ALGEBRA STUDY GUIDE ( INCLUDES SOLVED EXAMPLES AND PRACTICE PROBLEMS)

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SECTION 1.3 Graphs of linear equations Objectives At the end of this section you should be able to: z Plot points on graph paper given their coordinates. z Sketch a line by finding the coordinate ... s of two points on the line. z Solve simultaneous linear equations graphically. z Sketch a line by using its slope and intercept. Consider the two straight lines shown in Figure 1.1 . The horizontal line is referred to as the x axis and the vertical line is referred to as the y axis. The point where these lines intersect is known as the origin and is denoted by the letter O. These lines enable us to identify uniquely any point, P, in terms of its coordinates (x, y). The fi rst 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 axes indicate the positive direction in each case. Figure 1.1 Figure 1.2 shows the five points A(2, 3), B(−1, 4), C(−3, −1), D(3, −2) and E(5, 0) plotted on coordinate axes. The point A with coordinates (2, 3) is obtained by starting at the origin, moving 2 units to the right and then moving 3 units vertically upwards. Similarly, the point B with coordinates (−1, 4) is located 1 unit to the left of O (because the x coordinate is negative) and 4 units up. Note that the point C lies in the bottom left-hand quadrant since its x and y coordinates are both negative. It is also worth noticing that E actually lies on the x axis since its y coordinate is zero. Likewise, a point with coordinates of the form (0, y) for some number y would lie somewhere on the y axis. Of course, the point with coordinates (0, 0) is the origin, O.SECTION 1.3 GRAPHS OF LINEAR EQUATIONS 41 Practice Problem 1. Plot the following points on graph paper. What do you observe? (2, 5), (1, 3), (0, 1), (−2, −3), (−3, −5) In economics we need to do rather more than just plot individual points on graph paper. We would like to be able to sketch curves represented by equations and to deduce information from such a picture. We restrict our attention in this section to those equations whose graphs are straight lines, deferring consideration of more general curve sketching until Chapter 2 . In Practice Problem 1 you will have noticed that the fi ve points (2, 5), (1, 3), (0, 1), (−2, −3) and (−3, −5) all lie on a straight line. In fact, the equation of this line is −2x + y = 1 Any point lies on this line if its x and y coordinates satisfy this equation. For example, (2, 5) lies on the line because when the values x = 2, y = 5 are substituted into the left-hand side of the equation we obtain −2(2) + 5 = −4 + 5 = 1 which is the right-hand side of the equation. The other points can be checked similarly (See Table 1.1 ). Table 1.1 Point Check (1, 3) −2(1) + 3 = −2 + 3 = 1 ✓ (0, 1) −2(0) + 1 = 0 + 1 = 1 ✓ (−2, −3) −2(−2) − 3 = 4 − 3 = 1 ✓ (−3, −5) −2(−3) − 5 = 6 − 5 = 1 ✓ Figure 1.242 CHAPTER 1 LINEAR EQUATIONS Practice Problem 2. Check that the points (−1, 2), (−4, 4), (5, −2), (2, 0) all lie on the line 2x + 3y = 4 and hence sketch this line on graph paper. Does the point (3, −1) lie on this line? The general equation of a straight line takes the form a multiple of a multiple of a number x y + = that is, dx + ey = f for some given numbers d, e and f. Consequently, such an equation is called a linear equation. The numbers d and e are referred to as the coefficients. The coefficients of the linear equation, −2x + y = 1 are −2 and 1 (the coefficient of y is 1 because y can be thought of as 1 × y). In general, to sketch a line from its mathematical equation, it is sufficient to calculate the coordinates of any two distinct points lying on it. These two points can be plotted on graph paper and a ruler used to draw the line passing through them. One way of finding the coordinates of a point on a line is simply to choose a numerical value for x and to substitute it into the equation. The equation can then be used to deduce the corresponding value of y. The whole process can be repeated to fi nd the coordinates of the second point by choosing another value for x. Example Sketch the line 