FM101 Ch 1 Linear Equations PDF

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Chapter 1: Linear Equations In this Chapter we will provide the basic techniques in financial calculation and will concentrate mainly on forming and solving Linear Equations. We begin the chapter with a review of the basic background in algebra such as simplifying algebraic expression, the order of mathematical operation, rectangular coordinates, lines and solving basic algebra word problem. We then study linear equations and their properties and end with some application of linear equations.

1.1

Basic Algebra

1.1.1 ALGEBRAIC EXPRESSION In algebra we use algebraic quantities or variables (unknowns) which are commonly denoted by letters as well as numbers and arithmetic symbols such as +, -, =, etc to represent a relationship or to create a mathematical statements or expression. A mathematical statement that combines all the above is called an algebraic expression. For example 5x  2x2  15 is an algebraic expression that consists of three terms 5 x, 2 x 2 and -15. The numbers 5 and 2 are called coefficients of the terms x and x 2 respectively. In general coefficients are usually placed in front of an algebraic quantity. Note that the number -15 is called a constant term since it is not associated with any algebraic quantity.

The rules of algebra follows the rules of arithmetic, but these rules are written using letters. For example, 9  5  5  9 this shows that the order of addition does not matter. In algebra, we express such statement as: x  y  y  x . The letter x simply means the first number with any value, and y the second number. Letters are used to show that the statement is true for any number.

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1.1.2 Simplifying Expressions Algebraic expressions can be simplified and the following examples illustrate the rule. 1. When the terms are similar (like terms such as x with different coefficients) then they can be combined by adding the coefficients. Consider the expression5x2  3x  4x2  3 x . This expression can be simplified to x 2  6 x since 5x2 and 4x2 are like terms or similar terms so 5  4  1 (coefficient of x 2 ), likewise 3x and 3x , so 3  3  6 (coefficient of x ). Other examples of similar terms are 2

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2( x  3) and 5(x  3) , 2( x  x  2) and 5(x  x  2) .

2. When the terms are not similar, they cannot be combined by simply adding or subtracting the coefficients. The expression 12x  3 y  2 z cannot be simplified further.

3. Different powers of a variable cannot be treated as like terms. The expression 3x  4 x2 cannot be simplified since the terms x and x 2 are not considered as like terms. The expression can be written in a different form x(3  4 x) which has some benefits.

4. Multiplication and division follow the rules for manipulating numbers. For example,

c 7  ab 21 ab 7 ab 7abc 1   2    . 2 2 c c c 21ab 21abc 3c

1.1.3 Mathematical Operations When several arithmetic operations are involved there is an agree order of precedence with an acronym BEDMAS. The algebraic operations order of operations is as follows: 1. Evaluate the parentheses or Brackets, if there are any, and if they require evaluation, 2. Evaluate the powers, that is, the Exponents, 2

3. Divide or Multiply -- it does not matter, 4. Add or Subtract.

Example 1.

Evaluate 8 + 4(2 + 3)² − 7

Solution

8  4(2  3) 2  7  8  4(5) 2  7 The paranthesis is first evaluated ie 2+3 is replaced by 5  8  4(25) - 7 Next evaluate the exponent  8  100 - 7 Multiply =108-7 Add or subtract, it will not matter = 101

Example 2.

Let x 10, y  3, and z  3. Evaluate ( x  y) z.

Solution (10  3) 3  13 3  10.

1.1.4 Word problems Let’s have a look at a few word problems to summarize this section (basic algebra).

Example 4 A calculator has been marked up 15% and is being sold for $78.50. How much did the store pay the manufacturer of the calculator?

