Simplifying Algebraic Expressions- In the algebraic expression
[ad_1]x is called the variable, and 7 is known as the coefficient of x3. Expressions
consisting simply of a coefficient multiplying one or more variables raised to the
power of a positive integer are called monomials. Monomials can be added or
subtracted to form polynomials. Each of the monomials comprising a polynomial
is called a term. For example, the terms in the polynomial 3×2+2x+1
are 3×2, 2x, and 1. The coefficient of x2 is 3, the coefficient of x is 2, and the
constant term is 1.
To add or subtract two polynomials, we collect like terms and add or subtract
their coefficients. For example, if we wish to add 7x+2 and 5− 2x, then
we collect the terms in x and the constant terms:
(7x + 2) + (5 − 2x) = (7+(−2))x + (2 + 5) = 5x + 7.
Example 1.8
Simplify the following:
- (3×2 + 2x + 1) + (5×2 − x − 7),
- (9×4 + 12×3 + 6x + 1) − (x4 + 2×2 − 4),
- (x3 + 4x − 5) + (2×2 − x + 8).
Solution. - (3 + 5)x2 + (2 − 1)x + (1 − 7) = 8×2 + x − 6.
- (9 − 1)x4 + 12×3 − 2×2 + 6x + (1 + 4) = 8×4 + 12×3 − 2×2 + 6x + 5.
- x3 + 2×2 + (4 − 1)x + (−5 + 8) = x3 + 2×2 + 3x + 3.
1.7.1 Multiplying Brackets
There are occasions when mathematical expressions may be simplified by removing
any brackets present. This process, which is also known as expanding
the brackets or multiplying out the brackets, culminates in an equivalent expression
without brackets. The removal of brackets is based on the following
basic rule:
a(b + c) = ab + ac, (1.17) - Essential Skills 17
where a, b, and c are any three numbers. Since the order in which multiplication
is performed is not important, we also have
(b + c)a = ba + ca, (1.18)
The rules (1.17) and (1.18), which are examples of what is known as the distributive
law, may be generalized to include expressions involving polynomials.
For example,
3(x + 2y) = 3x + 6y,
and
−2(3×2 − 5y) = −6×2 + 10y.
It is important to take care multiplying out brackets when there is a negative
sign outside the brackets. In this case, the sign of each term inside the brackets
is changed when the brackets are removed. For example,
−(2×2 − 3x − 2y + 5) = −2×2 + 3x + 2y − 5.
We also have the following rule for multiplying two brackets:
(a + b)(c + d) = ac + bc + ad + bd, (1.19)
where a, b, c, and d are any three numbers. So to multiply out two brackets
we simply multiply each term in the second bracket by each term in the first
bracket and add together all contributions. For example,
(x + 2)(2x − 3) = (x)(2x) + (2)(2x) + (x)(−3) + (2)(−3)
= 2×2 + 4x − 3x − 6
= 2×2 + x − 6.
The rule (1.19) extends to brackets containing more than two terms. The important
thing to remember is that each term in the second bracket is multiplied by
each term in the first before all contributions are added together. For example,
(2x − y + 5)(x − 3) = (2x)(x) + (−y)(x) + (5)(x)
+(2x)(−3) + (−y)(−3) + (5)(−3)
= 2×2 − xy + 5x − 6x + 3y − 15
= 2×2 − xy − x + 3y − 15.
Example 1.9
Multiply out the brackets and simplify the following: - (2x + 3)(7 − 5x),
2.
(120 − 24x)
4.8
, - (x + 3y)(2x − 5y − 1).
18 Elements of Mathematics for Economics and Finance
Solution. - Using the rule (1.19), we have
(2x + 3)(7 − 5x) = (2x)(7) + (3)(7) + (2x)(−5x) + (3)(−5x)
= 14x + 21 − 10×2 − 15x
= 21− x − 10×2. - In this example, we just note that division of 120−24x by 0.48 is the same
as multiplication of 120−24x by 1/(4.8), and therefore we can use the rule
(1.17):
(120 − 24x)
4.8
1
4.8
(120 − 24x)
120
4.8
+
−24x
4.8
= 25− 5x.
- Using the generalization of rule (1.19), we have
(x + 3y)(2x − 5y − 1) = (x)(2x) + (3y)(2x) + (x)(−5y)
+(3y)(−5y) + (x)(−1) + (3y)(−1)
= 2×2 + 6xy − 5xy − 15y2 − x − 3y
= 2×2 + xy − 15y2 − x − 3y.
1.7.2 Factorization
Factorization is the reverse process to multiplying out the brackets. It involves
taking a mathematical expression and rewriting it by expressing it in terms of
a product of factors. There are a number of techniques that can be used to
factorize an expression: - The simplest technique is to identify a common factor in two or more
terms. The equivalent factorized expression can then be written in terms
of the common factor multiplying a bracketed expression. For example,
a) ab − ac = a(b − c),
b) 4×2 + 6x = 2x(2x + 3),
c) ax2 − a2x = ax(x − a),
d) −36×2 − 9x = −9x(4x + 1), - Essential Skills 19
e)
5x + 10y
10x − 5y
5(x + 2y)
5(2x − y)
= x + 2y
2x − y
.
