Department of Economics | Maths Revision Notes 3
1.
Equations
An equation is any algebraic expression that contains an equals sign.
Some equations express statements that are unconditionally true, and are called identities.
Example:
2(a + b) = 2a + 2b (true for any values of a and b)
Other equations express statements that are conditionally true. Example: x2 = 9. This is
true provided x = +3 or -3. We say that the values x = +3 or x = -3 satisfy the equation
x2 = 9
1.1
Rule for manipulating equations
We manipulate equations in order to extract useful information from them.
The rule is: Any algebraic operation may be performed on an equation, provided the
operation is performed on the whole of both sides of the equation. By "any algebraic
operation" we mean adding, subtracting, multiplying, dividing or raising to any power.
An equation is left unchanged (in the sense that the conditions upon which its truth
depends remain the same) by these operations.
Example:
x2 = 9
(true when x = +3 or -3)
if we add, say, 7 to both sides, we get: x2 + 7 = 9 + 7
which is also true provided x equals either +3 or -3. In other words, the conditional truth
expressed in the statement is left unchanged. That is why adding 7 to both sides is a
legitimate step. Obviously, adding 7 only to the right hand side changes the statement, to
one which is true provided x = +4 or -4.
------------------------------------------Examples of manipulating equations
We have stressed that any operation must be performed on the whole of each side of the
equation. Failure to do this is a very common mistake.
Here are some very simple examples to illustrate the techniques of manipulating
equations, and some of the common mistakes that beginners make.
Example 1. Adding a constant to both sides of an equation
if
x - 4 = 9
we can add 4 to both sides, giving
x - 4 + 4 = 9 + 4
Tidying this up, we are left with
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x = 13
Example 2. Multiplying both sides by a constant
x = 15
if
3
we can multiply both sides by 3, giving
3( 3x ) = 45
Multiplying out the brackets, this becomes
3x
3
= 45
and from this the 3 may be cancelled to give
x = 45
Example 3
if
x + 20
x
+ 4 = 9
we can multiply the whole of both sides by x and get
x( x +x20 + 4) = 9x
When we multiply out the brackets, this becomes
x( x + 20 )
x
+ 4x = 9x
(notice that the 4 becomes 4x)
In the first term on the left hand side, the x's cancel between numerator and denominator
(refer to Revision Notes 1 or 2 if you’re not sure on this point). This gives:
x + 20 + 4x = 9x
Collecting the x's on the left hand side, we get
20 + 5x = 9x
Then we can subtract 5x from both sides, giving
20 = 4x
Finally we divide both sides by 4 and get
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5 = x
Example 4
x 2 x −1
if
3
=
4
we can first multiply both sides by 4, giving
4( 4x ) = 4( 2 x3−1 )
Cancelling the 4's on the left hand side, and multiplying out the right hand side, gives
x =
8x −4
3
Then we can multiply both sides by 3 and get
3x = 8x - 4
from which, by adding 4 to both sides and subtracting 3x from both sides, we get
4 = 5x
Finally, we divide both side by 5 and get
4
5
= x
"Cross multiplying" (This section adds nothing new, hence may be skipped without loss).
The previous example can be used to illustrate a technique called "cross-multiplying".
Given
x
4
= 2 x3−1
we can multiply the x on the left hand side (LHS) by the 3 on the right hand side (RHS),
giving 3x, which we put on one side of a new equation:
3x =
Then we multiply the 4 on the LHS by the 2x - 1 on the RHS, giving 8x - 4, which we put
on the other side of our new equation:
3x = 8x - 4
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Comparing this with the previous example we see that only two more steps are now
necessary to get to the last line:
4
5
= x
Generalising the example above: if
a c
=
, then ad = bc
b d
Thus "cross-multiplying" consists of multiplying the numerator of each side of an
equation by the denominator of the other side, and setting the two results equal to one
another. It is a convenient short cut, nothing more.
