How to Solve Simultaneous Equations Using Elimination Where the Coefficients Differ

Solving Simultaneous Equations Using Elimination Where the Coefficients Differ

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.

Question

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

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

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.

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 = 12

x = 12 is a solution to the simultaneous equations.

Answer:

We have solved the simultaneous equations:

x = 12, y = 2 solves

4x + 3y = 8
6x + y = 5

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See Also

What are simultaneous equations? What is an equation? What is a coefficient? What is a variable? What is a constant? What is a term in algebra? Adding letters in algebra Subtracting letters in algebra Adding terms in algebra Subtracting terms in algebra What is a multiple? What is the least common multiple?