Simultaneous Linear Equations

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This pdf includes the following topics:-
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
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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.
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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.
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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
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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
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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.
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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)
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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
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