4x + 3y = 11 Solution For the first point, let us choose x = 5. Substitution of this number into the equation gives 4(5) + 3y = 11 20 + 3y = 11 The problem now is to solve this equation for y: 3y = −9 (subtract 20 from both sides) y = −3 (divide both sides by 3) Consequently, the coordinates of one point on the line are (5, −3).SECTION 1.3 GRAPHS OF LINEAR EQUATIONS 43 For the second point, let us choose x = −1. Substitution of this number into the equation gives 4(−1) + 3y = 11 −4 + 3y = 11 This can be solved for y as follows: 3y = 15 (add 4 to both sides) y = 5 (divide both sides by 3) Hence (−1, 5) lies on the line, which can now be sketched on graph paper as shown in Figure 1.3 . Figure 1.3 Practice Problem 3. Find the coordinates of two points on the line 3x − 2y = 4 by taking x = 2 for the fi rst point and x = −2 for the second point. Hence sketch its graph. In this example we arbitrarily picked two values of x and used the linear equation to work out the corresponding values of y. There is nothing particularly special about the variable x. We could equally well have chosen values for y and solved the resulting equations for x. In fact, the easiest thing to do (in terms of the amount of arithmetic involved) is to put x = 0 and find y and then to put y = 0 and find x.44 CHAPTER 1 LINEAR EQUATIONS Example Sketch the line 2x + y = 5 Solution Setting x = 0 gives 2(0) + y = 5 0 + y = 5 y = 5 Hence (0, 5) lies on the line. Setting y = 0 gives 2x + 0 = 5 2x = 5 x = 5/2 (divide both sides by 2) Hence (5/2, 0) lies on the line. The line 2x + y = 5 is sketched in Figure 1.4 . Notice how easy the algebra is using this approach. The two points themselves are also slightly more meaningful. They are the points where the line intersects the coordinate axes. Figure 1.4SECTION 1.3 GRAPHS OF LINEAR EQUATIONS 45 In economics it is sometimes necessary to handle more than one equation at the same time. For example, in supply and demand analysis we are interested in two equations, the supply equation and the demand equation. Both involve the same variables Q and P, so it makes sense to sketch them on the same diagram. This enables the market equilibrium quantity and price to be determined by fi nding the point of intersection of the two lines. We shall return to the analysis of supply and demand in Section 1.5 . There are many other occasions in economics and business studies when it is necessary to determine the coordinates of points of intersection. The following is a straightforward example which illustrates the general principle. Example Find the point of intersection of the two lines 4x + 3y = 11 2x + y = 5 Solution We have already seen how to sketch these lines in the previous two examples. We discovered that 4x + 3y = 11 passes through (5, −3) and (−1, 5), and that 2x + y = 5 passes through (0, 5) and (5/2, 0). These two lines are sketched on the same diagram in Figure 1.5 , from which the point of intersection is seen to be (2, 1). It is easy to verify that we have not made any mistakes by checking that (2, 1) lies on both lines. It lies on 4x + 3y = 11 because 4(2) + 3(1) = 8 + 3 = 11 ✓ and lies on 2x + y = 5 because 2(2) + 1 = 4 + 1 = 5 ✓ For this reason, we say that x = 2, y = 1 is the solution of the simultaneous linear equations 4x + 3y = 11 2x + y = 5 Practice Problem 4. Find the coordinates of the points where the line x − 2y = 2 intersects the axes. Hence sketch its graph. ➜46 CHAPTER 1 LINEAR EQUATIONS Practice Problem 5. Find the point of intersection of 3x − 2y = 4 x − 2y = 2 [Hint: you might fi nd your answers to Problems 3 and 4 useful.] Figure 1.5 Quite often it is not necessary to produce an accurate plot of an equation. All that may be required is an indication of the general shape together with a few key points or features. It can be shown that, provided e is non-zero, any equation given by dx + ey = f can be rearranged into the special form y = ax + b An example showing you how to perform such a rearrangement will be considered in a moment. The coefficients a and b have particular signifi cance, which we now examine. To be specific, consider y = 2x − 3 in which a = 2 and b = −3.SECTION 1.3 GRAPHS OF LINEAR EQUATIONS 47 When x is taken to be zero, the value of y is y = 2(0) − 3 = −3 The line passes through (0, −3), so the y intercept is −3. This is just the value of b. In other words, the constant term, b, represents the intercept on the y axis. In the same way it is easy to see that a, the coefficient of x, determines the slope of a line. The slope of a straight line is simply the change in the value of y brought about by a 1 unit increase in the value of x. For the equation y = 2x − 3 let us choose x = 5 and increase this by a single unit to get x = 6. The corresponding values of y are then, respectively, y = 2(5) − 3 = 10 − 3 = 7 y = 2(6) − 3 = 12 − 3 = 9 The value of y increases by 2 units when x rises by 1 unit. The slope of the line is therefore 2, which is the value of a. The slope of a line is fi xed throughout its length, so it is immaterial which two points are taken. The particular choice of x = 5 and x = 6 was entirely arbitrary. You might like to convince yourself of this by choosing two other points, such as x = 20 and x = 21, and repeating the previous calculations. A graph of the line y = 2x − 3 is sketched in Figure 1.6 . This is sketched using the information that the intercept is −3 and that for every 1 unit along we go 2 units up. In this example the coefficient of x is positive. This does not have to be the case. If a is negative then for every increase in x there is a Figure 1.648 CHAPTER 1 LINEAR EQUATIONS Figure 1.7 Practice Problem 6. Use the slope–intercept approach to sketch the lines (a) y = x + 2 (b) 4x + 2y = 1 corresponding decrease in y, indicating that the line is downhill. If a is zero then the equation is just y = b indicating that y is fi xed at b and the line is horizontal. The three cases are illustrated in Figure 1.7 . It is important to appreciate that in order to use the slope–intercept approach it is necessary for the equation to be written as y = ax + b If a linear equation does not have this form, it is usually possible to perform a preliminary rearrangement to isolate the variable y on the left-hand side. For example, to use the slope–intercept approach to sketch the line 2x + 3y = 12 we begin by removing the x term from the left-hand side. Subtracting 2x from both sides gives 3y = 12 − 2x and dividing both sides by 3 gives y x = − 4 2 3 This is now in the required form with a = −2/3 and b = 4. The line is sketched in Figure 1.8 . A slope of −2/3 means that, for every 1 unit along, we go 2/3 units down (or, equivalently, for every 3 units along, we go 2 units down). An intercept of 4 means that it passes through (0, 4).SECTION 1.3 GRAPHS OF LINEAR EQUATIONS 49 Figure 1.8 We conclude this section with two examples showing how linear graphs can be used in business. Example Two new models of a smartphone are launched on 1 January 2015. Predictions of sales are given by: Model 1: S1 = 4 + 0.5n Model 2: S2 = 8 + 0.1n where Si (in tens of thousands) denotes the monthly sales of model i after n months. (a) State the values of the slope and intercept of each line and give an interpretation. (b) Illustrate the sales of both models during the fi rst year by drawing graphs on the same axes. (c) Use the graph to fi nd the month when sales of Model 1 overtake those of Model 2. Solution (a) The intercept for Model 1 is 4. There are 40 000 sales of this phone when the product is launched. The slope is 0.5 so each month sales increase by 5000. The corresponding figures for Model 2 are 8 and 0.1, respectively. (b) The intercept for Model 1 is 4 so the line passes through (0, 4). For every one-unit increase in n the value of S1 increases by 0.5 so, for example, a two-unit increase in n results in a one-unit increase in S1. The line passes through (2, 5), (4, 6) and so on. The line is sketched in Figure 1.9 . For Model 2, the line passes through (0, 8) and since the slope is 0.1, it passes through (10, 9). (c) The graphs intersect at (10, 9) so sales of Model 1 overtake sales of Model 2 after 10 months. ➜50 CHAPTER 1 LINEAR EQUATIONS Figure 1.9 Example Three companies can supply a university with some mathematical