- The second technique is based on the following identity involving the
difference of two squares:
a2 − b2 = (a − b)(a + b).
An identity is a formula valid for all values of the variables; in this case, a
and b. The following are examples of the application of this identity:
a) x2 − 36 = (x − 6)(x + 6);
b) 9a2 − 16×2 = (3a)2 − (4x)2 = (3a − 4x)(3a + 4x);
c) 9 − 36×2 = 9(1 − 4×2) = 9(12 − (2x)2) = 9(1 − 2x)(1 + 2x).
An additional technique that can be used for factorizing quadratic expressions
of the form ax2 + bx + c or ax2 + bxy + cy2 will be discussed in Chapter 3.
EXERCISES
1.1. Evaluate
35 − 8 ÷ 22 + 5 + 23 × 4.
1.2. Express the following fractions using decimal notation:
a)
3
10
,
b)
5
16
,
c)
3
4
,
d)
3
13
,
e)
2
7
,
f)
1
19
.
1.3. Simplify the following fractions:
a)
2
5
+
3
8
,
b)
5
16
− 3
32
,
20 Elements of Mathematics for Economics and Finance
c)
15
54
× 18
35
,
d)
32
49
÷ 56
21
.
1.4. Find which is the larger of the two fractions: 11/32, 7/24 by expressing
the numbers as:
a) fractions with the same denominator;
b) decimals.
1.5. Write each of the following numbers correct to two decimal places:
a) 51.2361
b) 7.896
c) 362.275
1.6. Write each of the following numbers correct to three significant figures:
a) 5,889
b) 0.0002817
c) 72,961
d) 0.09274
1.7. Write each of the following numbers in standard form:
a) 495,200
b) 0.000000837
1.8. The computing equipment belonging to a company is valued at
$45,000. Each year, 12% of the value is written off for depreciation.
Find the value of the equipment at the end of two years.
1.9. Death duties of 20% are paid on a legacy to three children of
£180,000. The eldest child is bequeathed 50%, the middle child 30%,
and the youngest child the remainder. How much does each child receive?
What percentage of the original legacy does the youngest child
receive?
1.10. Simplify the following:
a) x2/3×7/3,
b) x5
x2 , - Essential Skills 21
c) (x2/3)6,
d) x3y2
x2y5 .
1.11. Write down the values of the following without using a calculator:
a) 165/4,
b) 811/4,
c)
27
125
2/3
,
d) 81−3/4.
1.12. Multiply out the brackets and simplify the following:
a) (2x + 9)(3x − 8),
b) (x + 4)(6x + 3),
c) (3x − 2)(11 − 4x),
d)
(15 − 24x + 18y)
0.75
,
e) (x − 4y + 7)(5x − 2y − 3).
1.13. Factorize the following expressions:
a) 96x − 32,
b) −21x + 49×2,
c) 4×2 − 49.
2
Linear Equations
2.1 Introduction
In this book, we will be concerned primarily with the analysis of the relationship
between two or more variables. For example, we will be interested in the
relationship between economic entities or variables such as
– total cost and output,
– price and quantity in an analysis of demand and supply,
– production and factors of production such as labour and capital.
If one variable, say y, changes in an entirely predictable way in terms of another
variable, say x, then, under certain conditions (to be defined precisely
in Chapter 4), we say that y is a function of x. A function provides a rule
for providing values of y given values of x. The simplest function that relates
two or more variables is a linear function. In the case of two variables, the
linear function takes the form of the linear equation y = ax + b for a = 0.
For example, y = 3x + 5 is an example of a linear function. Given a value of
x, one can determine the corresponding value of y using this functional relationship.
For instance, when x = 2, y = 3 × 2 + 5 = 11 and when x = −3,
y = 3 × (−3) + 5 = −4. We will say more about functions in Chapter 4. Linear
equations or functions may be portrayed by a straight line on a graph.
In this chapter, we introduce graphs and give a number of examples showing
how linear equations can be used to model situations in economics and how to
interpret properties of their graphs.
23
24 Elements of Mathematics for Economics and Finance
2.2 Solution of Linear Equations
A mathematical statement setting two algebraic expressions equal to each other
is called an equation. The ability to solve equations is one of the most important
algebraic techniques to master. Equipped with this skill, you will be able
to solve a range of economic problems. The simplest type of equation is the
linear equation in a single variable or unknown, which we will denote by x
for the moment. In a linear equation, the unknown x only occurs raised to the
power 1. The following are examples of linear equations: - 5x + 3 = 11,
- 1 − 4x = 3x + 7,
3.
2 + 3x
5
2x − 1
6 .
A linear equation may be solved by rearranging it so that all terms involving
x appear on one side of the equation and all the constant terms appear on
the other side. This is achieved by performing a series of algebraic operations.
The key is to remember that you must perform the same operations to both
sides of the equation. You must be completely impartial so that each stage of
the rearrangement process yields an equivalent equation. Two equations are
said to be equivalent if and only if when one holds then so does the other.
