Jacques I. Mathematics for Economics and Business

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The formula for the ﬁrst graph goes in cell B2. As the x value is in cell A2, we type

=3*A2+2

By clicking and dragging this formula down the second column (up to, and including, cell B8), the

values of y are calculated.

Similarly, the formula for calculating the y coordinates for the second line is entered into cell

C2 as

=−2*A2+2

and the formula for the third line is entered into cell D2 as

=A2 /2+2

To plot these points on a graph, we highlight all the cells in the table, including the column titles, and

click on the Chart Wizard button on the toolbar. The Chart Wizard box will appear:

From the list of chart types, we choose XY (Scatter), and then choose an appropriate sub-type. As

we are plotting straight lines, we have selected Scatter with data points connected by lines without

markers.

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The Next screen shows you a preview of the graph, with the option to change the range of the cells

that have been plotted. If the graph looks wrong, it is usually because the wrong cells have been high-

lighted before going into Chart Wizard, so go back and check this, rather than altering the range.

The third screen allows you to label your graph, and alter its gridlines. You should always label your

axes, but you could, for example, delete the Legend if you feel it is inappropriate. Adding gridlines can

make it easier to read values off the graph.

Finally, we click Next and Finish, to transfer the graph on to the spreadsheet, as shown in Figure 1.9.

Notice that Excel provides a key showing which line is which.

1.1 • Graphs of linear equations

Figure 1.9

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(b) Following the same procedure for the three lines

y = 2x

y = 2x − 3

y = 2x + 1

produces a graph as shown in Figure 1.10.

parallel, which means that they have the same gradient.

This illustrates the fact that every straight line has an equation of the form

y = mx + c

where m is the gradient, and c is the intercept on the y axis.

In (a), the value of c in each equation is equal to 2, so all of the lines cut the y axis at this point.

In (b), the value of m in each equation is equal to 2, so all of the lines have the same gradient and are

parallel.

It is very useful to be able to recognize these properties from the equations, as it means we have a

fairly good idea of what our graph should look like even before we draw it.

Linear Equations

Figure 1.10

Advice

This completes your first piece of mathematics. I hope that you have not found it quite

as bad as you first thought. There now follow a few extra problems to give you more

practice. If you have successfully tackled all of the problems given in the text so far, you

might prefer to move straight on to the next section. Problems 9–16 can be attempted

later as part of your revision. However, if you failed to obtain the correct answers to

Problems 1–8 then you should try the Practice Problems now. Not only will they help to

strengthen your mathematical skills, but also they should improve your overall confidence.

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1.1 • Graphs of linear equations

Coefﬁcient A numerical multiplier of the variables in an algebraic term, such as the

numbers 4 and 7 in the expression, 4x + 7yz

Coordinates A set of numbers that determine the position of a point relative to a set of

axes.

Intercept The point(s) where a graph crosses one of the coordinate axes.

Linear equation An equation of the form y = ax + b.

Origin The point where the coordinate axes intersect.

Simultaneous linear equations A set of linear equations in which there are (usually) the

same number of equations and unknowns. The solution consists of values of the unknowns

which satisfy all of the equations at the same time.

Slope of a line Also known as the gradient, it is the change in the value of y when x

increases by 1 unit.

x axis The horizontal coordinate axis pointing from left to right.

y axis The vertical coordinate axis pointing upwards.

Practice Problems

9 Plot the following points on graph paper:

P (4, 0), Q (−2, 9), R (5, 8), S (−1, −2)

Hence find the coordinates of the point of intersection of the line passing through P and Q, and the

line passing through R and S.

