Chapter 3: Linear relations and equations

Transcription

1 Chapter : Linear relations and equations Ex A: Substitution of values into a formula Algebra involves the use of symbols or pronumerals to simplify expressions and solve problems. A pronumeral is a symbol or letter that takes the place of a number or a variable. A variable is a quantity that can represent different values. A formula is an algebraic rule which describes a mathematical relationship between two or more variables. For example: A = lw is the formula (or rule) used to calculate the area of a rectangle. We say that A is the subject of the formula. Since a formula contains an equals sign, the value of the subject of a formula can be found by substitution. Calculate the area of a rectangle with dimension.5 m and m. A = lw =.5 = 10.5 m 2 Example 2 From a height h m above sea level, an observer can see a distance of d km to the horizon, h where d = 8. 5 What distance, correct to the nearest kilometre, can be seen from a tower 128 m above sea level? 128 d = 8 5 = = 40 km (to the nearest km) 2

2 Ex B: Constructing a table of values We can use a formula to construct a table of values. This can be done by substitution. 4 The formula to calculate the volume of a sphere is given by V = πr. Use this formula to construct a table of values for V using values of r in intervals of 0.5 between r = 0.5 to r =. Give your answers correct to 2 significant figures. r V Substitute all the r values into the formula and use a CAS calculator to evaluate. 4 When r = 0.5: V = π 0.5 = = (correct to 2 significant figures) 4 When r = 1: V = π 1 = = 4. 2 (correct to 2 significant figures) 4 When r = 1.5: V = π 1.5 = = 14 (correct to 2 significant figures) 4 When r = 2: V = π 2 = = 4 (correct to 2 significant figures) 4 When r = 2.5: V = π 2.5 = = 65 (correct to 2 significant figures) 4 When r = : V = π = = 110 (correct to 2 significant figures) So the completed table of values is: r V

3 Using CAS: Substituting into an expression & constructing a table of values A CAS calculator has the capability to substitute a value into an expression and then give the value of the expression. It also has the capability to construct a table of values. Substitution The substitution command can be found by pressing /= Find the value of the expression y 2 (2y + 8) if y = 5 Type y 2 (2y + 8) y= 5 then Constructing a table of values For the rule A = 180(n 2), construct a table of values for A using values of n between and 10 in intervals of Press /N and Add Lists & Spreadsheet 2. Name the first two columns n and a. Names can be inserted in the very top grey cell.. Enter the numbers to 10 in intervals of 1 into the column named n Find the value of the expression a(8 2b) if a = and b = 12 Type a (8 2b) a= and b= 12 then Important! You must insert a multiplication symbol between a pronumeral and a bracket. Ensure there is a space before and after the word and 4. Place the cursor in the grey formula cell in column B (this is the grey cell directly below the name a) 5. Type =180 (n 2) then, select Variable Reference then OK n A

4 Ex C: Solving linear equations with one unknown An equation is a mathematical statement that says that two things are equal. For example, these are all equations: x + 8 = 11 w 4 = 6 5m = 50 Linear equations come in many different forms in mathematics but are easy to recognise because the powers on the pronumerals are always 1. Finding the unknown value is called solving the equation. When solving an equation, opposite (or inverse) operations are used to make the unknown the subject. Operation + 2 x (power of 2) Opposite operation + x (square root) x (square root) 2 x (power of 2) Important! To solve an equation you must keep it balanced, that is, you must perform the same operation on both sides of the equation. m Solve the equation. 10 = 6 By inspection What number divided by 10 gives 6? The answer is 60. m = 60 Inverse operations m = 6 10 m 10 = m = 60 When solving equations it is necessary to simplify like terms and answers (if required). 5

5 Example 2 Solve the equation 2x 6 = 14 for x. Think! The x has been multiplied by 2 first, and then 6 has been subtracted from this. Inverse operations: add 6, then divide by 2. 2x 6 = 14 2x = x = 20 2x 20 = 2 2 x = 10 (add 6 to both sides) (simplify) (divide both sides by 2) (simplify) Example Solve the equation 8 4b = 16 for b. b has been multiplied by 4, then 8 has been added to this. Inverse operations: subtract 8, then divide by 4 8 4b = b 8 = b = 24 4b 24 = 4 4 b = + 6 6

6 Example 4 Solve (6 2x ) = 28 for x. Expand the brackets first, then solve the equation. Expand: 6 + 2x = x = 28 Inverse operations: subtract 18, then divide by x = x 18 = x = 10 6x 10 = x = Using CAS: Solving equations b1 The solve( command can be used to solve and transpose many equations, for instance: Linear equations, non-linear equations, literal equations and simultaneous equations to name a few. The command can be found by pressing b1 in the Calculator page. The syntax for the command is: solve(equation,variable) Example Use the solve( command to solve the following equations: a. 7x = 28 b. 12t + 7 = 115 a. Type solve(7 x=28,x) then press b. Type solve(12 t+7=115,t) then press 7