Equivalent equations, therefore, have precisely the same solutions if they have
any at all. However, it is important that you never multiply or divide through
an equation by 0. For example, take the equation 1 = 2, which is not valid, and
multiply both sides by 0. Then we obtain the equation 0 = 0, which is true. So
the two equations are not equivalent. If an equation contains a fraction, then
the equation may be simplified by multiplying through by the denominator.
Remember that the value of a fraction a/b is the same if the numerator and
denominator are multiplied (or divided) by the same nonzero number. That is,
a
b
= ta
tb
,
for any number t = 0. It is instructive to look at an example.
Example 2.1
Solve the equation
7x − 4
2
= 2x + 4.
- Linear Equations 25
Solution. To determine the value of x that satisfies this equation, we rearrange
the equation so that all terms involving the unknown x appear on the one side
of the equation and all the constant terms appear on the other. - Multiply both sides by 2, which is the denominator of the fraction on the
left-hand side of this equation:
7x −4 = 2× (2x + 4)
= (2× 2)x + (2 × 4)
= 4x + 8. - Subtract 4x from both sides so that all terms involving x are on the lefthand
side:
7x − 4 − 4x = 4x + 8 − 4x,
3x − 4 = 8. - Add 4 to both sides so that all the constant terms are on the right-hand
side:
3x − 4 + 4 = 8 + 4,
3x = 12. - Finally divide both sides by 3:
3x
3
12
3 ,
x = 4.
So the solution to this equation is x = 4.
We can check to see if this answer is correct by replacing x by 4 in the
original equation. If x = 4 is the correct solution, then the left- and right-hand
sides of the equation should give the same numerical value.
LHS =
(7 × 4) − 4
2
28 − 4
2
24
2
= 12
RHS = 2 × 4 + 4
= 12.
26 Elements of Mathematics for Economics and Finance
Example 2.2
Solve the equation
x
4
−3 = x
5
- 1. (2.1)
Solution. Again, we go through the solution step-by-step. The idea is to
rearrange the equation so that all terms involving x appear on the left-hand
side and all the constant terms appear on the right-hand side. Once this is
done, the terms involving fractions are simplified.
- Subtract x/5 from both sides:
x
4
− x
5
− 3 = 1 - Add 3 to both sides
x
4
− x
5
= 1+3 = 4 - Simplify the left-hand side by expressing it as a single fraction. This is
achieved by expressing each of the fractions in terms of their lowest common
denominator, 20. In the case of the first fraction, both the numerator and
denominator are multiplied by 5, and in the case of the second fraction
they are both multiplied by 4, i.e.,
x
4
5x
5 × 4
5x
20
and x
5
4x
4 × 5
4x
20 .
Therefore
5x
20
− 4x
20
= 4
5x − 4x
20
= 4
x
20
= 4.
- Finally multiply both sides by 20:
x = 80.
The solution to this equation is x = 80. Again we can check that this is the
correct solution by substituting x = 80 into the left- and right-hand sides of
(2.1). - Linear Equations 27
2.3 Solution of Simultaneous Linear Equations
A number of economic models are built on linear relationships between variables.
For example, the economic concept of equilibrium requires the solution
of a system of equations.
The next degree of difficulty is to solve two linear equations in two unknowns.
Suppose the two unknowns are denoted by x and y. The most general
form of system of simultaneous linear equations in the unknowns x and y is
a1x + b1y = c1, (2.2)
a2x + b2y = c2. (2.3)
where a1, b1, c1, a2, b2, and c2 are constants. In the first equation (2.2), the
coefficient of x is a1 and that of y is b1. We are going to describe the elimination
method for solving this system of equations. As its name suggests,
the method involves eliminating one of the variables from the system. This
allows us to determine the value of the unknown that remains by solving a
single linear equation in one unknown. The value of the eliminated unknown
is then determined by substituting the known value into either of the original
equations and solving another linear equation.
Suppose we wish to eliminate the variable y from (2.2)–(2.3). To do this, we
multiply (2.2) by b2 and (2.3) by b1 so that the coefficients of y in the equivalent
equations are the same:
b2a1x + b2b1y = b2c1, (2.4)
b1a2x + b1b2y = b1c2. (2.5)
Next we eliminate the variable y by subtracting (2.5) from (2.4):
(b2a1 − b1a2)x = b2c1 − b1c2, (2.6)
from which we deduce
x = b2c1 − b1c2
b2a1 − b1a2
. (2.7)
Note that we can only perform this last step provided that (b2a1 − b1a2) = 0.
The quantity (b2a1 − b1a2) is known as the determinant (see Chapter 10) of
the system of equations (2.2)–(2.3). The condition for this system to possess a
unique solution is that the determinant is nonzero.
Similarly, we can eliminate x from equations (2.2)–(2.3) to obtain
y = c2a1 − c1a2
b2a1 − b1a2
; (2.8)
or we can obtain y by substituting the value of x we have obtained (2.7) in
either (2.2) or (2.3) and solving the resulting linear equation.
28 Elements of Mathematics for Economics and Finance
There is no guarantee that a system of two or more simultaneous equations
will possess a unique solution. Consider the system of equations
2x + y = 10,
2x + y = 5.
