimination method. 5. Check your answer in both of the original equations.
2. Example Step 1: Multiply one or both
equations by constants.
Let us follow the steps through in an example.
The coefficients of x are different in both equations.
2x + 5y = 6 (1)
If we multiplied Equation (1) by 3 and Equation (2) by 2, the
3x + 2y = −2 (2) coefficient of x in both equations would be 6.
Remember, we must multiply every term in the equation by it.
Step 1: continued Step 2: Subtract the equations to
Multiply Equation (1) by 3: eliminate a variable.
3 × (2x + 5y) = 3 × 6 ,
3 × 2x + 3 × 5y = 18 ,
6x + 15y = 18
6x + 15y = 18 . (3)
6x + 4y = −4
Multiply Equation (2) by 2: + 11y = 22 .
Thus the equation is
2 × (3x + 2y) = 2 × −2 , 11y = 22 .
2 × 3x + 2 × 2y = −4 ,
6x + 4y = −4 (4)
3. Step 3: Solve the equation. Step 4: Substitute to find the other
variable.
2x + 5y = 6,
11y = 22 , 2x + 5 × 2 = 6,
y = 2. 2x = 6 − 10 ,
2x = −4 ,
x = −2 .
From our calculations the answers are x = −2 and y = 2 .
Step 5: Check the answer in both of Worded example:
the original equations.
Check: x = −2 and y = 2 .
Equation (1): Equation (2): A city bakery sold 1 500 bread rolls on Sunday, with sales receipts
2x + 5y = 6 , 3x + 2y = −2 , of \$1 460. Plain rolls sold for 90 cents each while gourmet rolls
sold for \$1.45 each. How many of each type of roll were sold?
LHS = 2 × −2 + 5 × 2 LHS = 3 × −2 + 2 × 2
= 6 = −2
= RHS. = RHS.
Both values substitute correctly into both equations so the answer
must be correct.
4. Solution: Solution (continued)
Firstly we need to develop the equations to solve simultaneously. Multiply Equation (5) by 0.9.
The two equations generated from these sentences are: 0.9P + 0.9G = 1 350 . (7)
P + G = 1 500 , (5) Subtract Equation (7) from Equation (6).
0.9P + 1.45G = 1 460 . (6)
0.9P + 1.45G = 1 460
Where P is the number of plain rolls sold and G is the number of 0.9P + 0.9G = 1 350
gourmet rolls sold. 0P + 0.55G = 110
Solution (continued) Solution (continued)
Therefore the equation is
0.55G = 110 ,
G = 110 ÷ 0.55 , Check:
G = 200 . In Equation (5), LHS = 200 + 1 300 = 1 500 = RHS.
In Equation (6), LHS = 0.9 × 1 300 + 1.45 × 200 = 1 460 = RHS.
To find the value of P substitute G = 200 into equation (5).
So at the end of the day the shop had sold 200 gourmet rolls and
P + G = 1 500 , 1 300 plain rolls.
P + 200 = 1 500 ,
P = 1 300 .
Thus the solution is G = 200 and P = 1 300 .
5. Note: Exercise
Solve the following sets of simultaneous equations.
Finally, we have solved equations where multiples of the variables 1.
are only added to or subtracted from each other.
3x − y = 12 ,
We call these linear equations. x + y = 8.
Situations do arise where the variables are related in other ways
(non-linear equations). 2.
The elimination method can only be used for linear equations.
3x − 4y = 5 ,
5x − 12y = 3 .
Solution: Question 1 Solution: Question 1 (continued)
Rearranging to give:
3x − y = 12 , (8) 4x = 20
x + y = 8. (9) x = 5.
As y has the same coefficient (1) and opposite signs, this is the Substituting this into Equation (9) gives
variable I will eliminate.
x+y = 8
Adding Equation (8) and Equation (9) gives:
5+y = 8
3x − y = 12 y = 8−5
x + y = 8 = 3.
4x = 20 .
From our calculations the answers are x = 5 and y = 3.
6. Check the answer in both of the Solution: Question 2
original equations.
Check: x = 5 and y = 3. 3x − 4y = 5 , (10)
5x − 12y = 3 . (11)
Equation (8): Equation (9):
3x − y = 12 , x + y = 8, Multiply Equation (10) by −3 gives:
−9x + 12y = −15 . (12)
LHS = 3 × 5 − 3 LHS = 5 + 3
Adding Equation (12) and Equation (11) gives:
= 12 = 8
= RHS. = RHS. −9x + 12y = −15
Both values substitute correctly into both equations so the answer 5x − 12y = 3
must be correct. − 4x = − 12
Solution: Question 2 (continued) Check the answer in both of the
Rearranging to give: original equations.
−4x = −12 Check: x = 3 and y = 1.
x = 3.
Equation (10): Equation (11):
Substituting this into Equation (8) gives 3x − 4y = 5 , 5x − 12y = 3 ,
3x − 4y = 5
3 × 3 − 4y = 5 LHS = 3 × 3 − 4 × 1 LHS = 5 × 3 − 12 × 1
9 − 4y = 5 = 5 = 3
− 4y = 5−9 = RHS. = RHS.
y = 1. Both values substitute correctly into both equations so the answer
From our calculations the answers are x = 3 and y = 1. must be correct.
7. Summary
This presentation covered the use of the elimination method to
solve a system of two equations with two unknown variables.