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Solving Simultaneous Equations Using Elimination Where the Coefficients Differ

(KS4, Year 10)

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The Test

Here is a -question, multi-choice test for the "Solving Simultaneous Equations Using Elimination Where the Coefficients Differ" lesson. The pass mark is 90%. Don't worry! All the information you need to pass is in the lesson section under the test.

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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: solve simultaneous equations coefficients differ 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.

Question

Solve the simultaneous equations shown below using the elimination method.
solve simultaneous equations elimination coefficients differ example 1 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: solve simultaneous equations elimination coefficients differ example 1 step 3 Now the coefficient of the x terms is 2 in both equations.

4

Subtract the new equations. Subtract Equation (1) from 2 × Equation (2). solve simultaneous equations elimination coefficients differ example 1 step 4

5

Subtract the x terms.
2x − 2x = 0
solve simultaneous equations elimination coefficients differ example 1 step 5 We have eliminated the x.

6

Subtract the y terms.
6y − y = 5y
solve simultaneous equations elimination coefficients differ example 1 step 6

7

Subtract the constants.
14 − 4 = 10
solve simultaneous equations elimination coefficients differ example 1 step 7

8

We have eliminated one unknown (the x). Solve for y. solve simultaneous equations elimination coefficients differ example 1 step 8

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.

Question

Solve the simultaneous equations shown below using the elimination method. solve simultaneous equations elimination coefficients differ example 2

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): solve simultaneous equations elimination coefficients differ example 2 step 3 multiples of 4
  • 6 is the coefficient of x in Equation (2): solve simultaneous equations elimination coefficients differ example 2 step 3 multiples of 6

4

Find the smallest multiple that appears in both lists. solve simultaneous equations elimination coefficients differ example 2 step 4 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.) solve simultaneous equations elimination coefficients differ example 2 step 5 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.) solve simultaneous equations elimination coefficients differ example 2 step 6 We must multiple Equation (2) by 2.

7

Multiply Equation (1) by 3 and Equation (2) by 2. solve simultaneous equations elimination coefficients differ example 2 step 7 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). solve simultaneous equations elimination coefficients differ example 2 step 8

9

Subtract the x terms.
12x − 12x = 0
solve simultaneous equations elimination coefficients differ example 2 step 9 We have eliminated the x.

10

Subtract the y terms.
9y − 2y = 7y
solve simultaneous equations elimination coefficients differ example 2 step 10

11

Subtract the constants.
24 − 10 = 14
solve simultaneous equations elimination coefficients differ example 2 step 11

12

We have eliminated one unknown (the x). Solve for y. solve simultaneous equations elimination coefficients differ example 2 step 12

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.

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 = 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.
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This page was written by Stephen Clarke.

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