This system of equations does not have a solution. In fact, the equations are
inconsistent. They cannot hold simultaneously since 10 = 5!We shall see later
in this chapter that the solution of a system of simultaneous linear equations
may be interpreted as the point of intersection of two straight lines. For the
example under consideration, the two lines are parallel and therefore never
intersect.
Next consider the system of equations
2x + y = 10,
−6x − 3y = −30.
At first sight this might seem to be an innocuous system of equations. However,
the second equation is just a multiple of the first; obtained by multiplying the
first equation by −3. In this case, the equations are not independent. The
second equation does not provide any additional information over the first
equation. Since there are two unknowns to be determined, there is no unique
solution – in fact there are infinitely many solutions. For the above system one
can verify that x = s and y = 10 − 2s is a solution for any number s.
To obtain a unique solution to a system of simultaneous linear equations,
the equations must be consistent and independent and there must be as many
equations as unknowns (variables).
Example 2.3
Solve the system of equations
3x + 2y = 1
−2x + y = 2.
Solution. We solve this system of equations using the elimination method in
which we eliminate the variable x. To do this, we arrange for the coefficients
of x in both equations to differ only in sign by multiplying the two equations
by appropriate factors. The variable can then be eliminated by adding or subtracting
the two equations. For example, suppose we multiply the first equation
by 2 and the second by 3:
6x + 4y = 2
−6x + 3y = 6. - Linear Equations 29
The variable x is eliminated by adding the two equations:
7y = 8,
which, after division by 7, gives
y =
8
7.
This value can now be substituted in either of the original two equations to
obtain the corresponding value of x. Let us use the first equation, then
3x + 2
8
7
= 1
3x +
16
7
= 1
3x = 1− 16
7
3x =
7 − 16
7
(since 1 = 7/7)
3x = −9
7
x =
1
3
×
−9
7
x = −3
7
Therefore, the solution is x = −3/7, y = 8/7. Of course, we can check that
we have the correct solution by substituting it back into the original set of
equations and checking that the equations are satisfied.
An alternative but equivalent method for solving simultaneous linear equations
is known as the substitution method. The idea is to rearrange one of
the equations in order to isolate one of the variables on the left-hand side. The
expression for this variable is then substituted into the second equation to yield
a linear equation for the other variable. We demonstrate this by means of an
example.
Example 2.4
At the beginning of the year, an investor had £50,000 in two bank accounts,
each of which paid interest annually. The interest rates were 4% and 6% per
annum, respectively. If the investor has made no withdrawals during the year
and has earned a total of £2,750 interest, what was the initial balance in each
of the two accounts?
30 Elements of Mathematics for Economics and Finance
Solution. Let x and y denote the initial balances in the accounts with interest
rates 4% and 6%, respectively. Since the total amount invested at the start of
the year was £50,000, we have
x + y = 50,000.
The amount of interest earned on the two bank accounts during the year is
given by
0.04x and 0.06y,
respectively. Since the total amount of interest earned during the year is £2,750,
0.04x + 0.06y = 2,750,
or, after multiplying through by 100
4x + 6y = 275,000.
Therefore, we have two equations with which to determine initial balances in
the two bank accounts:
x + y = 50,000 (2.9)
4x + 6y = 275,000. (2.10)
Multiplying (2.9) by 4, we obtain
4x + 4y = 200,000. (2.11)
Then subtracting (2.11) from (2.10) yields
2y = 75,000,
so that y = 37,500. Finally, it follows from (2.9) that x = 12,500. Therefore,
the initial balance in each of the two accounts was £12,500 and £37,500, respectively.
2.4 Graphs of Linear Equations
Consider the linear equation
y = 3x − 2.
Given a value of x, one can use this equation to determine the corresponding
value of y. For example, when x = 0, y = 3 × 0 − 2 = −2, and when x = 2,
y = 3× 2 − 2 = 6 − 2 = 4. The collection of all such pairs of values of x and y
that satisfy this linear equation can be represented on a graph. - Linear Equations 31
Consider the two perpendicular lines shown in Fig. 2.1. The horizontal line
is referred to as the x-axis and the vertical line as the y-axis. The point where
these lines intersect is known as the origin and is denoted by the letter O. At
this point, both variables take the value zero. Each axis is assigned a numerical
scale that is chosen appropriately for the situation being considered. On the
x-axis, the scale takes positive values to the right of the origin and negative
values to the left. Moreover, the further we move away from the origin, the
larger these values become. On the y-axis, the scale takes positive values above
the origin and negative values below. Again, the further we move away from
the origin in the vertical direction, the larger these values become. These axes
enable us to define uniquely any point, P, in terms of its coordinates, (x, y).
We write the coordinates (x, y) alongside the point P as in Fig. 2.1. The first
number, x, denotes the horizontal distance along the x-axis and the second
number y denotes the vertical distance along the y-axis. The arrows on the
axis denote the positive direction. The collection of all points (x, y) satisfying
a linear equation lie on a straight line. That is, any equation of the form
y = ax + b, (2.12)
where a and b are constants is a linear equation and can be represented by a
straight line graph. We sometimes say that y is a linear function of x since in
the equation defining y, the variable x only occurs linearly.