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Solution First, let’s define p to be the cost that the store paid for the calculator. The stores markup on the calculator is 15%. This means that 0.15p has been added on to the original price (p) to get the amount the calculator is being sold for. In other words, we have the following equation

p  0.15 p  78.50

that we need to solve for p. Doing this gives,

1.15 p  78.50



p

78.50  68.26087. 1.15

The store paid $68.26 for the calculator. Note that since we are dealing with money we rounded the answer down to two decimal places. Example 4 A tibuta, ladies top, is on sale for $15.00 and has been marked down 35%. How much was the tibuta being sold for before the sale? Solution This problem is pretty much the opposite of the previous example. Let’s start with defining p to be the price of the tibuta before the sale. It has been marked down by 35%. This means that 0.35p has been subtracted off from the original price. Therefore, the equation (and solution) is,

p  0.35 p  15.00 0.65 p  15.00 p

15.00  23.0769. 0.65

So, with rounding, it looks like the tibuta was originally sold for $23.08. 4

1.2

Linear Equations

Before we look at the main part of the chapter, let’s just revise our rectangular coordinate system which is very important in discussing linear equations. 1.2.1 Rectangular Coordinate System The rectangular coordinate system is illustrated in the diagram below. The vertical axis is called the y-axis, the horizontal axis is the x-axis. The axes divide the co-ordinate plane called the xyplane into four quadrants. The first quadrant, labeled as Quadrant I in the diagram is when x and y-values are both positive. The second quadrant QII is when the x-values are negative and yvalues are positive. The third quadrant QIII is when both x and y-values are negative and the fourth quadrant QIV is when the x-values are positive while the y-values are negative.

Rectangular coordinates are values we assign to every point on the xy -plane. The origin has coordinates (0,0) , any point on the x -axis has coordinates ( x,0) and on the y -axis has (0, y ) . If ( x, y ) are the coordinate of the point P, then x is called the x -coordinate and y is the y -

coordinate of P. For example, the point P on the diagram above is identified by the coordinate

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(55,16) which means that P is located 55 units along the x -axis to the right of the y-axis and 16

units upwards away from the x -axis and along the y -axis.

1.2.2 Definition An equation is a Mathematical statement which shows that two quantities are equal. It is usual for such an equation to contain at least one unknown quantity or variable. In the case of an equation with one unknown the equation will only be valid if the unknown is assigned one or more values, which are called solutions (or roots) of the equation. For example, the equation

x  5  2 has only one solution or root, namely x  3 . The value for x  3 is said to satisfy the equation. By “satisfied” we mean that if we represent the variables by their value in the equation and do the arithmetic we will get the same answer on both sides of the equation. The process of finding all solutions is called solving the equation. There are many types of equations but in this text we will concentrate on Linear equations. Let’s start off this section with the definition of a linear equation. Here are a couple of examples of linear equations. 

x  y  50 a linear equation with 2 variables.



8 y  3x  4 z  2 a linear equation with 3 variables.



2 5 x1  x2  4 x3  x4  9 a linear equation with 4 variables. 3

In order for an equation to be a linear equation, there are several main points to notice. First, the unknowns or variables only appear to the first power and there are no variables in the denominator of a fraction. Also notice that there are no products, quotients or root of variables and no variables involved in trigonometric, exponential, or logarithmic functions. All of these ideas are required for an equation to be a linear equation. The following are a few examples of non-linear equations: e x  2 y  8 and sin x1   x2  e2 .

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1.2.1 Straight Lines In this section we will study linear equation in 2 variables (and whenever we mention a linear equation we refer to a 2 variable linear equation). It is an equation that can be written in the form Ax  By  C , where A and B are both real numbers and together can not be equal to zero while

x and y are variables or unknowns. This form is sometimes called the standard form of a linear equation. The graph of such equations is always a line and this is the reason it is called a linear equation.

When the linear equation is given, there are several ways to draw its graph. One of the most common methods is by determining the x and y-intercepts:

GRAPHING A LINE BY DETERMINING THE X AND Y-INTERCEPTS

Steps 1 2

Example 3.

Determine the x and y-intercepts. Join the x and y-intercept to form the line.

Sketch the graph of y  3x  12

Solution To find the y  intercept: Let x  0, and solve for y y  3(0)   12 y  12

This implies that the y-intercept is at (0, 12).