10 Without using a calculator evaluate

(a)

10 × (−2) (b) (−1) × (−3) (c) (−8) ÷ 2

(d) (−5) ÷ (−5) (e) 5 − 6 (f) −1 − 2

(g) 7 − (−4) (h) −9 − (−9) (i)

11 If x = 2 and y =−3 evaluate

(a)

2x + y (b) x − y (c) 3x + 4y

(d) xy (e) 5xy (f) 4x − 6xy

12 Solve the following equations:

(a)

2x = 1 (b) 2x + 5 = 13 (c) 3 − x = 7

(d) 3 − 4x = 5 (e) 7x = 0 (f) + 7 = 10

13 If 4x + 3y = 24, complete the following table and hence sketch this line

(−3) × (−6) × (−1)

2 − 3

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Key Terms

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14 Solve the following pairs of simultaneous linear equations graphically:

(a)

−2x + y = 2 (b) 3x + 4y = 12 (c) 2x + y = 4 (d) x + y = 1

2x + y =−6 x + 4y = 84x − 3y = 36x + 5y = 15

15 Use the slope–intercept approach to sketch the lines

(a)

y =−x (b) x − 2y = 6

16 (a) Without using a calculator, work out the value of (−4)

(b) Press the following key sequence on your calculator:

Explain carefully why this does not give the same result as part (a) and give an alternative key

sequence that does give the correct answer.

4(−)

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section 1.2

Algebraic solution of

simultaneous linear

equations

In Section 1.1 a graphical method was described for the solution of simultaneous linear equa-

tions. Both lines are sketched on the same piece of graph paper and the coordinates of the point

of intersection are then simply read off from the diagram. Unfortunately this approach has

several drawbacks. It is not always easy to decide on a suitable scale for the axes. Even if the scale

allows all four points (two from each line) to ﬁt on the diagram, there is no guarantee that the

point of intersection itself also lies on it. You may have encountered this difﬁculty when solv-

ing Practice Problem 14(d) in the previous section. When this happens you have no alternative

but to throw away your graph paper and to start again, choosing a smaller scale in the hope that

the solution will now ﬁt. The second drawback concerns the accuracy of the graphical solution.

All of the problems in Section 1.1 were deliberately chosen so that the answers had nice numbers

in them; whole numbers such as −1, 2 and 5 or at worst simple fractions such as

/2, 2

/2 and

−

/4. In practice, the coefﬁcients of the equations may well involve decimals and we might

expect a decimal solution. Indeed, even if the coefﬁcients are whole numbers the solution itself

Objectives

At the end of this section you should be able to:

 Solve a system of two simultaneous linear equations in two unknowns using

elimination.

 Detect when a system of equations does not have a solution.

 Detect when a system of equations has inﬁnitely many solutions.

 Solve a system of three simultaneous linear equations in three unknowns using

elimination.

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could involve nasty fractions such as 7/8 or perhaps something like 231/571. A moment’s

thought should convince you that in these circumstances it is virtually impossible to obtain the

solution graphically, even if we use a really large scale and our sharpest HB pencil in the pro-

cess. The ﬁnal drawback concerns the nature of the problem itself. Quite frequently in eco-

nomics we need to solve three equations in three unknowns or maybe four equations in four

unknowns. Unfortunately, the graphical method of solution does not extend to these cases.

In this section an alternative method of solution is described which relies on algebra. It is

called the elimination method, since each stage of the process eliminates one (or more) of the

unknowns. This method always produces the exact solution and can be applied to systems of

equations larger than just two equations in two unknowns. In order to illustrate the method,

we return to the simple example considered in the previous section:

4x + 3y = 11 (1)

2x + y = 5 (2)

The coefﬁcient of x in equation (1) is 4 and the coefﬁcient of x in equation (2) is 2. If these

numbers had turned out to be exactly the same then we could have eliminated the variable x

by subtracting one equation from the other. However, we can arrange for this to be the case by

multiplying the left-hand side of the second equation by 2. Of course, we must also remember

to multiply the right-hand side of the second equation by 2 in order for this operation to be

valid. The second equation then becomes

4x + 2y = 10 (3)

We may now subtract equation (3) from (1) to get

y = 1

You may like to think of this in terms of the usual layout for the subtraction of two ordinary

numbers: that is,

4x + 3y = 11

4x + 2y = 10 −

y = 1

This number can now be substituted into one of the original equations to deduce x. From

equation (1)

4x + 3(1) = 11 (substitute y = 1)

4x + 3 = 11

4x = 8 (subtract 3 from both sides)

x = 2 (divide both sides by 4)

Hence the solution is x = 2, y = 1. As a check, substitution of these values into the other ori-

ginal equation (2) gives

2(2) + 1 = 5 ✓

The method of elimination can be summarized as follows.