Example 5
Given (x - 7)2 = 92
we can raise the whole of both sides to the power ½ to give
[(x - 7)2]½ = [92]½
(Raising both sides to the power ½ is of course the same as taking the square root of both
sides - see Revision Notes 2)
Using Rule 3 from Revision Notes 2, which says that (an)m = anm, this simplifies to
x - 7 = 9
from which
x = 16
---------------------------1.2
Transformations
The previous section showed how to manipulate equations. One frequent use of these
manipulative techniques is to "transform" (or, in simpler language, re-arrange) an
equation so that the information it contains is presented in a different way. These are also
sometimes called "formulae".
3x +4b
=1
For example, given
2c
we might be asked to "express x in terms of b and c". This simply means we have to
isolate x on one side of an equation. We can do this by performing elementary
operations. First, we multiply both sides by 2c, giving
3x + 4b = 2c
Then we subtract 4b from both sides, and divide both sides by 3. This gives
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x =
1.3
2c −4b
3
Solving linear equations with one variable (= unknown)
Variables and parameters
An equation may contain one unknown value (denoted by a letter of the alphabet) or it
may contain more than one. The unknown values in an equation are called (not
surprisingly!) unknowns or variables. The known values in an equation are called
constants, coefficients or parameters.
constants, coefficients or parameters
2x + 3y = 4z
variables or unknowns
Linear and non-linear equations
An equation is said to be linear if none of the variables that appear in it are raised to any
power other than the power 1.
Caution: an equation may look linear when in fact it isn't. For example: x + 3 = 1x .
When we multiply both sides by x, we get x2 + 3 = 1, and the x2 term tells us this is
not a linear equation.
Solving linear equations with one variable
If we are given a linear equation containing only one variable, we can solve it by using
the manipulative techniques explained earlier. For example, suppose we are given
1x
3
+ 4 = 7
This is a linear equation with one variable or unknown (x) and three constants or
parameters ( 13 , 4 and 7).
To solve it, we can multiply both sides by 3, then subtract 12 from both sides, giving
x = 9
Here x = 9 is the solution to the given equation in the sense that when x = 9, the equation
is a true statement. We say that x = 9 is the value of x that satisfies the given equation.
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Can every linear equation with one variable be solved in this way? The answer is yes,
and we can prove this as follows.
Every linear equation with one variable has the general form
ax + b = c
where x is the variable and a, b and c are unspecified parameters. We can find the
solution to this general form by first dividing both sides by the parameter a. Then we
subtract b from both sides, giving
a
x = ca − ba
which may also be written as
x = c-b
a
Notice that we can check that this solution is indeed correct by substituting our solution
for x into the given equation. Thus, in the general form, we put c-b in place of x, which
a
gives
a( c-b
) + b = c
a
which, after cancelling the a's, reduces to
c - b + b= c
that is, an identity. It is true whatever the values of
b and c.
2.
Linear equations with two variables
An equation such as
y = 2x + 1
is an example of a linear equation with two variables. Clearly, there are many pairs of
values of x and y (such as x = 3, y = 7) which satisfy this equation (that is, which are
consistent with it). There is no unique solution.
For this reason, an equation such as this which contains two or more variables is called a
relation, or a function. A relation, or function, does not pin down the variables involved
to any specific values, but defines a relationship, or mutual dependency, between them.
Dependent and independent variables
In any equation (or function) involving two variables, such as
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y = 2x + 1
The variable that appears on the right hand side of the equation (in this example, x) is
known as the independent variable. We view this variable as being free to take any
value we choose to assign to it.
The variable that stands alone on the left hand side of the equation (in this example, y) is
known as the dependent variable. We view it as being the variable that depends, for its
value, on the value assigned to x. Because of this dependency, we say that y is a function
of x. We use the concept of a function extensively in economics.
----------------------------------3.
Graphical representation
Returning to the equation (or function):
y = 2x + 1
we can show the relationship between x and y graphically.
