pl e pa ge s 02HVZ_SGM_En_TXT.fm Page 47 Thursday, May 28, 2009 3:26 PM O Sa m n Monday night, 7 January 2002, the asteroid, known as 2001 YB5, passed within 600 000 kilometres of Earth. In galactic terms, this is a very near miss. In 1908, a 60-metre-wide asteroid smashed into Siberia and released the same amount of energy as 600 nuclear bombs. 2001 YB5 was 500 metres wide! If 2001 YB5 had hit Earth near Melbourne, then nothing would have survived within a 150-kilometre radius of the impact point and the waves of destruction would have destroyed Sydney and Adelaide. Astronomers and physicists use complex algebraic mathematical models to predict the likelihood and location of impact of large asteroids. Thousands of small asteroids are destroyed each year entering Earth’s atmosphere, but large ones like 2001 YB5 have the potential for mass destruction. 47 47 02HVZ_SGM_En_TXT.fm Page 48 Thursday, May 28, 2009 3:26 PM Prepare for this chapter by attempting the following questions. If you have difficulty with a question, click on the Replay Worksheet icon on your Exam Café CD or ask your teacher for the Replay Worksheet. e Worksheet R2.1 e Worksheet R2.2 2 In the equations below, the equals sign is incorrectly positioned. Reposition the equals sign so that the left-hand side of the equation equals the right-hand side. (a) 7 + 4 + 6 = 5 (b) 6y + 8 + 5 + 3 = 5y + y (c) (10 ÷ 2) = (6 × 3) + (5 × 4) + (9 ÷ 3) (d) 5x + 6 − 2x + 8x = 9 − 3 − 5x e Worksheet R2.3 3 Expand each of the following expressions and simplify where possible. (a) 2(a + 4) (b) x(x + 5) (c) −3(p − 4) (d) x(y − x) (e) (y + 3)(y − 2) (f) (m + 3)(m − 3) e Worksheet R2.4 4 Solve each of the following equations. (a) x + 4 = 11 (b) 2x = 9 e Worksheet R2.5 5 Sketch each of the following on separate axes. (a) y = 3x − 2 (b) y = − 2--3- x + 4 (c) f − 2g (f) 6(e + 2f ) pa ge s 1 Given e = 6, f = −2 and g = 3, find the values of: (a) e + f (b) g − f (d) 6e + 2f (e) 2f 2 m (c) ---- = 3 5 m pl e (c) 2x + 3y = 6 Sa Order of operations: 1. Brackets. 2. Addition and subtraction (work from left to right). 3. Multiplication and division (work from left to right). Expanding: a(b + c) = a × b + a × c. y Sketching linear graphs: c Form: y = mx + c rise run rise y-intercept = c, gradient = m = --------- . run x Form: ax + by = d Substitute x = 0, to find y-intercept. Substitute y = 0, to find x-intercept. y y-intercept x x-intercept 48 Heine ma n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 49 Thursday, May 28, 2009 3:26 PM 2.1 Solving linear equations Operation + − × Opposite operation − + ÷ pa ge s Algebra enables the efficient summary of patterns and relationships. Algebraic techniques are the processes by which many mathematical and real-world problems are solved. A linear equation contains one or more variables raised only to the power of one. 1 Hence, a linear equation will not contain x2, x3, --- , x etc. x For example, 2x − 1 = 7 is a linear equation. An equation is a statement that is only true for certain values of the variable. The equation 2x − 1 = 7 is only true for x = 4. When we are solving any equation we are finding the value or values of the pronumeral that make the equation true; that is, the value on the left-hand side (frequently abbreviated to LHS) of the equation equals the value on the right-hand side (RHS) of the equation. A linear equation will have only one value for the pronumeral that makes the equation true. One method to solve equations is to perform opposite operations to both sides of the equals sign to ‘undo’ the equation and get back to the value of the pronumeral. Whatever operation we want to remove, we simply perform the opposite operation to both sides of the equals sign. Examples of the most common operations are given in the table below. ÷ × square ± square pl e Opposite operations are more formally called inverse operations. If we have built up the equation, the last operation performed is usually the first operation we undo in solving the equation. This undoing procedure is continued to find the value of the pronumeral that makes the equation true. To describe the steps involved in solving an equation we need to use the words ‘coefficient’ and ‘constant’. In the expression 2x − 1, the number multiplying the variable x is called the coefficient of x (in this case, 2) and the number added or subtracted is the constant or the term independent of x (in this case, −1). worked example 1 m Solve the equation 2x − 1 = 7. Sa Steps 1. Add the value of the constant to both sides. In this case, to undo subtract 1, we add 1 to both sides. 2. Divide both sides by the coefficient, in this case, 2. 3. State the answer. 