7 Ex D: Developing a formula Setting up linear equations in one unknown In many practical problems, mathematicians often need to set up a linear equation before finding the solution to a problem. Some practical examples are given below showing how a linear equation is set up and then solved. A rectangular paddock is 40 m long and x m wide. The perimeter of the paddock is 00 m. a. Find an expression for the perimeter of the paddock. b. Write an equation and calculate the width of the paddock. a. perimeter of a rectangle is: l + l + w + w = 2l + 2w x = x b. P = x (an equation must contain an equals sign) We are told that P is = x = x = 2x 220 2x = = x The width of the paddock is 110 m 8

8 Example 2 i. Write an equation to match the words. ii. Solve the resulting equation. a. twice a number less 8 is equal to 18. b. one third a number less 4 is equal to 1. Let n represent the unknown number. a. (i) 2n 8 = 18 n b. (i) 4 = 1 a. (ii) 2n = (add 8) 2n = 26 (divide by 2) n = 1 The unknown number is 1 b. (ii) n = 1+ 4 n = 5 n = 15 (add 4) (multiply by ) The unknown number is 15 Example Paula had n marbles and Stephen had 11 less than Paula. a. Write an expression for the number of marbles Stephen had. b. Write an expression for the total number of marbles. c. If they had a total of 61 marbles, write an equation and calculate the number of marbles Paula had. d. Determine the number of marbles Stephen had. 9

9 Ex E: Solving literal equations A literal equation is an equation whose solution will be expressed in terms of another variable rather than a number. For example: w 2 = 5 is a linear equation in one unknown, whose solution for w is 7; w = 7. x 5y = 9 is a literal equation in two unknowns, whose solution for x is 5y + 9; x = 5y + 9 Solving literal equations can be done using inverse operations. The solution for x has another pronumeral in it. This is a literal equation. Find an expression for x in the following: a. 2x 6y = 20 b. y 4x = 22 a. x has been multiplied by 2, then 6y has been subtracted from this. Inverse operations: add 6y, then divide by 2 2x 6y = 20 2x 6y + 6y = y 2x = y 2x 20 6y = x = 10 + y b. x has been multiplied by 4, then y has been added. Inverse operations: subtract y, then divide by 4 y 4x = 22 y 4x y = 22 y 4x = 22 y 4x 22 y = y x = Literal equations can also be solved using a CAS calculator, however, it is important you are able to solve them without the assistance of a calculator. 40

10 Example 2 d The average speed of a moving object in metres per second is s =, where d is the distance t travelled in metres and t is the time taken in seconds. a. Rearrange the formula to make d the subject. b. Calculate the distance travelled by a car in 20 seconds if its speed is 15 m/s. a. d has been divided by t. Inverse operations: multiply by t. d s t = t t st = d d = st b. d = = 00m Or using CAS: 41

11 Ex F: Developing a formula Setting up linear equations in two unknowns Pens cost $1.25 each and pencils cost $0.50 each. a. Construct a formula for the cost, C, of x pens and y pencils. b. Find the cost of 12 pens and 16 pencils. a. C = 1.25 x y C = 1.25x y Cost = Price of item number of items b. C = = = $2 Example 2 At the hardware store, a packet of nails can be bought for $5.60 and a packet of screws for $2.80. a. Using a (packet of nails) and b (packet of screws), construct a formula to show the total cost, C. b. John bought 6 packets of nails and packets of screws. How much did it cost? a. C = 5.60a b b. C = = = $42 42

12 Ex G: Setting up and solving simple non-linear equations Non-linear equations are equations where the powers of the pronumerals are not 1. In this course, they are best solved using a CAS calculator. Solve the following equations: 2 d a. 5x = 40 b. = 18 8 c. 2e = 6 d. y = 9 Use the solve( command on your CAS calculator. a. type solve(5xl=-40,x) x = 2 b. type solve(dl2p8=18,d) d = 12 or d = 12 c. type solve( 2e = 6,e) e = 108 d. type solve( y = 9,y) y = 144 4

13 Ex H: Transposition of formulas Sometimes the pronumeral we wish to find is not the subject of the formula. We thus rearrange the formula to make another variable the subject. This is called transposing the formula. The rules for transposing are similar to those for solving equations. Transpose the following formulas to make the variable (pronumeral) shown in brackets the subject. PRN a. I = ( N) b. V = AH ( H ) a. Think! N has been multiplied by P and R. N has also been divided by 100. Inverse operations: multiply by 100, then divide by P and R. PRN I = 100 PRN I 100 = I = PRN 100I PRN = PR PR 100I = N PR 100I N = PR b. H has be multiplied by A then divided by (or multiplied by one-third). Inverse operations: multiply by, then divide by A. 1 V = AH 1 V = AH V = AH V AH = A A V = H A V H = A Alternatively a CAS calculator could be used to transpose these equations. 44