Step 1

Add/subtract a multiple of one equation to/from a multiple of the other to eliminate x.

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Step 2

Solve the resulting equation for y.

Step 3

Substitute the value of y into one of the original equations to deduce x.

Step 4

Check that no mistakes have been made by substituting both x and y into the other original

equation.

1.2 • Algebraic solution of simultaneous linear equations

Advice

The following example involves fractions. If you have forgotten how to perform calcula-

tions with fractions, you might like to read through Section 1.4.3 (pp. 76–81) now, since

this section contains a review of their basic properties.

Example

Solve the system of equations

3x + 2y = 1 (1)

−2x + y = 2 (2)

Solution

Step 1

The coefﬁcients of x in equations (1) and (2) are 3 and −2 respectively. We can arrange for these to be the

same size (but of opposite sign) by multiplying equation (1) by 2 and multiplying (2) by 3. The new equa-

tions will then have x coefﬁcients of 6 and −6, so we can eliminate x this time by adding the equations

together. The details are as follows.

Doubling the ﬁrst equation produces

6x + 4y = 2 (3)

Tripling the second equation produces

−6x + 3y = 6 (4)

If equation (4) is added to equation (3) then

6x + 4y = 2

−6x + 3y = 6 +

7y = 8 (5)

Step 2

Equation (5) can be solved by dividing both sides by 7 to get

y = 8/7

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Step 3

If 8/7 is substituted for y in equation (1) then

3x + 2 = 1

3x +=1

3x = 1 − (subtract 16/7 from both sides)

3x = (put over a common denominator)

3x =−

x =×− (divide both sides by 3)

x =−

The solution is therefore x =−3/7, y = 8/7.

Step 4

As a check, equation (2) gives

−2 − +=+= = =2 ✓

6 + 8

7 − 16

Linear Equations

Advice

In the general description of the method, we suggested that the variable x is eliminated in

step 1. There is nothing special about x. We could equally well eliminate y at this stage and

then solve the resulting equation in step 2 for x.

You might like to solve the above example using this alternative strategy. You need to

double equation (2) and then subtract from (1).

Practice Problem

(a) Solve the equations

3x − 2y = 4

x − 2y = 2

by eliminating one of the variables.

(b) Solve the equations

3x + 5y = 19

−5x + 2y =−11

by eliminating one of the variables.

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The following examples provide further practice in using the method and illustrate some

special cases which may occur.

1.2 • Algebraic solution of simultaneous linear equations

Example

Solve the system of equations

x − 2y = 1

2x − 4y =−3

Solution

Step 1

The variable x can be eliminated by doubling the ﬁrst equation and subtracting the second:

2x − 4y = 2

2x − 4y =−3 −

0 = 5

The statement ‘0 = 5’ is clearly nonsense and something has gone seriously wrong. To understand what is

going on here, let us try and solve this problem graphically.

The line x − 2y = 1 passes through the points (0, −1/2) and (1, 0) (check this). The line 2x − 4y =−3 passes

through the points (0, 3/4) and (−3/2, 0) (check this). Figure 1.11 shows that these lines are parallel and so

they do not intersect. It is therefore not surprising that we were unable to ﬁnd a solution using algebra,

because this system of equations does not have one. We could have deduced this before when subtracting

the equations. The equation that only involves y in step 2 can be written as

0y = 5

and the problem is to ﬁnd a value of y for which this equation is true. No such value exists, since

×=

and so the original system of equations does not have a solution.

zero

any numberzero

Figure 1.11

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