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This pdf includes the following topics:-

Introduction

Solving simultaneous equations - method of substitution

Solving simultaneous equations - method of elimination Examples

Introduction

Solving simultaneous equations - method of substitution

Solving simultaneous equations - method of elimination Examples

1.
Simultaneous linear

The purpose of this section is to look at the solution of simultaneous linear equations. We will

see that solving a pair of simultaneous equations is equivalent to ﬁnding the location of the

point of intersection of two straight lines.

In order to master the techniques explained here it is vital that you undertake plenty of practice

exercises so that all this becomes second nature. To help you to achieve this, the unit includes

a number of such exercises.

After reading this text, and/or viewing the video tutorial on this topic, you should be able to:

• solve pairs of simultaneous linear equations

• recognise that this is equivalent to ﬁnding the point of intersection of two straight line

graphs

Contents

1. Introduction 2

2. Solving simultaneous equations - method of substitution 4

3. Solving simultaneous equations - method of elimination 6

4. Examples 6

1

c mathcentre August 7, 2003

The purpose of this section is to look at the solution of simultaneous linear equations. We will

see that solving a pair of simultaneous equations is equivalent to ﬁnding the location of the

point of intersection of two straight lines.

In order to master the techniques explained here it is vital that you undertake plenty of practice

exercises so that all this becomes second nature. To help you to achieve this, the unit includes

a number of such exercises.

After reading this text, and/or viewing the video tutorial on this topic, you should be able to:

• solve pairs of simultaneous linear equations

• recognise that this is equivalent to ﬁnding the point of intersection of two straight line

graphs

Contents

1. Introduction 2

2. Solving simultaneous equations - method of substitution 4

3. Solving simultaneous equations - method of elimination 6

4. Examples 6

1

c mathcentre August 7, 2003

2.
1. Introduction

The purpose of this section is to look at the solution of elementary simultaneous linear equations.

Before we do that, let’s just have a look at a relatively straightforward single equation. The

equation we are going to look at is

2x − y = 3

This is a linear equation. It is a linear equation because there are no terms involving x2 , y 2 or

x × y, or indeed any higher powers of x and y. The only terms we have got are terms in x, terms

in y and some numbers. So this is a linear equation.

We can rearrange it so that we obtain y on its own on the left hand side. We can add y to each

side so that we get

2x = 3 + y

Now let’s take 3 away from each side.

2x − 3 = y

This gives us an expression for y: namely y = 2x − 3.

Suppose we choose a value for x, say x = 1, then y will be equal to:

y = 2 × 1 − 3 = −1

Suppose we choose a diﬀerent value for x, say x = 2.

y =2×2−3=1

Supppose we choose another value for x, say x = 0.

y = 2 × 0 − 3 = −3

For every value of x we can generate a value of y.

We can plot these as points on a graph. We can plot the ﬁrst as the point (1, −1). We can plot

the second one as the point (2, 1), and the third one as the point (0, −3) and so on. Plotting

the points on a graph, as shown in Figure 1, we see that these three points lie on a straight line.

This is the line with equation y = 2x − 3. It is a straight line and this is another reason for

calling the equation a linear equation.

y

2

y = 2x − 3

1

0 1 2 x

-1

-2

-3

Figure 1. Graph of y = 2x − 3.

c mathcentre August 7, 2003 2

The purpose of this section is to look at the solution of elementary simultaneous linear equations.

Before we do that, let’s just have a look at a relatively straightforward single equation. The

equation we are going to look at is

2x − y = 3

This is a linear equation. It is a linear equation because there are no terms involving x2 , y 2 or

x × y, or indeed any higher powers of x and y. The only terms we have got are terms in x, terms

in y and some numbers. So this is a linear equation.

We can rearrange it so that we obtain y on its own on the left hand side. We can add y to each

side so that we get

2x = 3 + y

Now let’s take 3 away from each side.

2x − 3 = y

This gives us an expression for y: namely y = 2x − 3.

Suppose we choose a value for x, say x = 1, then y will be equal to:

y = 2 × 1 − 3 = −1

Suppose we choose a diﬀerent value for x, say x = 2.

y =2×2−3=1

Supppose we choose another value for x, say x = 0.

y = 2 × 0 − 3 = −3

For every value of x we can generate a value of y.

We can plot these as points on a graph. We can plot the ﬁrst as the point (1, −1). We can plot

the second one as the point (2, 1), and the third one as the point (0, −3) and so on. Plotting

the points on a graph, as shown in Figure 1, we see that these three points lie on a straight line.

