Solving Simultaneous Equations Using Elimination Where the Coefficients Differ
(KS4, Year 10)
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.
This is what we mean by the coefficient of each unknown differing.
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.
Solve the simultaneous equations shown below using the elimination method.
Because the coefficients of the
x and
y differ, we cannot
eliminate x or
y by
adding or
subtracting Equation (1) and
Equation (2).
The way forward is to
multiply one or both equations so that the coefficient of one of the unknowns are the same.
Step-by-Step:
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)
-
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.
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
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
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.
Solve the simultaneous equations shown below using the elimination method.
Step-by-Step:
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)
- The coefficient of x in Equation (1) is 4.
- The coefficient of x in Equation (2) is 6.
Because the coefficients of
x differ, we cannot eliminate it by addition or subtraction.
We make the coefficients the same by multiplying one or both of the equations by a number.
To find out which numbers to multiply the equations by, we will find the
least common multiple of the coefficients
4 and
6.
3
List the
multiples of
4 and
6, by multiplying them by successive
integers.
- 4 is the coefficient of x in Equation (1):
- 6 is the coefficient of x in Equation (2):
4
Find the smallest multiple that appears in both lists.
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.
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.
7
Multiply
Equation (1) by
3 and
Equation (2) by
2.
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
Lesson Slides
The slider below shows another real example of how to solve simultaneous equations using the elimination method where the coefficients differ.
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.
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 = 2y
By subtracting a negative letter, you are adding the positive letter.
-
Subtract the constants:
3 − 1 = 2
By subtracting the equations, we find
2y = 2, from which we can solve the equations.