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x  intercept: Lety  0, and solve for x, (0)  3 x   12 3x  12 dividing both sides by 3 we get x   4. This means that the x  intercept is (4, 0) . By joining the two intercepts we get the line as shown below. y

x

-4 -12

DETERMINING THE SLOPE OF A STRAIGHT LINE

The slope is denoted as m , where m 

rise change in y y y 2  y1    , x1  x2 . run change in x x x2  x1

y

( x2 , y2 )

y2

rise  y 2  y 1  y

y1

( x1 , y1 ) run  x2  x1  x

x1

x2

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x

The value of the slope m can be positive, negative, zero or undefined. CASE 1: Positive Slope

y  2x  1

CASE 2: Negative Slope y  2x  1 CASE 3: Slope of Zero

y  a (horizontal line)

CASE 4: Slope not defined x  a (vertical line)

Lets us now consider a few word problem related to linear equations.

Example 4

A server purchased at a cost of $60,000 in 2006 is depreciated linearly. The value

of the server at the end of 4 years is $12,000. Let y be the value of the server at the end of x years. Find the rate (slope) of depreciation.

Solution We are given that ( x1, y1)  (0, 60000) and ( x2 , y2 )  (4,12000). The rate of depreciation is m 

 y y1  y 2 60000  12000     12000. x x1  x2 0 4

EQUATION OF A LINE.

We will study the two types of equations, a point-slope form and a slope-intercept form.

Point–slope form

y  y  m (x  x ) 1 1

The line with equation above has a slope m and it passes through a point ( x1 , y1 ) .

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Example 5.

Find the equation of the line with slope 2 and containing the point(2, 3) .

Solution Since we are given the slope and a point on the line we use the point slope form

y  y  m (x  x ) 1 1 and use m  2 (since the slope is 2), x 1  2 and y 1  3 (the point has an x -cordinate of 2 and y as 3).

Then

y  3  2( x  2) y  2x  4  3 y  2x  1. Therefore the equation is y  2 x 1. Example 6.

Find the equation of the line passing through the points (1, 2) and (5, 2) .

Solution Find the slope m 

5 1 4    1 , and then use the point-slope form with one  2 2  4

point on the line to form an equation of the line. Let’s use point (1, 2).

y  y  m( x  x ) 1 1 y  2   1( x  1) y   x  1 2 y   x  3.

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Example 7

Referring to Example 4 above: (a) Find the linear function expressing y in terms of x. (b) What is the value of the server at the end of the second year?

Solution (a) The equation is

y  y  m(x  x ). 1 1

So, y  60000  12000( x  0) y

 12000x  60000.

(b) The value of the server at the end of the second year is y  12000(2)  60000  36000.

Lets us look now at the second form of linear equation. Slope-intercept form:

y  mx  c

An equation of a line with slope m and y-intercept (0,c) is y  mx  c.

Example 8

Find the slope m and the y-intercept of the line2 x  4 y  8 .

Solution To obtain the slope and y-intercept, we need to transform the equation into its slope-intercept form, that is, we need to solve for y.

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2 x 4 y 8 4 y  2 x  8 1 y  x2 2

1 The coefficient of x,  is the slope and the y-intercept is (0, 2). 2

1.2.3 Pairs of lines Parallel lines Two lines are called parallel if they have the same slope, that is m1 = m2 and different y-intercepts. Parallel lines do not have common points.

Lets consider the two lines with the following equations

y  x  1 y  x  2. The two lines are parallel since they have the same slope of 1 and have different y-intercept. Note that if they have the same y-intercepts as well, the two lines will be the same and therefore are not parallel but coincident (will be discussed next).

Example 9:

Find the equation of the line parallel to the line x - 2y + 4 = 0, and passing through

(2, 2). Solution We need to find the slope and y-intercept of the given line by rearranging its formula into a slope-intercept form, that is y  mx  c . By rearranging the formula we have 12

x  2y  4  0 x  4  2y

Making y the subject we get 2y  x  4 1 y  x  2. 2

So the slope of the given line is

1 , and the y-intercept is 2. 2

1 Hence the slope of the new line is also , therefore it should have a general 2 formula y 

1 x  c , where c  2 . 2

Since the new line passes through (2, 2), then the point (2, 2) should satisfy its equation. We determine the value of c as shown:

1 x c 2 1 (2)  (2)  c 2 2 1 c c 1 y

Therefore the equation of the new line is y 

1 x  1. 2

Coincident lines Two lines are coincident if they have the same slopes, same y-intercepts. They have all points in common.