Note also that the equation x = k, where k is any constant, is also represented
by a straight line graph: the ‘vertical’ line, parallel to the y-axis, through
the point (k, 0).
Example 2.5
Plot the following points A : (−2, 3), B : (−3,−4), C : (3, 5), D : (1,−4).
Solution. The position of A is determined by the pair of values x = −2 and
y = 3, and therefore it is located 2 units in the negative x-direction and 3 units
in the positive y-direction as shown in Fig. 2.2. The other points are plotted in
a similar way.
The general form of a linear equation is
cx + dy = e, (2.13)
where c, d, and e are constants. We assume that c and d are not both zero.
This equation contains multiples of x and y and a constant. These are the only
terms involving x that are present in a linear equation; otherwise the equation
32 Elements of Mathematics for Economics and Finance
x
y
P:(x,y)
x
y
0
Figure 2.1 The coordinate axes and the position of a general point P.
is said to be nonlinear. The values c and d are referred to as the coefficients
of x and y, respectively,. For example, the coefficients of the linear equation
2x − y = −3
are 2 and −1. More specifically, the coefficient of x is 2 and the coefficient of y
is −1.
Any equation of the form (2.13) can be rearranged into the form (2.12)
provided d = 0. First subtract cx from both sides of (2.13):
dy = −cx + e.
Then divide both sides by d provided d = 0:
y = −c
d
x + e
d
. (2.14)
If we now compare this equation with (2.12) by comparing the coefficients of x
and the constant terms in both equations, we see that (2.14) is just (2.12) with
a = −c
d
, b = e
d
. - Linear Equations 33
x
-4 -3 -2 -1 1 2 3 4
-5
-4
-3
-2
-1
0
1
2
3
4
5
y 6
A
B
C
D
Figure 2.2 The location of the points specified in Example 2.5.
Note that when d = 0, the linear equation (2.13) reduces to
cx = e or x = e
c
.
This is represented by a straight line parallel to the y-axis passing through the
point (e/c, 0) on the x-axis.
To sketch the graph of a straight line, it is sufficient to draw a line through
any two points lying on it.
Example 2.6
Sketch the graph of the straight line
y = 2x + 3,
for values of x lying between 0 and 4.
Solution. We determine the coordinates of two points on the line. When
x = 0, we have that y = 3 and when x = 4, we have y = 11. Therefore,
the points (0, 3) and (4, 11) lie on the line. The graph is formed by drawing a
straight line through these points as shown in Fig. 2.3.
34 Elements of Mathematics for Economics and Finance
x
1 2 3 4
0
2
4
6
8
10
y 12
(0,3)
(4,11)
Figure 2.3 The graph of the equation y = 2x + 3.
Example 2.7
Sketch the straight line
2x + y = 5.
Solution. Setting x = 0 gives y = 5. Hence (0, 5) lies on the line. Setting
y = 0 gives 2x = 5 or x = 5/2. Hence (5/2, 0) lies on the line.
2.4.1 Slope of a Straight Line
The coefficients a and b in the linear equation y = ax+b of (2.12) have special
significance and can be related to features of its graph. When x = 0, y = b and
therefore the constant b represents the intercept on the y-axis, i.e., it is the
value of y corresponding to the point of intersection of the straight line with
the y-axis. The value of x for which y = 0 is the solution of the linear equation
ax + b = 0.
This equation has solution x = −b/a, provided a = 0. - Linear Equations 35
x
-1 1 2 3
1
2
3
4
5
y
0
Figure 2.4 The graph of the equation 2x + y = 5.
The coefficient a in the equation y = ax + b defines the slope or gradient
of the straight line with that equation. The slope of a straight line provides
important information about the behaviour of the relationship between the
variables x and y. Let A : (x1, y1) and B : (x2, y2) be any two distinct points
lying on a straight line as shown in Fig. 2.5. The slope or gradient of the line
measures the ratio of the change in the vertical direction with respect to the
change in the horizontal direction as one moves from A to B. We illustrate this
with reference to Fig. 2.5. Since y1 = ax1 + b and y2 = ax2 + b, then
y2 − y1 = ax2 − ax1 = a(x2 − x1).
Therefore,
BC
AC
= y2 − y1
x2 − x1
= a(x2 − x1)
x2 − x1
= a,
i.e.
a = y2 − y1
x2 − x1
= BC
AC
. (2.15)
The value of a is independent of the choice of points A, B on the line. Positive
values of a correspond to straight lines where y increases as x increases, while
negative values of a correspond to straight lines where y decreases as x increases.
Larger values of a correspond to straight lines with steeper slopes. For example,
36 Elements of Mathematics for Economics and Finance
A:(x ,y )
B:(x ,y )
C
y
0 x
1 1
2 2
Figure 2.5 The graph of a linear equation and its slope.
the slope of the straight line y = 6x−3 is steeper than that of y = x+3. Another
way of viewing the slope a is that it is the change in y when x increases by one
unit, as then x2 − x1 = 1 and therefore a = y2 − y1.