14 Ex I: Finding the point of intersection of two linear graphs The point where two linear graphs cross each other (intersect) can be found by sketching both graphs on the same set of axes and reading off the coordinates at the point of intersection. When we find the point of intersection, we are said to be solving the equations simultaneously. When giving the coordinates of the point of intersection they should be written in the form (x, y). That is, the x value is written first followed by the y value. The graphs of y = 2x + 5 and y x = 25 are shown. Write their point of intersection. Answer: (, ) Example 2 The graphs of 2x + 9y = 5 and x 2y = 12 are shown. Write their point of intersection. Answer: (, ) 45

15 Using CAS: finding the point of intersection of two linear graphs Example Find the point of intersection of the two lines y = 2x + 5 and y x = 25 using a CAS calculator. 1. Open the Graphs application 2. Type 2x+5 into f1(x) followed by The other equation is in standard form not y = mx + c form. To correctly enter this equation into the calculator:. Press b21 and type 1 into the box next to x and type into the box next to y. Type 25 into the box after the equals sign. Press To find the intersection point: 5. Press b64 For lower bound move the cursor to a point that is left of the point of intersection and press The point of intersection is not visible on the screen so the Window Settings must be changed. 4. Press b41 and set the following: For upper bound move the cursor to a point that is right of the point of intersection and press Intersection point is (2, 9) 46

16 Ex J: Solving simultaneous linear equations algebraically When solving simultaneous equations algebraically, there are two methods that can be used: substitution or elimination: both methods will be demonstrated here. Method 1: Substitution To solve simultaneous equations using the method of substitution: 1. Check that one of the equations is transposed so that one of the variables has been made the subject. {e.g. x = or y = } 2. Substitute the transposed equation into the second equation.. Solve for the unknown variable 4. Find the value of the other variable Use the method of substitution to solve the following pair of simultaneous equations: y = 2x + and 4x y = 5. The first equation y = 2x + has been transposed so that y is the subject. This equation tells us that y has the same value as (2x + ). They are interchangeable. Substitute the first equation into the second equation, that is, replace y with (2x + ) 4x y = 5 becomes 4x (2x + ) = 5 4x 2x = 5 2x = 5 2x = 8 x = 4 (expand the brackets) Now substitute x = 4 into y = 2x + to find the value of y. y = = 11 The solution is x = 4 and y = 11 or (4, 11) 47

17 Example 2 Use the method of substitution to solve the following pair of simultaneous equations: 2y 6 = x and 7x + y = 25. The first equation 2y 6 = x has been transposed so that x is the subject. This equation tells us that x has the same value as (2y 6). They are interchangeable. Substitute the first equation into the second equation, that is, replace x with (2y 6) 7x + y = 25 becomes 7(2y 6) + y = 25 (expand the brackets) 14y 42 + y = 25 17y 42 = 25 17y = 17 y = 1 Now substitute y = 1 into 2y 6 = x to find the value of x. x = = 4 The solution is x = 4 and y = 1 or ( 4, 1) 48

18 Method 2: Elimination 1. Choose the variable you want to eliminate. 2. Make the coefficients (the number in front of the letter) of that variable equal in both equations. This can be done by multiplication.. Eliminate the variable by either adding or subtracting the two equations. (SSS) If the coefficients have the Same Sign i.e. both + or both Subtract the two equations to eliminate one unknown. (DSA) If the coefficients have a Different Sign i.e. one + and one Add the two equations to eliminate one unknown. 4. Solve for the unknown variable 5. Find the value of the other variable Example Use the elimination method to solve the following: 2x + y = 4 and x y = 2 Let s choose to eliminate y because the coefficients are the same. They are both. The y terms have different signs so use DSA to eliminate them. We only add like terms: x s with x s, y s with y s and constants with constants. 2x + y = 4 x y = 2 Adding gives: 2x + x = x = 6 x = 6 2 = Substitute x = 2 into 2x + y = y = 4 (adding the equations automatically eliminates the y terms) 4 + y = 4 (subtract 4 from both sides) y = 0 (divide both sides by ) y = 0 The solution is x = 2 and y = 0 or (2, 0) (choose the equation with a + y term) 49