This is the line with equation y = 2x − 3. It is a straight line and this is another reason for

calling the equation a linear equation.

y

2

y = 2x − 3

1

0 1 2 x

-1

-2

-3

Figure 1. Graph of y = 2x − 3.

c mathcentre August 7, 2003 2

3.
Suppose we take a second linear equation 3x + 2y = 8 and plot its graph on the same ﬁgure. A

quick way to achieve this is as follows.

When x = 0, 2y = 8, so y = 4. Therefore the point (0, 4) lies on the line.

8 2

When y = 0, 3x = 8, so x = = 2 . Therefore the point (2 23 , 0) lies on the line.

3 3

Because this is a linear equation we know its graph is a straight line, so we can obtain this by

joining up the points. Both straightline graphs are shown in Figure 2.

y

4

3 3x + 2y = 8

2

y = 2x − 3

1

0 1 2 x

-1

-2

-3

Figure 2. Graphs of y = 2x − 3 and 3x + 2y = 8

When we solve a pair of simultaneous equations what we are actually looking for is the intersec-

tion of two straight lines because it is this point that satisﬁes both equations at the same time.

From Figure 2 we see that this occurs at the point where x = 2 and y = 1.

Of course it could happen that we have two parallel lines; they would never meet, and hence

the simultaneous equations would not have a solution. We shall observe this behaviour in one

of the examples which follows.

Key Point

When solving a pair of simultaneous linear equations we are, in fact, ﬁnding a common point

- the point of intersection of the two lines.

3

c mathcentre August 7, 2003

quick way to achieve this is as follows.

When x = 0, 2y = 8, so y = 4. Therefore the point (0, 4) lies on the line.

8 2

When y = 0, 3x = 8, so x = = 2 . Therefore the point (2 23 , 0) lies on the line.

3 3

Because this is a linear equation we know its graph is a straight line, so we can obtain this by

joining up the points. Both straightline graphs are shown in Figure 2.

y

4

3 3x + 2y = 8

2

y = 2x − 3

1

0 1 2 x

-1

-2

-3

Figure 2. Graphs of y = 2x − 3 and 3x + 2y = 8

When we solve a pair of simultaneous equations what we are actually looking for is the intersec-

tion of two straight lines because it is this point that satisﬁes both equations at the same time.

From Figure 2 we see that this occurs at the point where x = 2 and y = 1.

Of course it could happen that we have two parallel lines; they would never meet, and hence

the simultaneous equations would not have a solution. We shall observe this behaviour in one

of the examples which follows.

Key Point

When solving a pair of simultaneous linear equations we are, in fact, ﬁnding a common point

- the point of intersection of the two lines.

3

c mathcentre August 7, 2003

4.
2. Solving simultaneous equations - method of substitution

How can we handle the two equations algebraically so that we do not have to draw graphs? We

are going to look at two methods of solution. In this Section we will look at the ﬁrst method -

the method of substitution.

Let us return to the two equations we met in Section 1.

2x − y = 3 (1)

3x + 2y = 8 (2)

By rearranging Equation (1) we ﬁnd

y = 2x − 3 (3)

We can now substitute this expression for y into Equation (2).

3x + 2(2x − 3) = 8

3x + 4x − 6 = 8

7x − 6 = 8

7x = 14

x = 2

Finally, using Equation (3), y = 2 × 2 − 3 = 1. So x = 2, y = 1 is the solution to the pair of

simultaneous equations.

This solution should always be checked by substituting back into both original equations to

ensure that the left- and right- hand sides are equal for these values of x and y. So, with x = 2,

y = 1, the left-hand side of Equation (1) is 2(2) − 1 = 3, which is the same as the right-hand

side. With x = 2, y = 1, the left-hand side of Equation (2) is 3(2) + 2(1) = 8, which is the same

as the right-hand side.

Let’s have a look at another example using this particular method.

The example we are going to use is

7x + 2y = 47 (1)

5x − 4y = 1 (2)

Now we need to make a choice. We need to choose one of these two equations and re-arrange it

to obtain an expression for y, or if we wish, for x

The choice is entirely ours and we have to make the choice based upon what we feel will be

the simplest. Looking at a pair of equations like this, it is often diﬃcult to know which is the

c mathcentre August 7, 2003 4

How can we handle the two equations algebraically so that we do not have to draw graphs? We

are going to look at two methods of solution. In this Section we will look at the ﬁrst method -

the method of substitution.

Let us return to the two equations we met in Section 1.

2x − y = 3 (1)

3x + 2y = 8 (2)

By rearranging Equation (1) we ﬁnd

y = 2x − 3 (3)

We can now substitute this expression for y into Equation (2).