Example 2.8
Determine the slope and intercept of the straight line 9x + 3y = 4.
Solution. We need to write this equation in the form y = ax + b.
9x + 3y = 4
3y = −9x + 4
y = −3x +
4
3
One can say immediately that the slope of this straight line is −3 and the
intercept is 4/3. - Linear Equations 37
Example 2.9
Find the slope of the straight line that passes through the points (2,−1) and
(−2,−11).
Solution. The slope of a straight line passing through the points (x1, y1),
(x2, y2) is
a = y2 − y1
x2 − x1
.
Therefore the required slope is
a =
−11 − (−1)
−2 − 2
−10
−4
5
2.
2.5 Budget Lines
Suppose that a company or an individual has a given budget, B, that can
be used to purchase two goods. If the cost or price of each of these goods is
known, then it is possible to determine the different combinations of the two
goods that can be bought with the given budget. Suppose that the two goods
are denoted by X and Y , and their respective prices are PX and PY. The
quantities purchased of these goods is also denoted by X and Y . Then the
equation of the budget line is
PXX + PY Y = B. (2.16)
Example 2.10
An electrical company has a budget of £6,000 a week to spend on the manufacture
of toasters and kettles. It costs £5 to manufacture a toaster and £12 to
manufacture a kettle. Write down the equation of the budget line and sketch
its graph.
Solution. Let T and K denote the number of toasters and kettles that are
manufactured each week. Then the cost of manufacture and the available budget
means that the budget line has the equation
5T + 12K = 6,000.
38 Elements of Mathematics for Economics and Finance
0 100 200 300 400 500
0
200
400
600
800
1000
1200
T
K
Figure 2.6 The graph of the budget line 5T + 12K = 6,000.
To sketch the graph of this budget line, it is sufficient to determine the coordinates
of two points on the line. When T = 0, 12K = 6,000 and therefore
K = 500. Similarly, when K = 0, 5T = 6,000 and therefore T = 1,200. The
graph of the budget line is given by the straight line joining the points T = 0,
K = 500 and T = 1,200, K = 0. The graph of the budget line is sketched in
Fig. 2.6.
Example 2.11
A person has £120 to spend on two goods (X, Y ) whose respective prices are
£3 and £5.
- Draw a budget line showing all the different combinations of the two goods
that can be bought with the given budget (B). - What happens to the original budget line if the budget falls by 25%?
- What happens to the original budget line if the price of X doubles?
- What happens to the original budget line if the price of Y falls to £4?
Draw the new budget lines in each case. - Linear Equations 39
Solution. - The general equation of a budget line is
PXX + PY Y = B
where PX is the price of X and PY is the price of Y . Now if PX = 3, PY =
5, B = 120, then the equation of the budget line is
3X + 5Y = 120.
We can rearrange this equation to give
Y = −3
5X + 24.
The graph of this budget line is represented by the solid line in Fig. 2.7. - If the budget falls by 25% it is reduced by 25% of £120, i.e., £30. The new
budget B = £120 − £30 = £90. The equation for the new budget line is
3X + 5Y = 90,
which, after rearrangement, can be written in the form
Y = −3
5X + 18.
This line has the same slope as the original budget line but lies to the left
of it. This is the dashed line in Fig. 2.7. - If PX = 6 the budget equation becomes
6X + 5Y = 120
or
Y = −6
5X + 24.
This time the intercept remains the same as the original budget line but
the slope is steeper – the slope is −6/5 compared with the slope of −3/5
of the original budget line. The graph of this budget line is represented by
the long dashed line in Fig. 2.7. - If PY = 4 , then the budget equation is
3X + 4Y = 120,
or
Y = −3
4X + 30.
This time both the slope and the intercept change. See the dash-dot line
in Fig. 2.7.
40 Elements of Mathematics for Economics and Finance
10 20 30 40
0
10
20
30
40
(1)
(2)
(3)
(4)
Y
X
Figure 2.7 The graph of the budget lines in Example 2.11.
2.6 Supply and Demand Analysis
Microeconomics is concerned with the analysis of the economic theory and
policy of individual firms and markets. The mathematics we have introduced
so far can be used to calculate the market equilibrium in which the demand
and supply of a particular good balance.
The quantity demanded, Q, of a particular good depends on the market
price, P. We shall refer to the way Q depends on P as the demand equation
or demand function. Functions will be defined in more detail later in the
book (Chapter 4). Economists normally plot the relationship between price
and quantity with Q on the horizontal axis and P on the vertical axis. We
assume that this relationship is linear, i.e.,
P = aQ + b,
for some appropriate constants (parameters) a and b. A graph of a typical linear
demand function is the dashed line in Fig. 2.8. Elementary theory shows that
demand usually falls as the price of the good rises so the slope of the line is
negative, i.e., a < 0. We say that P is a decreasing function of Q.
Similarly, the supply equation or supply function is the relation between
the quantity, Q, of a good that producers plan to bring to the market and - Linear Equations 41
supply equation
demand equation
P
Qo Q
Po
Figure 2.8 The graph of typical linear demand and supply equations. The
point of intersection provides the point of equilibrium for the model.
the price, P, of the good. A typical linear supply curve is the solid line in
Fig. 2.8. Economic theory indicates that as the price rises, so does the supply.