software. Each company has a diff erent pricing structure: Company 1 provides a site licence which costs $130 000 and can be used by anyone at the university; Company 2 charges $1000 per user; Company 3 charges a fi xed amount of $40 000 for the fi rst 60 users and $500 for each additional user. (a) Draw a graph of each cost function on the same set of axes. (b) What advice can you give the university about which company to use? Solution (a) If there are n users then the cost, C, from each supplier is: Company 1: C = 130 000. The graph is a horizontal line with intercept 130 000 Company 2: C = 1000n. The graph is a line passing through the origin with a slope 1000 Company 3: If n ≤ 60 then C = 40 000 If n > 60 then C = 40 000 + 500(n − 60) = 500n + 10 000 The graph for company C is a horizontal line with intercept 40 000 until n = 60 after which the line bends upwards with a slope 500. The graphs are sketched in Figure 1.10 .SECTION 1.3 GRAPHS OF LINEAR EQUATIONS 51 (b) The cheapest supplier depends on the number of users: If n ≤ 40 company 2 is the cheapest; If 40 ≤ n ≤ 240 company 3 is the cheapest; If n ≥ 240 company 1 is the cheapest. Figure 1.10 Key Terms Coefficient A numerical multiplier of the variables in an algebraic term, such as the numbers 4 and 7 in the expression 4x + 7yz2. 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 dx + ey = f. 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.52 CHAPTER 1 LINEAR EQUATIONS Exercise 1.3 1. On graph paper draw axes with values of x and y between −3 and 10, and plot the following points: 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. 2. An airline charges $300 for a fl ight of 2000 km and $700 for a flight of 4000 km. (a) Plot these points on graph paper with distance on the horizontal axis and cost on the vertical axis. (b) Assuming a linear model estimate (i) the cost of a flight of 3200 km (ii) the distance travelled on a flight costing $400. 3. By substituting values into the equation, decide which of the following points lie on the line, x + 4y = 12: A(12, 0), B(2, 2), C(4, 2), D(−8, 5), E(0, 3) 4. For the line 3x − 5y = 8, (a) Find the value of x when y = 2. (b) Find the value of y when x = 1. Hence write down the coordinates of two points which lie on this line. 5. If 4x + 3y = 24, complete the following table and hence sketch this line. x y 0 0 3 6. 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 = 8 4x − 3y = 3 6x + 5y = 15 7. State the value of the slope and y-intercept for each of the following lines: (a) y = 5x + 9 (b) y = 3x − 1 (c) y = 13 − x (d) −x + y = 4 (e) 4x + 2y = 5 (f) 5x − y = 6 8. Use the slope–intercept approach to produce a rough sketch of the following lines: (a) y = −x (b) x − 2y = 6 9. A taxi firm charges a fixed cost of $4 plus a charge of $2.50 a mile. (a) Write down a formula for the cost, C, of a journey of x miles. (b) Plot a graph of C against x for 0 ≤ x ≤ 20. (c) Hence, or otherwise, work out the distance of a journey which costs $24.SECTION 1.3 GRAPHS OF LINEAR EQUATIONS 53 ➜ 10. The number of people, N, employed in a chain of cafes is related to the number of cafes, n, by the equation: N = 10n + 120 (a) Illustrate this relation by plotting a graph of N against n for 0 ≤ n ≤ 20. (b) Hence, or otherwise, calculate the number of (i) employees when the company has 14 cafes; (ii) cafes when the company employs 190 people. (c) State the values of the slope and intercept of the graph and give an interpretation. 11. Monthly sales revenue, S (in $), and monthly advertising expenditure, A (in $), are modelled by the linear relation, S = 9000 + 12A. (a) If the firm does not spend any money on advertising what is the expected sales revenue that month? (b) If the firm spends $800 on advertising one month what is the expected sales revenue? (c) How much does the fi rm need to spend on advertising to achieve monthly sales revenue of $15 000? (d) If the firm increases monthly expenditure on advertising by $1 what is the corresponding increase in sales revenue? Exercise 1.3* 1. Which of the following points lie on the line 3x − 5y = 25? (5, −2), (10, 1), (−5, 0) (5, 10), (−5, 