3x + 2(2x − 3) = 8

3x + 4x − 6 = 8

7x − 6 = 8

7x = 14

x = 2

Finally, using Equation (3), y = 2 × 2 − 3 = 1. So x = 2, y = 1 is the solution to the pair of

simultaneous equations.

This solution should always be checked by substituting back into both original equations to

ensure that the left- and right- hand sides are equal for these values of x and y. So, with x = 2,

y = 1, the left-hand side of Equation (1) is 2(2) − 1 = 3, which is the same as the right-hand

side. With x = 2, y = 1, the left-hand side of Equation (2) is 3(2) + 2(1) = 8, which is the same

as the right-hand side.

Let’s have a look at another example using this particular method.

The example we are going to use is

7x + 2y = 47 (1)

5x − 4y = 1 (2)

Now we need to make a choice. We need to choose one of these two equations and re-arrange it

to obtain an expression for y, or if we wish, for x

The choice is entirely ours and we have to make the choice based upon what we feel will be

the simplest. Looking at a pair of equations like this, it is often diﬃcult to know which is the

c mathcentre August 7, 2003 4

5.
Let’s choose Equation (2) and rearrange it to ﬁnd an expression for x.

5x − 4y = 1

5x = 1 + 4y by adding 4y to each side

1 + 4y

x = by dividing both sides by 5

5

We now use this expression for x and substitute it in Equation (1).

1 + 4y

7 + 2y = 47

5

Now multiply throughout by 5. Why? Because we want to get rid of the fraction and the way

to do that is to multiply everything by 5.

7(1 + 4y) + 10y = 235

Now we need to multiply out the brackets

7 + 28y + 10y = 235

Gather the y’s and subtract 7 from each side to get

38y = 228

228

y= =6

38

So we have established that y = 6. Having done this we can substitute it back into the equation

that we ﬁrst had for x.

1 + 4y 1 + 24

x= =

5 5

and so

x=5

So again, we have our pair of values - our solution to the pair of simultaneous equations. In order

to check that our solution is correct these values should be substituted into both equations to

ensure they balance. So, with x = 5, y = 6, the left-hand side of Equation (1) is 7(5)+2(6) = 47,

which is the same as the right-hand side. With x = 5, y = 6, the left-hand side of Equation (2)

is 5(5) − 4(6) = 1, which is the same as the right-hand side.

1. Solve the following pairs of simultaneous equations:

y = 2x + 3 y = 3x − 1 6x + y = 4

a) b) c)

y = 5x − 3 2x + 4y = 10 5x + 2y = 1

x − 3y = 1 2x + 13 y = 1 4x + 3y = 5

d) e) f)

2x + 5y = 35 3x + 5y = 6 2x − 34 y = 1

5

c mathcentre August 7, 2003

5x − 4y = 1

5x = 1 + 4y by adding 4y to each side

1 + 4y

x = by dividing both sides by 5

5

We now use this expression for x and substitute it in Equation (1).

1 + 4y

7 + 2y = 47

5

Now multiply throughout by 5. Why? Because we want to get rid of the fraction and the way

to do that is to multiply everything by 5.

7(1 + 4y) + 10y = 235

Now we need to multiply out the brackets

7 + 28y + 10y = 235

Gather the y’s and subtract 7 from each side to get

38y = 228

228

y= =6

38

So we have established that y = 6. Having done this we can substitute it back into the equation

that we ﬁrst had for x.

1 + 4y 1 + 24

x= =

5 5

and so

x=5

So again, we have our pair of values - our solution to the pair of simultaneous equations. In order

to check that our solution is correct these values should be substituted into both equations to

ensure they balance. So, with x = 5, y = 6, the left-hand side of Equation (1) is 7(5)+2(6) = 47,

which is the same as the right-hand side. With x = 5, y = 6, the left-hand side of Equation (2)

is 5(5) − 4(6) = 1, which is the same as the right-hand side.

1. Solve the following pairs of simultaneous equations:

y = 2x + 3 y = 3x − 1 6x + y = 4

a) b) c)

y = 5x − 3 2x + 4y = 10 5x + 2y = 1

x − 3y = 1 2x + 13 y = 1 4x + 3y = 5

d) e) f)

2x + 5y = 35 3x + 5y = 6 2x − 34 y = 1

5

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6.
3. Solving simultaneous equations - method of elimination

We illustrate the second method by solving the simultaneous linear equations:

7x + 2y = 47 (1)

5x − 4y = 1 (2)

We are going to multiply Equation (1) by 2 because this will make the magnitude of the coeﬃ-

cients of y the same in both equations. Equation (1) becomes

14x + 4y = 94 (3)

If we now add Equation (2) and Equation (3) we will ﬁnd that the terms involving y disappear:

5x − 4y = 1

+

14x + 4y = 94

19x = 95

and so

95

x= = 5

19

Now that we have a value for x we can substitute this into Equation (2) in order to ﬁnd y.