Mathematically, P is then said to be an increasing function of Q. Note that the
supply Q is zero when P = b. It is only when the price exceeds this threshold
level that the producers decide that it is worth supplying any good whatsoever.
We are interested in the interplay between supply and demand. Of particular
significance is the point of intersection of the demand and supply curves (see
Fig. 2.8). At this point, the market is said to be in equilibrium because the
quantity demanded is equal to the quantity supplied. The corresponding price,
P0, and quantity, Q0, are called the equilibrium price and quantity. It is
also of interest to observe the effect of a shift of the market price away from
its equilibrium price.
Example 2.12
The demand and supply equations of a good are given by
4P = −Qd + 240,
5P = Qs + 30.
42 Elements of Mathematics for Economics and Finance
Determine the equilibrium price and quantity.
Solution. At market equilibrium, we have
Qd = Qs = Q, say ,
where Q is the equilibrium quantity. In this case, the demand and supply
equations become
4P = −Q + 240,
5P = Q + 30.
This is a system of two simultaneous equations in the unknowns P and Q. We
can eliminate Q from the system by adding the two equations. This gives
9P = 270.
Then, dividing both sides by 9 gives the equilibrium price
P = 30.
Finally, the equilibrium quantity Q is determined by substituting this value
into either of the demand or supply equations. The supply equation gives
5 × 30 = Q + 30,
which, after rearrangement yields the equilibrium quantity
Q = 120.
Example 2.13
The demand and supply functions of a good are given by
P = −Qd + 125,
2P = 3Qs + 30.
Determine the equilibrium price and quantity. Determine also the effect on the
market equilibrium if the government decides to impose a fixed tax of £5 on
each good. Who pays the tax? - Linear Equations 43
Solution. At market equilibrium, we have
Qd = Qs = Q, say ,
where Q is the equilibrium quantity. In this case, the demand and supply
equations become
P = −Q + 125, (2.17)
2P = 3Q + 30. (2.18)
This is a system of two simultaneous equations in the unknowns P and Q. We
can eliminate Q from the system by multiplying the demand equation (Eq.
(2.17)) by 3:
3P = −3Q + 375, (2.19)
and adding the resulting equation (2.17) to the supply equation (2.18). This
gives
5P = 405,
which, after dividing both sides by 5 gives the equilibrium price
P = 81.
Finally, the equilibrium quantity Q is determined by substituting this value
into either of the demand or supply equations. The demand equation gives
81 = −Q + 125,
which, after rearrangement yields the equilibrium quantity
Q = 125 − 81 = 44.
If the government imposes a fixed tax of £5 on each good, then the original
supply equation needs to be modified. This is because the amount the supplier
receives as a result of each sale is the amount that the consumer pays (P) less
the tax (£5), i.e., P −5. Thus, the new supply equation is obtained by replacing
P by P − 5 in the original supply equation:
2(P − 5) = 3Qs + 30. (2.20)
This equation can be simplified by multiplying out the bracket on the lefthand
side and taking the constant term to the right-hand side. The new supply
equation becomes
2P − 10 = 3Qs + 30,
or
2P = 3Qs + 40. (2.21)
44 Elements of Mathematics for Economics and Finance
We then proceed as before to determine the equilibrium price and quantity for
the new situation. At market equilibrium, we have
Qd = Qs = Q, say ,
where Q is the equilibrium quantity. In this case, the demand and supply
equations become
P = −Q + 125, (2.22)
2P = 3Q + 40. (2.23)
We can eliminate Q from the system by multiplying the demand equation (Eq.
(2.22)) by 3:
3P = −3Q + 375, (2.24)
and adding the resulting equation (2.24) to the supply equation (2.23). This
gives
5P = 415,
which, after dividing both sides by 5 gives the equilibrium price
P = 83.
Finally, the equilibrium quantity Q is determined by substituting this value
into either of the demand or supply equations. The demand equation gives
83 = −Q + 125,
which, after rearrangement yields the equilibrium quantity
Q = 125 − 83 = 42.
The influence of government taxation on the equilibrium price is to increase
it from £81 to £83. Therefore, not of all of the tax is passed on to the consumer.
The consumer pays an extra £2 per good after tax has been imposed. The
remaining part of the tax is borne by the supplier.
2.6.1 Multicommodity Markets
At the beginning of this section, we looked at supply and demand analysis for a
single good. We extend these ideas now to a multicommodity market. Suppose
that there are two goods in related markets, which we call good 1 and good 2. - Linear Equations 45
The demand for either good depends on the prices of both good 1 and good 2.
If the corresponding demand functions are linear, then
Qd1 = a1 + b1P1 + c1P2
Qd2 = a2 + b2P1 + c2P2
where Pi and Qdi denote the price and demand for the ith good, and ai, bi,
and ci are constants depending on the model. For the first equation a1 > 0,
because there is a positive demand when the prices of both goods are zero.