4. Check the solution by substituting into the original equation. Solution 2x − 1 = 7 2x − 1 + 1 = 7 + 1 ∴ 2x = 8 2x ------ = 8 --2 2 ∴x=4 2×4−1=7 ∴ The answer is correct. Often, the individual steps in solving an equation involve simple arithmetic; therefore, the steps shown in red in the above example are frequently omitted. Many of the more complicated equations require only one or two steps before they look like Worked Example 1. It is a useful practice to look for the steps required to get back to an equation of the form of Worked Example 1; that is, an equation that looks the same but has different numbers. The table on the next page gives a useful guide to the steps required to solve many common equations to the stage of being in the same form as Worked Example 1. If you can solve equations like this, look for the steps required to get to this stage. This table does not strictly follow the rule of undoing operations in the opposite order. It takes an approach designed to make the solution process as simple as possible. 2 ● al ge brai c TECHNIQUES 49 02HVZ_SGM_En_TXT.fm Page 50 Thursday, May 28, 2009 3:26 PM First steps in equation solving Equation First step Result after first step 2x – 1 -------------- = 4 3 Multiply both sides by 3. 2x − 1 = 12 2 4(3p + 2) = 7 Expand brackets. 12p + 8 = 7 3 6m − 7 = 4m + 2 Subtract 4m from both sides. 2m − 7 = 2 4 5(2y − 5) = 3(y − 1) Expand brackets. 10y − 25 = 3y − 3 Now, proceed as for equation 3. 5 3(2a – 1) --------------------- = 4 7 Expand brackets. 6a – 3 -------------- = 4 7 Now, proceed as for equation 1. 6 2t – 1 5t – 7 ------------- = ------------3 4 To avoid working in fractions multiply by the LCD, which is 12. 4(2t − 1) = 3(5t − 7) Now, proceed as for equation 4. 7 5(2x – 1) x + 2 --------------------- = ----------6 5 Expand bracket. 10x – 5 x + 2 ----------------- = ----------6 5 Now, proceed as for equation 6. 8 2x + 3 x + 1 -------------- = ----------- + 2 4 3 To avoid working in fractions multiply by the LCD, which is 12. worked example 2 12(2x + 3) 12(x + 1) ------------------------ = --------------------- + 2 × 12 4 3 3(2x + 3) = 4(x + 1) + 24 Expand (equation 4) and collect like terms before proceeding. pl e Solve 3(2 − 3a) = 2(a − 8). pa ge s 1 Sa m Steps 1. Expand the brackets on both sides of the equation. 2. ‘Remove’ the pronumeral from one side. (Either add 9a to both sides, or subtract 2a from both sides. In this case, adding 9a is better because you then work in positives rather than negatives.) 3. If necessary, ‘reverse’ the equation to get the pronumeral on the LHS. 4. Eliminate the constant from the LHS. 5. Divide both sides by the coefficient of the pronumeral. 6. Verify the solution by substituting the value into both sides of the equation and checking that LHS = RHS. Note: You don’t need to actually write step 6, as you can usually check your answer mentally or on your calculator. 50 Solution 3(2 − 3a) = 2(a − 8) 6 − 9a = 2a − 16 6 − 9a + 9a = 2a − 16 + 9a 6 = 11a − 16 11a − 16 = 6 11a = 22 11a 22 ---------- = -----11 11 a=2 LHS = 3(2 − 3 × 2) = 3(−4) = −12 RHS = 2(2 − 8) = 2(−6) = −12 As LHS = RHS the answer is correct. Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 51 Thursday, May 28, 2009 3:26 PM When you reverse the equation, as in step 3 in Worked Example 2, be careful not to change the sign of any term. If you reverse 6 = 16 − 11a, you get 16 − 11a = 6 not 11a − 16 = 6. worked example 3 2e – 4 4e – 1 Solve the equation ---------------- = ---------------- . 3 7 Solution 2e – 4 4e – 1 ---------------- = ---------------3 7 7(2e − 4) = 3(4e − 1) 14e − 28 = 12e − 3 14e − 28 − 12e = 12e − 3 − 12e 2e − 28 = −3 pl e Steps 1. To eliminate the fractions, cross-multiply. pa ge s 2t – 1 5t – 7 In equation 6 in the table on page 46, -------------- = -------------- our first objective is to eliminate the 3 4 fractions. To achieve this we can multiply both sides by the lowest common denominator, (3 × 4 = 12), and then ‘cancel down’. 12(2t – 1) 12(5t – 7) ------------------------- = ------------------------ simplifies to 4(2t − 1) = 3(5t − 7) 3 4 Alternatively, we can convert both sides to a common denominator, and then multiply by the denominator. 4(2t – 1)- --------------------3(5t – 7)--------------------= also simplifies to 4(2t − 1) = 3(5t − 7). 12 12 These steps can be shortened by a process known as cross-multiplying. Whenever an equation is of the form fraction = fraction, we can simply multiply the LHS numerator by the RHS denominator and multiply the RHS numerator by the LHS denominator. Sa m 2. Expand the brackets. 3. Subtract the smaller value of the variable (in this case, 12e) from both sides to eliminate the variable from one side, and collect like terms. 4. Eliminate the constant from the side involving the pronumeral. (Add 28 to both sides.) 5. Divide both sides by the constant (2). 