First, some discussion of the layout and terminology of graphs is necessary. By
convention the independent variable, x, is measured along the horizontal axis, and the
dependent variable, y, along the vertical axis. The point at which the two axes cut is
defined as the point where both x and y have the value zero, and this point is called the
origin. Positive values of x are measured to the right of the origin and negative values to
the left. Positive values of y are measured above the origin and negative values below it.
Every point on the surface of the graph paper (the x - y plane) corresponds to a unique
pair of values of x and y. These values are called the co-ordinates of the point A.
We can now plot the graph of y = 2x + 1. First we draw up a table of values, then
transfer the values of x and y to the graph paper, and finally join up the points (next
page).
x
y = 2x + 1
-3
2(-3) + 1 = -5
-2
2(-2) + 1 = -3
Table of values for y = 2x + 1, for x = -3 to +3
-1
2(-1) + 1 = -1
(See graph, next page)
0
2(0) + 1 = 1
1
2(1) + 1 = 3
2
2(2) + 1 = 5
3
2(3) + 1 = 7
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y=2x+1
(-2, 3)
7
y
6
5
4
3
2
1
x
0
-3
-2
-1
-1
0
1
2
3
-2
-3
-4
-5
y = 0, so 2x + 1 = 0
The slope and intercept of a linear function
The graph of y = 2x + 1 has two important features. First, the graph is a straight line.
Further, this straight line has a gradient (or slope) of 2 - meaning that every increase in x
by one unit is associated with an increase in y by 2 units. For example when x increases
from 1 to 2, y increases from 3 to 5.
Second, the graph cuts the y-axis at y = 1. This point is called the y-intercept. At this
point, x = 0.
Generalisation. These two properties are true of every linear function. Hence, if we are
given the linear function:
y = ax + b
where a and b are given constants (parameters), we can say straight away that the graph
of the function is a straight line, with slope (gradient) determined by the parameter a, and
intercept on the y-axis determined by the parameter b. (We’ll come back to this later)
Positive and negative gradients
For any linear function:
y = ax + b
if the parameter a is positive, the graph slopes upwards from left to right, and we say that
the gradient is positive. Conversely, if a is negative, the graph slopes downward from left
to right, and we say that the gradient is negative (an increase in x results in a decrease in
y).
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The case where the graph is horizontal can also be handled. If the coefficient a is zero,
the function y = ax + b degenerates into y = 0(x) + b = b. Thus x disappears from
the equation, and therefore variation in x has no effect on y. We are left with y = b, and
since the parameter b gives the intercept on the y axis, the graph is therefore a horizontal
line at a distance b units from the x axis.
The case of a vertical line is a little harder to grasp. In y = ax + b, with a assumed
positive, we know that as a increases in value so the graph slopes upward more and more
steeply. The extreme or “limiting” case of this is when a goes off to infinity, and the
graph becomes a vertical straight line. (We return to this later).
Implicit linear functions
We now know that the graph of y = ax + b is a straight line with gradient given by a
and y-intercept given by b.
Sometimes, however, a linear function is presented to us the form:
Ax + By + C = 0
where A, B, and C are parameters (constants).
This is known as an implicit linear function, because x and y both appear on the same
side of the equation.
But, by subtracting Ax + C from both sides, and dividing both sides by B, we can
transform the function into the explicit form
y = −BA x − CB
This reformulation of the relationship between the two variables is called an explicit
function. By isolating y on one side of the equation, it shows us explicitly how y depends
on x.
This function when graphed will therefore have a gradient of −BA and an intercept on the
y-axis of − C
. Whether the gradient or the intercept are positive or negative will depend,
B
of course, on whether A, B and C are positive or negative.
4.
Graphical solution of linear equations
Suppose we are asked to solve the linear equation
2x + 1 = 0
We can solve this equation graphically. First, we plot the graph of
y = 2x + 1
(see graph on previous page)
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Then we look for the point on the graph where y = 0. At this point, it will also be true
that 2x + 1 = 0, since by definition y = 2x + 1.
