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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.**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.**2**x + 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

Now the coefficient of the

*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.**4**x + 3y = 8 ... (1)

**6**x + 3y = 5 ... (2)

- The coefficient of
**x**in*Equation (1)*is**4**. - The coefficient of
**x**in*Equation (2)*is**6**.

**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

## 4

Find the smallest multiple that appears in both lists.

**12**is the least common multiple.## 5

Find which number we have to multiply

We must multiple

*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

We must multiple

*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

Now the coefficient of the

*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 =**is a solution to the simultaneous equations.^{1}⁄_{2}## Answer:

We have solved the simultaneous equations:**x = ^{1}⁄_{2}**,

**y = 2**solves

4x + 3y = 8

6x + y = 5

## Top Tip

## Add or Subtract?

- If the unknown you wish to eliminate has the
**s**ame sign,**s**ubtract the equations. - If the unknown you wish to eliminate has
**d**ifferent signs, a**dd**the equations.

## Beware

## Be Careful When Subtracting Equations

Consider the simultaneous equations shown below:x + y = 3 ... (1)

x − y = 1 ... (2)

*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

- Subtract the constants:
3 − 1 = 2

**2y = 2**, from which we can solve the equations.

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