6. Check the answer by substituting into the original equation. 2e = 25 e = 12 1--22(12 1--2 ) – 4 25 – 4 21 - = ---------------- = ------ = 7 LHS = ------------------------3 3 3 1 4(12 --2 ) – 1 50 – 1 49 - = ---------------- = ------ = 7 RHS = ------------------------7 7 7 ∴ LHS = RHS 2x + 5- = 3x – 3- + 2 ------------You cannot use cross-multiplying when the equation is of the form ------------2 6 2 ● al ge brai c TECHNIQUES 51 02HVZ_SGM_En_TXT.fm Page 52 Thursday, May 28, 2009 3:26 PM worked example 4 A rectangle has a perimeter of 72 cm. If the length is 2 cm more than the width, find the dimensions of the rectangle. Steps 1. Introduce a pronumeral to represent an unknown quantity. 2. Express other quantities in terms of this unknown. 3. If possible, draw a diagram. You can sometimes define the unknown quantities on the diagram and omit steps 1 and 2. Solution Let x represent the width of the rectangle. Length will be x + 2 x x+2 6. Solve the equation. 7. Interpret and verify the solution. 8. State the final answer in words. 2x + 2(x + 2) = 72 2x + 2x + 4 = 72 4x + 4 = 72 4x + 4 − 4 = 72 − 4 4x = 68 x = 17 If x = 17, then x + 2 = 19 2(17) + 2(19) = 72 The dimensions of the rectangle are width 17 cm and length 19 cm. pa ge s 4. Express the information as an equation. Twice the width plus twice the length equals the perimeter. 5. If possible, simplify the equation. pl e Solving equations using CAS Your CAS has a Solve function that enables you to check the solutions you have found. 2e – 4 4e – 1 The equation from Worked Example 3, -------------- = --------------, will be used to illustrate the process. 3 7 Sa m Using the TI-Nspire CAS Solve is found by pressing b > Algebra > Solve and we would be wise to use the fraction template, / p , to make correct entry easier. Notice that you need to put the variable for which you are solving after a comma. 52 Using the ClassPad Solve is found by tapping Action > Equation/ Inequality > solve and we would be wise to use the fraction template, found in the ) key pad, to make entry easier. You should use the V key pad to enter the variables. The variable of interest is included at the end after a comma. Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 53 Thursday, May 28, 2009 3:26 PM Solving linear equations Short answer 1 Solve the following equations. (a) 3m − 2 = 7 (b) (k) 4(3p + 2) = 7 (l) 5 − 2--7- d = 3 (b) 5 − 2a = 5a + 2 (e) 4 − 3x = 6 − 2x (h) 8 − 2(3 − 4e) = 5(e − 4) (k) −2(3 − 5d) + 1 = 2d − 1 (c) 5e + 7 = 2e − 5 (f) 3(3 − 2f ) = f + 5 (i) −4(3m + 1) = 5(2 − 3m) (l) 4 − 3(2b − 1) = 3(2b − 1) (b) (e) (h) (k) e 10x – 5 x + 2 ----------------- = ----------6 5 6 – 2y y + 1 --------------- = -----------3 5 3(5 – f ) 3f + 2 ------------------- = -------------2 6 7(4 – 2h) h ----------------------- = --3 4 Hint Worked Example 1 Worked Example 2 Worked Example 3 3m – 3 3m + 2 (c) ---------------- = ----------------5 3 − 2(2x + 4) 5(3x – 5) (f) ----------------------- = ------------------------3 7 − 4(g + 3) g (i) ---------------------- = --5 2 3(2a – 1) a (l) ----------------------- = --2 5 Worked Example 4 Sa m 5 – 2y (g) --------------- = 2 3 3(2a – 1) (j) ----------------------- = 4 7 2 Solve the following equations. (a) 6m − 7 = 4m + 2 (d) 5(2y − 5) = 3(y − 1) (g) 5(2y + 1) = 3y + 8 (j) −(7 − 2a) = 7 − 2a 3 Solve the following equations. 2x + 1 3x – 2 (a) --------------- = -------------3 4 5(2x – 1) x + 2 (d) ----------------------- = ----------6 5 2(3 – 4x) x + 1 (g) ----------------------- = ----------5 2 − 7(2 – y) y (j) ---------------------- = --3 4 4 The perimeter of a rectangular number plate is 92 cm. If the width is 6 cm less than the length, find the dimensions of the number plate. (c) 3 − 2x = 5 2(g – 3) (f) ------------------- = 6 5 2d (i) 3 − ------ = 7 5 pa ge s (d) 6d = 2d + 4 − 5 = −8 2x – 1 (e) -------------- = 4 3 − 4(1 – x) (h) ---------------------- = 8 3 −2e pl e exercise 2.1 5 Solve the following. Hint: Why can’t you cross-multiply these equations? 2x + 3 x + 1 2(4y + 3) 3y + 2 2e + 3 –4 (a) --------------- = ----------- + 2 (b) ----------------------- = --------------- − 1 (c) 3e -------------- + 2--- = --------------4 3 5 3 3 7 3 n c 3m – 5 3m 3 – n 2 2 – 3c 1 (d) ---------------- – ------- = 2m + 1 (e) ------------ + --- = --(f) -------------- + --- = --6 4 2 5 2 4 5 2 Multiple choice − 5(x – 2) 6 The solution to the equation ---------------------- = 3 is: 4 14 − − -----B ----C 22 A 2 5 5 3x – 4 2x + 1 7 The solution to the equation -------------- = --------------- − 1 is: 3 5 20168-----A ----B C 3 3 9 e Hint − 22 -----5 D − 2--5 E D 22----9 E 4 2 ● al ge brai c TECHNIQUES 53 02HVZ_SGM_En_TXT.fm Page 54 Thursday, May 28, 2009 3:26 PM e Hint pa ge s Extended answer 8 The repair charges to Jo’s hot water heater were a fixed service call fee of $55 