Also b1 < 0, because the demand of a good falls as its price rises. The sign of
c1 depends on the nature of the two goods. If the goods are substitutable,
then an increase in the price of good 2 would mean that consumers would
switch from good 2 to good 1, causing Qd1 to increase. Substitutable goods
are therefore characterized by a positive value of c1. On the other hand, if the
goods are complementary, then a rise in the price of either good would see
the demand fall so c1 is negative. Similar results apply to the signs of a2, b2
and c2.
Example 2.14
The demand and supply functions for two interdependent commodities are
given by
Qd1 = 145 − 2P1 + P2
Qs1 = −45 + P1
Qd2 = 30+P1 − 2P2
Qs2 = −40 + 5P2
where Qdi , Qsi, and Pi denote the quantity demanded, quantity supplied, and
price of good i, respectively. Determine the equilibrium price and quantity for
this two-commodity model. Are these goods substitutable or complementary?
Give reasons for your answer.
Solution. At equilibrium, the quantity supplied is equal to the quantity demanded
for each good, so that
Qd1 = Qs1 and Qd2 = Qs2 .
Let us write these respective common values as Q1 and Q2. Then for good 1
we have
Q1 = 145 − 2P1 + P2
Q1 = −45 + P1
46 Elements of Mathematics for Economics and Finance
Therefore
145 − 2P1 + P2 = −45 + P1
which simplifies to give
3P1 − P2 = 190.
Similarly for good 2 we have
Q2 = 30+P1 − 2P2
Q2 = −40 + 5P2
Therefore
30 + P1 − 2P2 = −40 + 5P2
which simplifies to give
−P1 + 7P2 = 70.
We have therefore shown that the equilibrium prices satisfy the simultaneous
equations
3P1 − P2 = 190 (2.25)
−P1 + 7P2 = 70 (2.26)
These equations can be solved by elimination. Multiply (2.26) by 3. This gives
3P1 − P2 = 190
−3P1 + 21P2 = 210
Adding these two equations yields
20P2 = 400,
and so P2 = 20. Substituting this value of P2 back into (2.25):
3P1 = 190 + 20 = 210,
which gives P1 = 70. Finally, substituting these values of P1 and P2 back into
the original supply equations, we obtain
Q1 = 145 − 140 + 20 = 25
and
Q2 = −40 + 100 = 60.
On inspection of the demand equation for good 1, we see that the demand
for this good increases when the price of good 2 increases. This is characterized
by a positive coefficient of P2 in this equation. Therefore, the two goods are
substitutable. - Linear Equations 47
EXERCISES
2.1. Solve the following linear equations:
a) 3x − 4 = 2,
b)
2x − 1
3
3x − 1
4
- 1.
2.2. Solve the system of equations
3x − 2y = 4
x − 2y = 2.
2.3. Solve the system of equations
3x + 5y = 19
−5x + 2y = −11.
2.4. Sketch the graph of the straight line y = −x + 2 for −1 ≤ x ≤ 5.
2.5. Sketch the graph of the straight line y = 2x − 3 for 0 ≤ x ≤ 4.
2.6. Find the slope of the straight line passing through the points
(−1,−3) and (4, 2).
2.7. Find the slope of the straight line passing through the points (0, 0)
and (2, 1).
2.8. A person has e60 to spend on two goods, X and Y , whose respective
prices are e6 and e4.
a) Draw a budget line showing all the different combinations of the
two goods that can be bought within the given budget.
b) What happens to the original budget line if the budget is increased
by 20%?
c) What happens to the original budget line if the price of X is
halved?
2.9. The demand and supply equations for a good are given by
2P = −Qd + 125,
8P = Qs + 45,
where P, Qd, and Qs denote the price, quantity demanded, and
quantity supplied, respectively.
48 Elements of Mathematics for Economics and Finance
a) Determine the equilibrium price and quantity.
b) Determine the effect on the market equilibrium if the government
decides to impose a fixed tax of £2.50 on each good. Who
pays the tax?
2.10. The demand and supply functions of a good are given by
P + 2Qd = 144
4P − 3Qs = 136
where P, Qd, and Qs, denote the price, quantity demanded, and
quantity supplied, respectively.
a) Determine the equilibrium price and quantity.
b) Determine the effect on the market equilibrium if the government
decides to impose a fixed tax of $11 on each good. Who
pays the tax?
2.11. The demand and supply functions of a good are given by
4P = −Qd + 102
5P = Qs + 6
where P, Qd, and Qs denote the price, quantity demanded, and
quantity supplied, respectively.
a) Determine the equilibrium price and quantity.
b) Determine the effect on the market equilibrium if the government
decides to impose a fixed tax of £9 on each good. Who
pays the tax?
2.12. The demand and supply equations for two complementary goods,
trousers (T) and jackets (J), are given by
QdT = 410 − 5PT − 2PJ
QsT = −60 + 3PT
and
QdJ = 295 − PT − 3PJ
QsJ = −120 + 2PJ
respectively, where QdT , QsT, and PT denote the quantity demanded,
quantity supplied, and price of trousers, and QdJ , QsJ ,
and PJ denote the quantity demanded, quantity supplied, and price
of jackets. Determine the equilibrium price and quantity for this
two-market model.
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