5x − 4y = 1

5 × 5 − 4y = 1

25 = 4y + 1

24 = 4y

y = 6

The solution is x = 5, y = 6.

4. Examples

Solve the simultaneous equations

3x + 7y = 27 (1)

5x + 2y = 16 (2)

We will multiply Equation (1) by 5 and Equation (2) by 3 because this will make the coeﬃcients

of x in both equations the same.

15x + 35y = 135 (3)

15x + 6y = 48 (4)

If we now subtract Equation (4) from Equation (3) we can eliminate the terms involving x.

c mathcentre August 7, 2003 6

We illustrate the second method by solving the simultaneous linear equations:

7x + 2y = 47 (1)

5x − 4y = 1 (2)

We are going to multiply Equation (1) by 2 because this will make the magnitude of the coeﬃ-

cients of y the same in both equations. Equation (1) becomes

14x + 4y = 94 (3)

If we now add Equation (2) and Equation (3) we will ﬁnd that the terms involving y disappear:

5x − 4y = 1

+

14x + 4y = 94

19x = 95

and so

95

x= = 5

19

Now that we have a value for x we can substitute this into Equation (2) in order to ﬁnd y.

5x − 4y = 1

5 × 5 − 4y = 1

25 = 4y + 1

24 = 4y

y = 6

The solution is x = 5, y = 6.

4. Examples

Solve the simultaneous equations

3x + 7y = 27 (1)

5x + 2y = 16 (2)

We will multiply Equation (1) by 5 and Equation (2) by 3 because this will make the coeﬃcients

of x in both equations the same.

15x + 35y = 135 (3)

15x + 6y = 48 (4)

If we now subtract Equation (4) from Equation (3) we can eliminate the terms involving x.

c mathcentre August 7, 2003 6

7.
15x + 35y = 135

−

15x + 6y = 48

29y = 87

from which

87

y= = 3

29

If we substitute this result in Equation (1) we can ﬁnd x.

3x + 7y = 27

3x + 21 = 27

3x = 6

x = 2

As before, the solution should be checked by substitution into the original equations. So, with

x = 2, y = 3, the left-hand side of Equation (1) is 3(2) + 7(3) = 27, which is the same as the

right-hand side. With x = 2, y = 3, the left-hand side of Equation (2) is 5(2) + 2(3) = 16, which

is the same as the right-hand side.

All the examples that we have looked at so far have all had whole number coeﬃcients; let’s have

a look at a couple that don’t look like the ones we have just done.

Solve the simultaneous equations

x = 3y

x

− y = 34

3

First of all let us rearrange the ﬁrst equation so that x and y terms are on the left. We will also

multiply the second equation by 3 to remove the fraction. These operations give

x − 3y = 0

x − 3y = 102

Notice that the terms on the left in both equations are exactly the same. If we subtract the

equations we will ﬁnd 0 = −102. This does not make sense. Remember right at the beginning

of this unit we explained that if two lines are parallel they will not intersect. This is the case

here. There are no solutions.

x y

− = 0 (1)

5 4

1

3x + y = 17 (2)

2

Observe that if both sides of Equation (1) are multiplied by 20 we can remove the fractions:

4x − 5y = 0 (3)

7

c mathcentre August 7, 2003

−

15x + 6y = 48

29y = 87

from which

87

y= = 3

29

If we substitute this result in Equation (1) we can ﬁnd x.

3x + 7y = 27

3x + 21 = 27

3x = 6

x = 2

As before, the solution should be checked by substitution into the original equations. So, with

x = 2, y = 3, the left-hand side of Equation (1) is 3(2) + 7(3) = 27, which is the same as the

right-hand side. With x = 2, y = 3, the left-hand side of Equation (2) is 5(2) + 2(3) = 16, which

is the same as the right-hand side.

All the examples that we have looked at so far have all had whole number coeﬃcients; let’s have

a look at a couple that don’t look like the ones we have just done.