covering the first 30 minutes of service, plus an additional charge for each 15 minutes of service beyond the initial 30 minutes. (a) If the additional charge was $20 per each 15 minutes beyond the initial 30 minutes, what would be the charge for a repair that took 1 hour? (b) If the service call took 1 hour 15 minutes and the bill was $122.50, what was the charge for each 15 minutes beyond the initial 30 minutes? (c) Another plumber charges a fixed service call of $35, covering the first 20 minutes of the service, plus an additional charge of $20 for each additional 15 minutes, or part of 15 minutes. How much will this plumber charge for a service call that takes 1 hour 15 minutes? pl e 2.2 Formulae and substitution Formulae Sa m A formula is an equation or rule that mathematically expresses the relationship between two or more quantities. A formula is therefore a shorthand form of expressing a relationship, thus enabling us to more easily calculate solutions and analyse and solve practical problems. Some common formulae that you might recognise include: A = --12- bh area (A) of triangle of base b and height h A = πr2 area (A) of circle of radius r 2 F = 9--5- C + 32 converts degrees Celsius to V = πr h volume (V ) of a cylinder of radius r and height h degrees Fahrenheit PrT I = --------calculates the simple interest earned (I ) when an amount ($P) is invested at a rate of r% for T years 100 T = 2π -l- period (T) in seconds of a swinging pendulum of length l metres g d v = --average speed (v) of a moving object that travels d metres in t seconds t The letters used in equations are called pronumerals, as they represent numbers. If the base of a triangle is 6 cm and the height is 8 cm the area is given by: A = 1--2- bh, then A = 1--2- × 6 × 8 = 24 cm2 Note: A pronumeral can represent either an unknown constant, which is a fixed value, or a variable, which can take different values. In the equation y = 2x + 3, 2 and 3 are constants and x and y are variables. l The formula for calculating the period of a pendulum is: T = 2π -g 54 Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 55 Thursday, May 28, 2009 3:26 PM The pronumeral, g, represents the gravitational constant, which is assumed at the Earth’s surface to have a value of 9.8, and π = 3.141 592 6…; hence, π and g are constants and the period, T, and pendulum length, l, are variables. If we were conducting an experiment to examine the period of a pendulum, we would vary the length, l, of the pendulum to determine its effect on T; hence, the values of l and T would vary. Conventions Conventions exist for choosing which letters to use for pronumerals and for the formatting of formal typed or word-processed formulae. • Letters at the beginning of the alphabet (a, b, c, …) are usually used to denote constants. • Letters at the end of the alphabet (… x, y, z) are usually used to denote variables. • Where possible, letters are chosen that help to relate to the quantity they represent, such as A for area and V for volume. • This textbook follows the convention of presenting all pronumerals in italic. Substitution pa ge s Substitution is the process of replacing one quantity or number by another. The most common mathematical use of substitution is to replace a pronumeral by the number that it represents. pl e Before substituting values into a formula, check to make sure the units are compatible. For example, if one given side of a rectangle was in centimetres and the other in metres, you would have to convert them both to centimetres or both into metres before proceeding to calculate perimeter, area, volume etc. Common sense and practicality usually dictate the sensible choice of units. We wouldn’t, for example, measure the distance from Melbourne to Perth in millimetres! worked example 5 Sa m This formula is used to convert temperature in degrees Fahrenheit (°F) to temperature in degrees Celsius (°C): C = 5-9 (F − 32) Use the formula to convert 100°F to degrees Celsius. Steps 1. Write the formula. Solution C = 5-9 (F − 32) 2. Substitute known values. 3. Evaluate, using a calculator if necessary. (Calculator shows 37.777 777 78, but we would never write down all these digits.) 