19 Example 4 Use the elimination method to solve the following: 5y 2x = 4 and 6x + y = 4 Let s choose to eliminate the y term. The coefficients are different 5y and 1y. Make them the same by multiplication. Leave the first equation alone. Multiply the second equation by 5. [2] 5 : 0x + 5y = 20 [1]: 5y 2x = 4 (unchanged) The y terms have the same signs so use SSS to eliminate them. 0x ( 2x) = x + 2x = 16 2x = 16 (divide by 2) x = 1 2 or x = 0.5 Substitute x = 0.5 into 6x + y = y = 4 + y = 4 y = 4 y = 1 1 The solution is x = 0.5 and y = 1 or (0.5, 1) or,

20 Ex K Solving simultaneous linear equations using a CAS calculator To solve simultaneous equations we make use of the solve( command. The syntax for solving simultaneous equations is as follows: solve(equation1 and equation2,variable) Note: Insert a space either side of the word and. The word and can simply be typed using the alphabetic keys. Solve the following simultaneous equations using a CAS calculator: 5y 2x = 4 and 6x + y = 4 Type solve(5y 2x = 4 and 6x + y = 4, x) Answer: x = 0.5 and y = 1 Example 2 Solve the following simultaneous equations using a CAS calculator: m + n = 17 and 2n m = 4 Type solve(m +n = 17 and 2n m = 4, m) Answer: m = 2 and n = 5 51

21 Ex L Practical applications of simultaneous equations Simultaneous equations are used to solve a variety of problems. Here is a simple algorithm which can be applied to any of them: 1. Identify the variables. 2. Set up simultaneous equations by transferring written information into algebraic equations.. Solve the equations by using the substitution, elimination or CAS calculator methods. 4. Interpret your answer by referring back to the original problem. The word sum in mathematics means addition The word difference in mathematics means subtraction The word product in mathematics means multiplication To finish a project, Genevieve buys a total of 25 nuts and bolts from a hardware store. If each nut costs 12 cents, each bolt costs 25 cents and the total purchase price is $4.0, how many nuts and how many bolts does Genevieve buy? Let x represent the number of nuts Let y represent the number of bolts Buys a total of 25 nuts and bolts turns into x + y = 25 If each nut costs 12 cents, each bolt costs 25 cents and the total purchase price is $4.0 turns into 0.12x y = 4.0 (units must be consistent: divide cents by 100) Type solve(x + y = 25 and 0.12x y = 4.0, x) Number of nuts is 15 and the number of bolts is

22 Example 2 The length of a rectangle is 5 cm longer than its width. If the perimeter of the rectangle is 8 cm, find its dimensions. Let l represent the length of the rectangle Let w represent the width of the rectangle l = w + 5 P = 8 Recall: perimeter of a rectangle is P = 2l + 2w 8 = 2l + 2w and l = w + 5 Using substitution we have: 8 = 2(w + 5) + 2w 8 = 2w w 8 = 4w = 4w w = 7 and l = = 12 (expand brackets) (collect like terms) Dimensions are: width = 7 cm and length = 12 cm Example Rachel is 4 times as old as her brother Nathan. If she will be twice as old as him in two years time, how old is she now? Let R represent Rachel s current age Let N represent Nathan s current age R = 4N R + 2 = 2(N + 2) the +2 represents in two years time Using substitution: 4N + 2 = 2(N + 2) 4N + 2 = 2N + 4 2N = 2 N = 1 Since Rachel is 4 times as old as Nathan, Rachel is currently 4 years old 5

23 Ex M Problem solving and modelling For each of the following, form the relevant linear equations and solve it for x. a. The length of the side of a square is (x 6) cm. Its perimeter is 52 cm. b. The perimeter of a square is (2x + 8) cm. Its area is 100 cm 2. a. The perimeter = 4 length of a side 4 (x 6) = 52 4x 24 = 52 4x = 76 x = 19 b. The perimeter of a square is 2x The length of one side = x + 4 Area of a square is l x 4 = 100 Solve this non-linear equation using CAS. x = 16 as the side length must be a positive number. 54

24 Example 2 Three girls are playing Scrabble. At the end of the game, the total of their scores adds up to 504. Annie scored 10% more than Belinda, while Cassie scored 60% of the combined scores of the other two. What did each player score? A + B + C = % more = an increase of 10% = (100% + 10%) = 110% = 1.1 A = 1. 1B C = 0.6(A + B) Hence, A + B + C = 504 becomes: 1.1B + B + 0.6(A + B) = B + 0.6(1.1B + B) = B + 0.6(2.1B) = 504 (expand brackets) 2.1B B = 504.6B = 504 (divide by.6) B = 150 A = = 165 C = 0.6( ) = 189 Annie scored 165 points Belinda scored 150 points Cassie scored 189 points Example A chemical manufacturer wishes to obtain 700 litres of a 24% acid solution by mixing a 40% solution with a 15% solution. How many litres of each solution should be used? Answer: 252 litres of the 40% solution and 448 litres of the 15% solution 55