Since y is positive everywhere above the x-axis and negative everywhere below it, the
point where y = 0 must be at the point where the graph cuts the x-axis. From inspection
of the graph, we can see that the value of x at this point is x = - 12 .
This approach is valid in general, for any linear function y = ax + b. The point at which
the graph cuts the x-axis is the solution to the equation ax + b = 0.
In fact, the reasoning is valid for any function of x, including even a non-linear function.
We will return to this point later.
-----------------------------------------------------------5.
Simultaneous linear equations
Example 1
Suppose we are told that two variables x and y are related to one another by the function
(or equation)
y = 3x
(equation (1))
and also by the function
y = x + 10
(equation (2))
Our task is to find the pair (or pairs) of values of x and y that satisfy both of these
equations simultaneously. This is what "solving simultanously equations" means.
The way to solve this pair of simultaneous equations is to make use of the fact that, if
they both hold simultaneously, the value of y in equation (1) will be the same as the value
of y in equation (2). Therefore, the left hand side of equation (1) and the left hand side of
equation (2) will be equal to one another. But, if the two left hand sides are equal, then
the two right hand side must be equal too. So we will have
3x = x + 10
This is a linear equation with only one variable or unknown, x. Its solution is
x = 5
This is the only value of x which satisfies both equations simultaneously. If x takes on
any other value, then the value of y in equation (1) will differ from the value of y in
equation (2).
To complete our solution, we need to find the value of y when x = 5. We can find this
by substituting x = 5 into either equation (1) or equation (2). Substituting x = 5 into
equation (1) gives
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y = 3(5) = 15
So x = 5, y = 15 is the solution.
Example 2
y = 3x + 4
Equation (i)
2y + 6 = -x
Equation (ii)
Here we cannot immediately proceed as we did in the previous example, because neither
the left hand side nor the right hand side of the two equations are equal to one another.
We have to produce this equality by performing elementary operations on one or both
equations. The easiest thing to do in this case is to subtract 6 from both sides of equation
(ii), and divide both sides by 2. This gives
−x −6
y=
Equation (iii)
2
Since equation (iii) is simply a rearrangement of equation (ii), there is no reason why we
can't simply replace equation (ii) with equation (iii). Thus our pair of simultaneous
equations become
y = 3x + 4
Equation (i)
and
−x −6
y=
Equation (iii)
2
Now we can proceed as in the previous example. When these two equations are true
simultaneously, their left hand sides will be equal, and so too therefore will their right
hand sides be. Thus we can set the right hand sides equal to one another, and get
−x −6
3x + 4 =
2
By elementary operations this can be rearranged to get
6x + 8 = -x - 6
from which
7x = -14, and therefore x = -2. Finally, y = 3x + 4 = 3(-2) + 4 = -2
This method of solution is called the elimination method. Given two equations involving
two variables, we eliminated one variable (y) and hence were left with one equation
involving one variable (x). In some examples it may be more convenient to eliminate x
rather than y; this is equally valid. All techniques for solving simultaneous equations are
variants of this method.
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6.
Graphical solution of simultaneous linear equations
The first example of simultaneous equations that we just considered was
y = 3x
(1)
y = x + 10
(2)
If we plot the graphs of these two functions on the same diagram, we see that the two
graphs intersect at the point where x = 5, y = 15 (see Fig. below). These values were, of
course, our solution to the simultaneous equations. On reflection we should not find this
surprising. For the point of intersection lies, by definition, on both functions. Therefore
the co-ordinates of this point satisfy both equations simultaneously.
y 35
30
25
y = 3x
20
15
y = x + 10
10
5
-5
0
-3
-5
1
3
5
7
9
x
-10
-15
-20
This conclusion is valid for any pair of simultaneous equations, whether linear or not.
(Non-linear cases will be considered later.)
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