4. State the final answer, including units, to an appropriate degree of accuracy. C = 5-9 (100 − 32) Display shows 37.777… C = 37.8°C 2 ● al ge brai c TECHNIQUES 55 02HVZ_SGM_En_TXT.fm Page 56 Thursday, May 28, 2009 3:26 PM Defining a function to aid substitution Your CAS can be very helpful with substitutions, especially if you need to substitute a lot of numbers into the same expression. One of the ways your CAS can help you is by defining the expression. The degree conversion from Worked Example 5 will be used to illustrate this. Once you have defined the rule you can just enter whatever value you like for f. If you don’t want the exact answer you can get an approximation by tapping D . . pl e Once you have defined the rule you can just enter whatever value you like for f. If you don’t want the exact answer you can get an approximation by pressing / · . Using the ClassPad You find Define by tapping Action > Command > Define. Notice the need to include the variable, f, on the LHS of the definition. Of course, you could use whatever variable you liked as long as you were consistent. You do need to be careful to use c from the 0 key pad. pa ge s Using the TI-Nspire CAS You find Define by pressing b > Actions > Define. Notice the need to include the variable, f, on the LHS of the definition. Of course, you could use whatever variable you liked as long as you were consistent. Sa m Don’t forget to include a multiplication between the fraction and the bracket. Degree of accuracy Depending on the situation, an appropriate degree of accuracy is determined either by common sense in the particular situation or, if we are referring to anything measured with an instrument, by the accuracy limitation of the particular instrument. You would not normally need to know your height in millimetres—to the nearest centimetre would usually be accurate enough. For a scientific application, an appropriate degree of accuracy would usually be determined by the accuracy limitations of the measuring instruments. With a ruler, for example, we could probably measure to an accuracy of 0.5 mm. If we needed greater accuracy we would require a more accurate measuring instrument. If you are going to perform a calculation on measured figures, your answer cannot be more accurate than the figures you used to calculate the answer. Therefore, if the given figures in the same units were to two decimal places, an answer you calculated from these figures could not be to more than to two decimal places. 56 Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 57 Thursday, May 28, 2009 3:26 PM Sometimes you will be instructed to state your final answer to a given number of decimal places. When such a situation occurs, assume this instruction overrides any other considerations. worked example 6 The period of a pendulum, T seconds, which is the time taken for a pendulum of length, l metres, to swing one to-and-fro motion, is given by T = 2π --l- , where g = 9.8. Find the period of a pendulum of length 6 cm. g Solution 1. Write the formula. 2. Convert units if necessary. The rule states that l is measured in metres, so we must convert the length into metres. 3. Substitute known values. 4. Evaluate and state the answer correct to appropriate accuracy (one decimal place here, as this is the accuracy of the input values). exercise 2.2 l T = 2π --g l = 6 cm 6 = ---------- m 100 = 0.06 m 0.06 T = 2π ----------9.8 T = 0.5 seconds pa ge s Steps Formulae and substitution pl e Short answer 1 The formula C = --59- (F − 32) is used to convert temperature in Fahrenheit to temperature in Celsius. Use the formula to convert 68.0°F to degrees Celsius. l 2 The period of a pendulum, T seconds, of length, l metres, is given by T = 2π -g where g = 9.8. Find the period of a pendulum of length 15.4 cm. Worked Example 5 e Hint Worked Example 6 Sa m 3 Write each of the following statements as a mathematical formula. (a) The area (A) of a rectangle is the product of its length (l) and width (w). (b) The circumference (C) of a circle is the product of its radius (r) and 2π. (c) The average speed (v) of a moving body is the distance (d) travelled divided by the time (t) taken. (d) The charges (C) to repair a video camera are $35 per service, plus $12 per 15 minutes of repair time (n). (e) In a right-angled triangle, the square of the hypotenuse (c) is equal to the sum of the squares of the other two sides. Let the other two sides be a and b. (f) The potential difference (V) across the terminals of an ohmic conductor is given by the product of the current (I) flowing through the device and the resistance (R) of the conductor. 2 ● al ge brai c TECHNIQUES 57 02HVZ_SGM_En_TXT.fm Page 58 Friday, July 6, 2012 11:46 AM ge s (g) The daily wage (w) Laura earns selling ﬂowers at a ﬂorist shop is $16 per day plus $1.20 per bunch (n) of ﬂowers she sells. (h) The cost (C) per day of hiring a car is a ﬁxed charge of $22 per day, plus 40 cents per kilometre (d ) travelled. (i) The volume (V ) of a prism is the product of the cross-sectional area (A) and the height (h). (j) The volume (V ) of a sphere is the product of 4--3- π and the cube of the radius (r). e Hint e Hint pl e pa 4 For each of the following, ﬁnd the value of the ﬁrst named pronumeral by substituting the stated values into the given formula. (a) v = u + at, u = 4, a = 3, t = 7 (b) A = lw, l = 4.2, w = 6.7 (c) V = 4--3- πr 3 , r = 3 (d) S = ut + 1--2- at 2 , u = 3.2, t = 3, a = 2 (e) E = mc2, m = 5, c = 3 × 108 (f) V = IR, I = 2.4, R = 400 (g) S = 2πr (r + h), r = 4.6, h = 8.9 (h) F = ma, m = 8.7, a = 4.7 n (i) t = ar , a = 3, r = 2, n = 4 (j) F = --95- C + 32, C = 30 2 πd (k) A = --------, d = 6 (l) T = 2π -l-, l = 16.5, g = 9.8 4 g m Multiple choice l 5 The period, in seconds, of a swinging pendulum of length, l metres, is given by T = 2π -g where g = 9.80. The period, in seconds, of a pendulum of length 8 cm is closest to: A 0.05 B 0.18 C 0.57 D 5.13 E 5.68 Sa S – 2πr 2 6 Substituting S = 64 and r = 3 into the formula h = -------------------- produces a value of h closest to: 2πr A 0.395 B 2.395 C 3.902 D 23.641 E 34.391 Extended answer 7 Bernie, the organic dairy farmer, wants to fence off a new rectangular paddock to quarantine his new bull, named Dunny Doo, from the cows. He only needs to fence three sides, as the new paddock will be adjoining an existing fence. Bernie has a 200-metre roll of fencing wire. The wire is multi-strand hinge-joint, so Bernie doesn’t need to consider the strands of wire in determining his fencing requirements. Let x = the length of the longer side of the paddock. 58 H e i n e ma n n V CE Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 82 Thursday, May 28, 2009 3:26 PM The diagram opposite shows a quadrilateral drawn on a grid of square dots. For the given shape, the number of enclosed or inside dots (I) is 4, and the number of perimeter dots (P) is 9. As the shape is a trapezium the enclosed area (A) is 1--2- (3 + 2)3 = 7 1--2- squares. (i) Using 1 cm graph paper, taking the points of intersection of the 1 cm lines as representing ‘dots’, construct squares of increasing size, for example, (1 × 1), (2 × 2), up to (5 × 5), and record the results in a table as shown below. Square size Area (A) Perimeter dots (P) Inside dots (I) 1×1 1 4 0 2×2 4 pa ge s (a) 1 3×3 4×4 5×5 8 Sa m pl e (ii) Find a relationship connecting all three variables, the perimeter dots (P), inside dots (I) and area (A). (b) Systematically construct rectangles of varying sizes and find a relationship between P, I and A for rectangles. (c) Find the relationship between P, I and A if the shape is a trapezium. (d) Generalise the relationship between P, I and A for any quadrilateral drawn on a square grid of dots. 82 Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 83 Thursday, May 28, 2009 3:26 PM Westland pa ge s Part A At the Westland Shopping Centre, there are two florist shops. Both have offered Laura a part-time Saturday morning job, standing outside the florist shop selling cut flowers. The main difference between the two jobs is the way in which Laura will be paid. At the ‘Poise and Ivy’ florist shop she will be paid $16 every Saturday morning, plus a bonus of $1.20 for each bunch of cut flowers she sells. At the ‘Friendly Triffid’ florist shop she will be paid $32 every Saturday morning, plus a bonus of 40 cents for each bunch of cut flowers she sells. 1 Let w represent Laura’s wage, and n represent the number of bunches of flowers sold. Express the given information as two simultaneous equations. 2 How many bunches of flowers would Laura need to sell so that she makes the same amount of money whichever job she accepts? 3 How much money would Laura make if she did sell this many bunches of flowers? 4 What information should Laura consider when deciding which job to accept? Sa m pl e Part B Next door to the ‘Friendly Triffid’ at Westland Shopping Centre is ‘Mario’s Wholesale Green-grocers’, which is having a sale on boxes of fruit and vegetables.You can buy two different boxes for the following combinations and prices. • A box of apples and a box of bananas for $21. • A box of bananas and a box of carrots for $24. • A box of carrots and a box of dill for $32. • A box of dill and a box of eggplant for $37. • A box of eggplant and a box of figs for $31. • A box of figs and a box of garlic for $25. • A box of garlic and a box of apples for $26. 1 If Mario allows you to buy the boxes individually for the same comparative price as the sale price, find the cost of each individual box. Part C Mario sells apples in a number of different-sized boxes. Five different-sized boxes of apples are now weighed two at a time, in all possible combinations. The weights obtained are 16, 18, 19, 20, 21, 22, 23, 24, 26 and 27 kilograms. 1 Find the weight of each individual box of apples. 2 ● al ge brai c TECHNIQUES 83 02HVZ_SGM_En_TXT.fm Page 84 Thursday, May 28, 2009 3:26 PM Summary Solving simultaneous equations • Simultaneous equations can be solved in three ways: 1. Graphically: Sketch the two lines on the same set of axes and then read off the point of intersection. This method is good for analysis, but is tedious and possibly inaccurate. 