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For example, the lines represented by x  y  3 and -2x  2 y  6 are coincident. Both equations have the same slope-intercept form which is y  x  3. Thus both have the same slope and same y-intercepts and have all points in common.

Intersection lines Lines with one common point and different slopes are called Intersection lines.

For example, the two lines with the equations below intersect. 2x  y 5 x y 3

When we put each equation into a slope-intercept form, respectively we have y  2x  5 y  x  3

The slope of the first equation is 2 while the second is 1 , so the lines intersect. Note that the y-intercept for the lines could either be the same or not, it does not matter.

1.3

Applications of Linear Algebra

Linear equations have a wide range of applications. Many real life situations can be represented or described as linear relationships. The following situations may be linear: 

The number of items produced by a manufacturer and the sale price;



The cost of production and the number of items produced by a company.

The following are few examples of the application of linear algebra.

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PREDICTION Linear equations are sometimes used as predictors of the future results. Example 9 In the year 2000, the cost of a Rewa powder milk averaged $3.85 a packet. Now, in 2011 if you go to any supermarket in Suva you will get that same milk for average of $5.80. Assuming that the relationship between time and cost is linear, develop a formula for predicting the average cost of a packet of Rewa powder milk in the future. What will be the average cost of the same milk in 2015 assuming it follows the same linear trend? Solution Let the year 2000 be t  0 , then we have to give an equation of a line joining the point (0,3.85) and (11,5.80) . The slope m 

 y 5.80  3.85   0.177 and the y-intercept is 3.85. So the equation of the line or x 11 0

the relationship between time and cost is given byC (t )  0.177t  3.85 . Thus in 2015 the cost of the Rewa powder milk will beC (15)  0.177(15)  3.85  $6.51 .

Break-Even Point In many businesses the cost of production of x number of any item, C ( x ) as well as the revenue generated from the sale of x number of the item, R( x) can be expressed as linear equations.

Cases: The following can be interpreted as shown: When C ( x)  R( x) then there is a Loss in the business, C( x)  R( x) then there is a Gain or Profit , and C( x)  R( x), the business reach a Break-even point i.e no loss nor gain .

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Example 10 Mary’s restaurant in Kiribati has daily fixed costs of $250. Each serve of food on average costs $4.00 and is sold for $5.00. i)

Find the cost of production for x serves of food.

ii)

Find the revenue from selling x serves of food.

iii)

What is the break-even point?

Solution i)

The cost of production, C (x )  4x  250.

ii)

The revenue, R ( x)  5x.

iii)

The break even point is where R( x)  C ( x)

5x  4x  250 x  250. This means that Mary’s restaurant needs to produce and sell more than 250 serves daily in order to gain.

Financial Planning

Example 11 Marian has $100,000 to invest and requires an overall rate of return of 5% per year. She can invest in a safe, government-insurance company, but it only pays 3% per year. To obtain 5%, she agrees to invest some of her money in a non-insured corporate bond paying 8% per year. How much should she place in each investment to achieve her goal?

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Solution Let x be the amount to be invested in a (3% pay) government-insurance company, and let y be the amount to be invested in the (8% pay) non-insured corporate bonds. We can write

x  y  100, 000.

Solving this for y we get y  100, 000  x.

Since the total interest from the investments is0.05(100,000)  5000 , then the equation relating the interest earned on the accounts is:

interest earned on the bonds + interest earned on the certificate = total interest earned 0.08 x  0.03(100, 000  x)  5000 0.08 x 3000 0.03 x 5000 0.05 x  2000 x  40, 000.

So Marian should invest $40,000 in corporate bonds and$100,000  $40,000  $60,000 in the Certificate of Deposit.

ECONOMICS A supply equation specifies the amount of any item the sellers are willing to offer in the market at various prices. A demand equation specifies the amount of any item the buyers are willing to buy at various prices. Market price or market equilibrium is defined as the price at which the supply and demand are EQUAL.

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Example 12 For a certain commodity D   0.5 p  1.6 and S  0.7 p  0.4 . Find the market price. What quantity is demanded at this market price? Solution The market price is where D  S .

 0.5 p 1.6  0.7 p 0.4 0.7 p  0.5 p 1.6 0.4 1.2 1.2 p p 1

At a price of $1...