Solve the simultaneous equations

x = 3y

x

− y = 34

3

First of all let us rearrange the ﬁrst equation so that x and y terms are on the left. We will also

multiply the second equation by 3 to remove the fraction. These operations give

x − 3y = 0

x − 3y = 102

Notice that the terms on the left in both equations are exactly the same. If we subtract the

equations we will ﬁnd 0 = −102. This does not make sense. Remember right at the beginning

of this unit we explained that if two lines are parallel they will not intersect. This is the case

here. There are no solutions.

x y

− = 0 (1)

5 4

1

3x + y = 17 (2)

2

Observe that if both sides of Equation (1) are multiplied by 20 we can remove the fractions:

4x − 5y = 0 (3)

7

c mathcentre August 7, 2003

8.
If Equation (2) is multiplied by 2 we can remove the fraction there too.

6x + y = 34 (4)

Now multiply Equation (4) by 5:

30x + 5y = 170 (5)

We can now add (3) and (5) to obtain

34x = 170

170

x= =5

34

Substituting this value into Equation (1) gives

x y

− = 0

5 4

y

1− = 0

4

from which y = 4.

So the solution is: x = 5, y = 4. As before, this should be checked by substitution into the

original equations. So, with x = 5, y = 4, the left-hand side of Equation (1) is 55 − 44 = 0, which

is the same as the right-hand side. With x = 5, y = 4, the left-hand side of Equation (2) is

3(5) + 12 (4) = 17, which is the same as the right-hand side.

To summarise:

A pair of simultaneous equations represent two straight lines. In eﬀect when we solve them we

are looking for the point where the two straight lines intersect. The method of elimination is

much better to use than the ﬁrst method.

Remember the answer you get can always be checked by substituting the pair of values into the

original equations.

2. Use elimination to solve the following pairs of simultaneous equations.

5x + 3y = 9 2x − 3y = 9 x + 7y = 10

a) b) c)

2x − 3y = 12 2x + y = 13 3x − 2y = 7

5x + y = 10

1

3

x+ y = 10

3

3x − 2y = 52

d) e) f)

7x − 3y = 14 2x + 1

4

y = 11

4

1

3

x + 3y = − 43

3. Solve the following pairs of simultaneous equations by a method of your choice.

x = 3y x = 13 y 7x + 3y = −15

a) b) c)

4x − 5y = 35 2y − 6x = 9 12y − 5x = 39

a) x = 2, y = 7 b) x = 1, y = 2 c) x = 1, y = −2

1.

d) x = 10, y = 3 e) x = 1/3, y = 1 f) x = 3/4, y = 2/3

a) x = 3, y = −2 b) x = 6, y = 1 c) x = 3, y = 1

2.

d) x = 2, y = 0 e) x = 1, y = 3 f) x = 1/2, y = −1/2

3. a) x = 15, y = 5 b) no solution c) x = −3, y = 2

c mathcentre August 7, 2003 8

6x + y = 34 (4)

Now multiply Equation (4) by 5:

30x + 5y = 170 (5)

We can now add (3) and (5) to obtain

34x = 170

170

x= =5

34

Substituting this value into Equation (1) gives

x y

− = 0

5 4

y

1− = 0

4

from which y = 4.

So the solution is: x = 5, y = 4. As before, this should be checked by substitution into the

original equations. So, with x = 5, y = 4, the left-hand side of Equation (1) is 55 − 44 = 0, which

is the same as the right-hand side. With x = 5, y = 4, the left-hand side of Equation (2) is

3(5) + 12 (4) = 17, which is the same as the right-hand side.

To summarise:

A pair of simultaneous equations represent two straight lines. In eﬀect when we solve them we

are looking for the point where the two straight lines intersect. The method of elimination is

much better to use than the ﬁrst method.

Remember the answer you get can always be checked by substituting the pair of values into the

original equations.

2. Use elimination to solve the following pairs of simultaneous equations.

5x + 3y = 9 2x − 3y = 9 x + 7y = 10

a) b) c)

2x − 3y = 12 2x + y = 13 3x − 2y = 7

5x + y = 10

1

3

x+ y = 10

3

3x − 2y = 52

d) e) f)

7x − 3y = 14 2x + 1

4

y = 11

4

1

3

x + 3y = − 43

3. Solve the following pairs of simultaneous equations by a method of your choice.

x = 3y x = 13 y 7x + 3y = −15

a) b) c)

4x − 5y = 35 2y − 6x = 9 12y − 5x = 39

a) x = 2, y = 7 b) x = 1, y = 2 c) x = 1, y = −2

1.

d) x = 10, y = 3 e) x = 1/3, y = 1 f) x = 3/4, y = 2/3

a) x = 3, y = −2 b) x = 6, y = 1 c) x = 3, y = 1

2.

d) x = 2, y = 0 e) x = 1, y = 3 f) x = 1/2, y = −1/2

3. a) x = 15, y = 5 b) no solution c) x = −3, y = 2

c mathcentre August 7, 2003 8