2. Substitution: Substitute one equation into the other equation and then solve. This is a versatile technique that can be applied to non-linear equations and gives exact answers. 3. Elimination: The coefficients of the variable being eliminated must be the same in both equations. This variable is then eliminated by either adding or subtracting the equations. If the coefficients of the variable being eliminated are the same sign in both equations, then subtract the equations; if they are opposite signs, then add the equations. • Simultaneous equations can be checked on the CAS. pl e Transposition and substitution • Transposing an equation is the same process as solving an equation. We perform the same steps on pronumerals as we do with numbers. • It is more efficient and usually easier to transpose an equation to make the desired quantity the subject, before substituting known values. • Substitution is the process of replacing one number or pronumeral by another. Before substituting, check to ensure that all quantities are measured in the same or appropriate units. Checking algebraic processes • Algebraic simplifications and factorisations can be checked by making up values for the pronumerals and substituting to determine if LHS = RHS. • Solutions to equations can be checked by: 1. Substituting the answer and determining if LHS = RHS. 2. Using solve on a CAS. pa ge s Solving linear equations • When solving linear equations we often expand brackets first. • If fractions are present, multiply the equation by the lowest common denominator. • If the equation is of the form fraction = fraction, then to eliminate the fractions you can cross-multiply. • Collect all the unknowns on one side and then undo or reverse the operations. Sa m Developing formulae from description • To develop a formula, first define the variables. • If the situation allows, draw a large, clear diagram and, if possible, connect your variables on the diagram so that the number of variables is reduced. • An even number can be represented by 2n and an odd number by 2n + 1. 84 Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 85 Thursday, May 28, 2009 3:26 PM Use the following to check your progress. If you need more help with any questions, turn back to the section given in the side column, look carefully at the explanation of the skill and the worked examples, and try a few similar questions from the Exercise provided. Short answer 1 Solve each of the following equations. 2y – 1 (a) --------------- = 6 (b) 2(3m + 4) = 8 4 2.1 (c) 2(3e − 4) = 2e + 1 2(3x – 1) (d) ----------------------- = 2x + 1 5 m pl e pa ge s PrT 2 The formula to calculate the simple interest earned is given by I = --------- where I is the interest on a 100 sum of money, P, invested for T years at an interest rate of r% per annum. (a) Find the interest earned if $850 is invested for 3 years at a simple interest rate of 3% per annum. (b) Find the interest rate if the simple interest earned on $620 invested for 4 years is $62. (c) How long will it take $500 to earn $50 interest if the interest rate is 3.5%? Round your answer up to the next month. 3 The volume of a cylinder is given by V = πr2h. Find the radius, correct to two decimal places, of a cylinder with a height of 2.5 metres and a volume of 25 cubic metres. 4 Transpose each of the following to make the letter in brackets the subject. m(v – u) GMm (a) F = --------------------(u) (b) F = -------------(m) t r2 5 An isosceles right-angled triangle has one side length of (x + 2) cm and the length of the hypotenuse is 12 cm. Find the perimeter of the triangle. 6 Solve each of the following equations using your CAS, giving your answers to three decimal places. (b) 5x2 − 3x = 8 (a) 5x3 + 2x − 3 = 0 h 7 The distance to the visible horizon, in kilometres, is given by d = 8 --- where h metres is the height 5 above sea level. 2.2 2.3 2.3 2.4 2.5 2.6 (a) Sa Using a graphics calculator, generate a table of values and, hence, complete the following tables, giving your values correct to one decimal place. h metres 4 7 10 13 16 0.6 1.5 8 22 40 d kilometres (b) h metres d kilometres 8 Solve each of the following sets of simultaneous equations. (a) 2x + y = 4 (b) 2x + y = 7 (c) 3x − 2y = 10 y=x−1 3x − y = 3 5x − 2y = 18 (d) + 3y = 6 3x − 4y = 8 2.7 −2x 2 ● al ge brai c TECHNIQUES 85 02HVZ_SGM_En_TXT.fm Page 86 Thursday, May 28, 2009 3:26 PM Sa m pl e pa ge s 9 The cost of two adult and two children’s tickets to attend the Australian Open Tennis Tournament is $124, while the cost of purchasing one adult and three children’s tickets is $112. Find the cost of purchasing two adult and four children’s tickets. 