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Solving Simultaneous Equations Using Elimination Where the Coefficients Differ
(KS4, Year 10)
The Lesson
Simultaneous equations are a set of several equations with several unknowns. We can use the elimination method to find the values of the unknowns which solve both equations at the same time. In this lesson, we will be looking at simultaneous equations where the coefficients of each unknown differ in each equation:In the simultaneous equations above,
 The coefficient of the x is 2 in the top equation and 1 in the bottom equation (Note: if a letter does not have number written in front of it, it has a coefficient of 1.)
 The coefficient of the y is 1 in the top equation and 3 in the bottom equation.
How to Solve Simultaneous Equations Using the Elimination Method Where the Coefficients Differ
Solving simultaneous equations using the elimination method where the coefficients differ is easy.Question
Solve the simultaneous equations shown below using the elimination method.StepbyStep:
1
Decide which unknown you wish to eliminate.
We will eliminate the x.
2
Look at the coefficients of the x term in each equation. The coefficient of x is the number in front of it.
2x + 3y = 4 ... (1)
2x + 3y = 7 ... (2)
2x + 3y = 7 ... (2)
 The coefficient of x in Equation (1) is 2.
 The coefficient of x in Equation (2) is 1. Don't forget: if there is no number in front of the unknown, the coefficient is 1, but there is no need to write it.
3
Make the coefficients the same by multiplying one or both of the equations by a number.
If we multiply Equation (2) by 2, then the coefficients of x will be the same:
Now the coefficient of the x terms is 2 in both equations.
Now the coefficient of the x terms is 2 in both equations.
4
Subtract the new equations.
Subtract Equation (1) from 2 × Equation (2).
5
Subtract the x terms.
2x − 2x = 0
We have eliminated the x.
6
Subtract the y terms.
6y − y = 5y
7
Subtract the constants.
14 − 4 = 10
8
We have eliminated one unknown (the x).
Solve for y.
5y = 10
5y ÷ 5 = 10 ÷ 5
y = 2
y = 2 is a solution to the simultaneous equations.
9
Substitute the variable we have just found (y = 2) into one of the equations.
Solve for x.
2x + y = 4
2x + 2 = 4
2x + 2 − 2 = 4 − 2
2x = 2
2x ÷ 2 = 2 ÷ 2
x = 1
x = 1 is a solution to the simultaneous equations.
Answer:
We have solved the simultaneous equations:
x = 1, y = 2 solves
2x + y = 4
x + 3y = 7
x + 3y = 7
Using the Least Common Multiple to Solve Simultaneous Equations Using the Elimination Method Where the Coefficients Differ
In the previous example, we multiplied one equation by 2 to ensure that the coefficients of the x was the same. In the next example, we will see that if we wish to eliminate an unknown, we should use the least common multiple of that unknown's coefficients to tells us what we should multiply each equation by.Question
Solve the simultaneous equations shown below using the elimination method.StepbyStep:
1
Decide which unknown you wish to eliminate.
We will eliminate the x.
2
Look at the coefficients of the x term in each equation. The coefficient of x is the number in front of it.
4x + 3y = 8 ... (1)
6x + 3y = 5 ... (2)
6x + 3y = 5 ... (2)
 The coefficient of x in Equation (1) is 4.
 The coefficient of x in Equation (2) is 6.
3
4
Find the smallest multiple that appears in both lists.
12 is the least common multiple.
12 is the least common multiple.
5
Find which number we have to multiply Equation (1) by.
The coefficient of x in Equation (1) is 4.
Refer back to the list of multiples of 4 and look for the multiplier above 12 (the least common multiple.)
We must multiple Equation (1) by 3.
We must multiple Equation (1) by 3.
6
Find which number we have to multiply Equation (2) by.
The coefficient of x in Equation (2) is 6.
Refer back to the list of multiples of 6 and look for the multiplier above 12 (the least common multiple.)
We must multiple Equation (2) by 2.
We must multiple Equation (2) by 2.
7
Multiply Equation (1) by 3 and Equation (2) by 2.
Now the coefficient of the x terms is 12 in both equations.
Now the coefficient of the x terms is 12 in both equations.
8
Subtract the new equations.
Subtract 2 × Equation (2) from 3 × Equation (1).
9
Subtract the x terms.
12x − 12x = 0
We have eliminated the x.
10
Subtract the y terms.
9y − 2y = 7y
11
Subtract the constants.
24 − 10 = 14
12
We have eliminated one unknown (the x).
Solve for y.
7y = 14
7y ÷ 7 = 14 ÷ 7
y = 2
y = 2 is a solution to the simultaneous equations.
13
Substitute the variable we have just found (y = 2) into one of the equations.
Solve for x.
12x + 2y = 10
12x + 2( 2 ) = 10
12x + 2 × 2 = 10
12x + 4 = 10
12x + 4 − 4 = 10 − 4
12x = 6
12x ÷ 12 = 6 ÷ 12
x = ^{1}⁄_{2}
x = ^{1}⁄_{2} is a solution to the simultaneous equations.
Answer:
We have solved the simultaneous equations:
x = ^{1}⁄_{2}, y = 2 solves
4x + 3y = 8
6x + y = 5
6x + y = 5
Top Tip
Add or Subtract?
 If the unknown you wish to eliminate has the same sign, subtract the equations.
 If the unknown you wish to eliminate has different signs, add the equations.
Beware
Be Careful When Subtracting Equations
Consider the simultaneous equations shown below:
x + y = 3 ... (1)
x − y = 1 ... (2)
You decide to eliminate the x by subtracting Equation (2) from Equation (1).

The x's cancel:
x − x = 0

Be careful with the y terms:
y − (−y) = y −− y y − (−y) = y + y = 2yBy subtracting a negative letter, you are adding the positive letter.

Subtract the constants:
3 − 1 = 2
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