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Algebra: Solving Simultaneous Equations -
Elimination Method
Overview
This presentation will cover the use of the elimination method to solve a system of two equations with two unknown variables.
Simultaneous equations
I (^) A set of simultaneous equations is a set of equations for which common solutions are sought for a number of variables.
I (^) You need as least the same number of equations as variables to be able to find a solution.
I (^) This presentation will only focus on two equations with two unknown variables.
I (^) There are a number of different ways that you can solve a set of simultaneous equations. All methods are equally valid. It is up to you to choose the method that is easiest for you to use.
I (^) This presentation will cover the elimination method.
Elimination method
In this method we eliminate one variable and from one equation. The steps involved are:
- Multiply one or both equations by constants so that one of the variables has the same coefficient.
- Add or subtract one equation from the other so that the variable with the same coefficient is eliminated.
- Solve this equation to find the value of the variable.
- Substitute the value of this variable into one of the equations to find the value of the other variable.
- Check your answer in both of the original equations.
Example
Let us follow the steps through in an example.
2 x + 5y = 6 (1) 3 x + 2y = โ 2 (2)
Step 1: Multiply one or both
equations by constants.
The coefficients of x are different in both equations. If we multiplied Equation (1) by 3 and Equation (2) by 2 , the coefficient of x in both equations would be 6. Remember, we must multiply every term in the equation by it.
Step 1: continued
Multiply Equation (1) by 3 :
3 ร (2x + 5y) = 3 ร 6 , 3 ร 2 x + 3 ร 5 y = 18 , 6 x + 15y = 18. (3)
Multiply Equation (2) by 2 :
2 ร (3x + 2y) = 2 ร โ 2 , 2 ร 3 x + 2 ร 2 y = โ 4 , 6 x + 4y = โ 4 (4)
Step 2: Subtract the equations to
eliminate a variable.
6 x + 15 y = 18 6 x + 4 y = โ 4
- 11 y = 22. Thus the equation is 11 y = 22.
Solution:
Firstly we need to develop the equations to solve simultaneously.
The two equations generated from these sentences are:
P + G = 1 500 , (5)
- 9 P + 1. 45 G = 1 460. (6)
Where P is the number of plain rolls sold and G is the number of gourmet rolls sold.
Solution (continued)
Multiply Equation (5) by 0.9.
- 9 P + 0. 9 G = 1 350. (7)
Subtract Equation (7) from Equation (6).
- 9 P + 1. 45 G = 1 460
- 9 P + 0. 9 G = 1 350 0 P + 0. 55 G = 110
Solution (continued)
Therefore the equation is
- 55 G = 110 , G = 110 รท 0. 55 , G = 200.
To find the value of P substitute G = 200 into equation (5).
P + G = 1 500 , P + 200 = 1 500 , P = 1 300.
Thus the solution is G = 200 and P = 1 300.
Solution (continued)
Check: In Equation (5), LHS = 200 + 1 300 = 1 500 = RHS. In Equation (6), LHS = 0. 9 ร 1 300 + 1. 45 ร 200 = 1 460 = RHS. So at the end of the day the shop had sold 200 gourmet rolls and 1 300 plain rolls.
Note:
Finally, we have solved equations where multiples of the variables are only added to or subtracted from each other.
We call these linear equations.
Situations do arise where the variables are related in other ways (non-linear equations).
The elimination method can only be used for linear equations.
Exercise
Solve the following sets of simultaneous equations.
3 x โ y = 12 , x + y = 8.
3 x โ 4 y = 5 , 5 x โ 12 y = 3.
Solution: Question 1
3 x โ y = 12 , (8) x + y = 8. (9)
As y has the same coefficient (1) and opposite signs, this is the variable I will eliminate.
Adding Equation (8) and Equation (9) gives:
3 x โ y = 12 x + y = 8 4 x = 20.
Solution: Question 1 (continued)
Rearranging to give: 4 x = 20 x = 5. Substituting this into Equation (9) gives
x + y = 8 5 + y = 8 y = 8 โ 5 = 3.
From our calculations the answers are x = 5 and y = 3.
Summary
This presentation covered the use of the elimination method to solve a system of two equations with two unknown variables.