86 Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 2.7 02HVZ_SGM_En_TXT.fm Page 87 Thursday, May 28, 2009 3:26 PM Multiple choice 10 The solution to 2--3- x − 1 = 7 is: A 9 B 12 2y – 4 11 The solution to --------------- = 2y is: 5 A 4--5B − 1--3- 2.1 C 4 D 8 E 11 2.1 C −2 1--2 = −2, then 12 Given s = ut + 1--2- at 2 , when u = 4, t = 3 and a A 21 B 9 C 3 D − 1--2 s is equal to: D 15 E 1 --3 E −2 2.2 m pl e pa ge s 13 In the equation S = 2(x + 16) 2x 2 + 2x + 1 + 2x(2x + 17), when x = 7, the value of S, to the nearest whole number, is: A 923 B 5632 C 15 593 D 489 E 212 220 14 When v = u + at is transposed to make a the subject, a is equal to: v v u v–u A -- − u B v−t−u C --- − t D v − --E -----------t u t t 15 When S = 2πr(r + h) is transposed to make h the subject, h is equal to: S S – 2πr S S A s−r B --------- − r C -----------------D --E ----------2- − 2πr 2πr r r 2πr 16 Bronwyn is three times as old as her sister, Jessica. In 4 years time, Bronwyn will only be twice as old as Jessica. If j represents Jessica’s age now, this information can be represented algebraically as: A 2j + 4 = 3j + 4 B 3j + 4 = 2j C 6j + 4 = j + 4 D 2( j + 4) = 3j + 4 E j + 4 = 2(3j + 4) x 2 – 4x + 4 17 The expression -------------------------- is equivalent to: 2–x D x2 − 2x + 4 E 1 A 2−x B x−2 C x2 − 2x + 2 Sa 18 Given W = πd(r2h + --23- r 3 ), if d = 3 and h = 2, then the value of W when r = 5.7 is closest to: A 312 B 736 C 817 D 1388 E 1776 19 The solution to the pair of simultaneous equations y = 2x − 7 and 3x + y = 8 is: A (−3, −13) B (−7, 8) C (−15, −37) D (3, 1) E (3, −1) − 20 The solution to the pair of simultaneous equations 2x − y = 4 and x − 2y = 1 is: A (9, 5) B (1, −1) C (4, −1) D (3, 2) E (−1, −6) 21 The points (2, −3) and (4, 3) lie on the straight line with equation ax + by = 6. The values of a and b, respectively, are: A 1, − 2--3B 2, − 2--3C 0, 2 D 3, 0 E 1 1--2- , 0 2 ● al ge brai c 2.2 2.3 2.3 2.4 2.5 2.6 2.7 2.7 2.7 TECHNIQUES 87 02HVZ_SGM_En_TXT.fm Page 88 Thursday, May 28, 2009 3:26 PM Extended answer 22 w x x x l pa ge s x 21 cm x x x x 30 cm 23 The range, R, of a projectile, which is the horizontal distance travelled, is given by the formula: V 2 sin 2θ R = -----------------------, 9.8 where V = the speed of projection, and θ = the angle projected to the horizontal. (a) Find the range of a projectile with projection speed 10 m/s and angle to the horizontal of: (i) 30° (ii) 45° (iii) 60° (iv) 70° (b) Explain why the maximum range occurs when the object is projected at an angle of 45° to the horizontal. (c) Find the maximum range of an object projected with an initial speed of 25 m/s. (d) Find the speed of projection of a projectile with a maximum range of 50 metres. Sa m pl e The corners are cut from a 21 cm by 30 cm sheet of paper as shown in the diagram above. The sides are folded up and the shaded rectangle is folded over to form the lid of a box. The volume of a box is given by: V = length × width × height (a) Express the length, l, and width, w, of the box in terms of x. (b) Show that the volume of the box is given by V = (21 − 2x)(15 − x)x. (c) Find the volume of the box in cm3, to two decimal places, when x = 4 cm. (d) State, with reasons, the largest possible value for x. Use your CAS to sketch the graph of v = (21 − 2x)(15 − x)x and use the graph to find answers to the next two questions. (e) Find the value for x that produces the box of maximum volume. (f) Find the maximum volume of the box. 88 Heinem a n n V C E Z O N E : S T A N D A R D G E N E R A L M A T H E M A T I C S E N H A N C E D 02HVZ_SGM_En_TXT.fm Page 89 Thursday, May 28, 2009 3:26 PM exam focus 3 VCAA 2004 Further Mathematics Units 3 & 4, Exam 1, Module 3, Question 6 pa ge s The cost, $C, of hiring a boat for x hours is given by the equation C = ax + b where a is the hourly rate and b is a fixed booking fee. When the boat is hired for 4 hours the cost is $320. When the boat is hired for 6 hours the cost is $450. When the boat is hired for one hour the cost is A. $65 B. $75 C. $77 D. $80 E. $125 exam focus 4 VCAA 2004 Further Mathematics Units 3 & 4, Exam 2, Module 1, Question 2 pl e The purchase and installation of a basic heating system with five outlets costs $3500. Each additional outlet costs an extra $80. a. Determine the cost of installing a heating system with eight outlets. 1 mark 2 marks c. Australian Heating recommends that a house with 20 squares of living area should have 12 heating outlets. Using this recommended ratio, determine the cost of installing a heating system for a house having 35 squares of living area. 2 marks Sa m b. A customer has $4400 to spent on a heating system and outlets. Determine the greatest number of outlets that can be bought with this heating system. 2 ● al ge brai c TECHNIQUES 89

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