10), (0, −5) 2. Solve the following pairs of simultaneous equations graphically: (a) y = 3x − 1 (b) 2x + y = 6 (c) 2x + 3y = 5 (d) 3x + 4y = −12 y = 2x + 1 x − y = −3 5x − 2y = −16 −2x + 3y = 25 3. State the value of the slope and y-intercept for each of the following lines: (a) y = 7x − 34 (b) y = 1 − x (c) 3x − 2y = 6 (d) −4x + 2y = 5 (e) x − 5y = 0 (f) y = 2 (g) x = 4 4. Identify the two lines in the following list which are parallel: (a) 3x + 5y = 2 (b) 5x − 3y = 1 (c) 5x + 3y = 13 (d) 10x − 6y = 9 (e) y = 0.6x + 2 5. (a) The Wonderful Mobile Phone Company charges $70 per month, and calls cost $0.50 per minute. If I use my phone for x minutes in a month, write down an expression for the total cost in terms of x. (b) Repeat part (a) for the Fantastic Mobile Phone Company, which charges $20 per month and $1 per minute. (c) Plot both graphs on the same axes and hence fi nd the call time per month which gives the same total cost for these two companies.54 CHAPTER 1 LINEAR EQUATIONS 6. A bakery discovers that if it decreases the price of its birthday cakes by $1, it sells 12 more cakes each month. (a) Assuming that monthly sales, M, are related to prices, P, by a linear model, M = aP + b, state the value of a. (b) If the bakery sells 240 cakes in a month when the price of the cake is $14 work out the value of b. (c) Use this model to estimate monthly sales when the price is $9. (d) If the bakery can only make 168 cakes in a month work out the price that it needs to charge to sell them all. 7. (1) Show that the lines ax + by = c and dx + ey = f are parallel whenever ae − bd = 0. (2) Use the result of part (1) to comment on the solution of the following simultaneous equations: 2x − 4y = 1 −3x + 6y = 7 8. Write down the coordinates of the points where the line ax + by = c intercepts the axes.SECTION 1.4 Algebraic solution of simultaneous linear equations Objectives At the end of this section you should be able to: z Solve a system of two simultaneous linear equations in two unknowns using elimination. z Detect when a system of equations does not have a solution. z Detect when a system of equations has infinitely many solutions. z Solve a system of three simultaneous linear equations in three unknowns using elimination. In Section 1.3 a graphical method for the solution of simultaneous linear equations was described. 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 fi t on the diagram, there is no guarantee that the point of intersection itself also lies on it. 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 fi t. The second drawback concerns the accuracy of the graphical solution. All of the problems in Section 1.3 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 1/2, 21/2 and −1/4. In practice, the coeffi cients 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 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 process. The fi nal drawback concerns the nature of the problem itself. Quite frequently in economics 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)56 CHAPTER 1 LINEAR EQUATIONS 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, 4 3 11 4 3 10 1 x y x y y + = + = = − 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 original equation ( 2 ) gives 2(2) + 1 = 5 ✓ The method of elimination can be summarised as follows. Step 1 Add/subtract a multiple of one equation to/from a multiple of the other to eliminate x. 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.SECTION 1.4 ALGEBRAIC SOLUTION OF SIMULTANEOUS LINEAR EQUATIONS 57 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 equations 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 6 4 2 6 3 6 7 8 x y x y y + = − + = = + (5) Step 2 Equation ( 5 ) can be solved by dividing both sides by 7 to get y = 8/7 Step 3 If 8/7 is substituted for y in equation ( 1 ) then 3 2 8 7 1 3 16 7 1 3 1 16 7 x x x + ⎛⎜⎝ ⎞⎟⎠ = + = = − (subtract 16 /7 from both sides) 3 7 16 (put over a common denom 7 x = − inator) 3 9 7 1 3 9 7 x x x = − = × − ⎛⎜⎝ ⎞⎟⎠ = − (divide both sides by 3) 37 The solution is therefore x = −3/7, y = 8/7. [Show More]

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