Solving equations where the unknown appears on both sides

Transcription

1 Solving equations where the unknown appears on both sides 7.4 So far, we have solved equations with the unknown (variable) on only one side of the equals sign. However, many equations have the unknown on both sides of the equation; for example, 5x + 5 = 4x + 8. Backtracking cannot be used to solve these equations, but we can solve them using algebra. Graphs can also be used to solve these equations. Using algebra Consider the scales on the right. The left-hand side of the scales shows five blocks of mass x g and five blocks of mass 1 g. This can be written in algebraic form as 5x + 5. The right-hand side of the scales shows four blocks of mass x g and eight blocks of mass 1 g. This can be written in algebraic form as 4x + 8. How can we find the value of x? Because the scales are balanced, we can represent the situation with the equation 5x + 5 = 4x x + 5 = 4x + 8 If all four blocks of mass x are removed from the right side of the balance, we will also need to remove four blocks of mass x from the left side to keep the scales balanced. This will leave one block of mass x on the left side of the equation and no blocks of mass x on the right. Writing in algebra: 5x 4x + 5 = 4x 4x + 8 x + 5 = 8 We have used the inverse operation ( 4x) on both sides x + 5 = 8 of the equation to remove the unknown from the right side of the equation. We can now remove the five blocks of mass 1 g from both sides of the scales. They will stay balanced. One block of mass x will remain on the left side of the scales and three blocks of mass 1 g will remain on the right side. Writing in algebra: x = 8 5 x = 3 x = 3 7 Linear equations 425

2 7.4 We have used the inverse operation ( 5) on both sides of the equation to remove the number from the left side of the equation. We have found that the unknown mass x is equal to 3 g by solving the equation 5x + 5 = 4x + 8 and finding that x = 3 is the solution to the equation. We can check by substitution that the value you have found, x = 3, is the solution. LHS = 5x + 5 RHS = 4x + 8 = = = = = 20 = 20 LHS = RHS, so x = 3 is the solution. To solve equations where the unknown appears on both sides of the equation: 1 Perform the same inverse operation on both sides of the equation to remove the variable from the RHS of the equation. This keeps the unknown on the LHS (left-hand side.) 2 Perform the same operation on both sides of the equation to remove any number on the LHS of the equation. 3 Continue to use inverse operations on both sides of the equation until a solution is found. 4 Always check by substitution that you have found the solution. With some equations, this process will give a negative coefficient of x after the first step. Consider 3x + 4 = 5x + 2 Subtracting 5x from both sides: 3x 5x + 4 = 5x 5x + 2-2x + 4 = 2 We prefer to keep the coefficient of the unknown positive. We can do this by interchanging (swapping) the sides of the equation before we start. 3x + 4 = 5x + 2 5x + 2 = 3x + 4 Now, we subtract 3x from both sides of the equation: 5x 3x + 2 = 3x 3x + 4 2x + 2 = 4 This equation is much easier to solve than -2x + 4 = 2. If we think about the balanced scales, this is the same as turning the scales around so the lefthand side and the right-hand side of the scales are reversed. If the coefficient of x is greater on the RHS of the equation than on the LHS, interchange the sides before beginning the solution process. Worked Example Solve the following equations using algebra. (a) 5x 2 = 3x + 8 (b) 2x + 7 = 5x PEARSON mathematics 8

3 7.4 Thinking (a) 1 Write the equation. (a) 5x 2 = 3x Use inverse operations to remove the variable from the RHS of the equals sign. ( 3x) 5x 3x 2 = 3x 3x + 8 2x 2 = 8 3 Use inverse operations to solve for x. (+ 2 first, 2 next) 4 Check by substitution that you have found the solution. 2x = x = 10 2x = x = 5 LHS = RHS = = 23 = 23 x = 5 is the solution (b) 1 Write the equation. (b) 2x + 7 = 5x 8 2 As the coefficient of the variable on the LHS is less than the coefficient of the variable on the RHS, interchange sides. 3 Use inverse operations to remove the variable from the RHS of the equals sign. ( 2x) 5x 8 = 2x + 7 5x 2x 8 = 2x 2x + 7 3x 8 = 7 4 Use inverse operations to solve for x. 3x = x = 15 3x = x = 5 5 Check by substitution that you have found the solution. LHS = RHS = = 17 = 17 x = 5 is the solution Worked Example Solve the following equations using algebra. (a) 3(x + 4) = 2(1 x) (b) 3x x = Thinking (a) 1 Write the equation. (a) 3(x + 4) = 2(1 x) 2 Expand the brackets first, then simplify. 3 Use inverse operations to collect the variable onto the left-hand side of the equals sign. (+ 2x to both sides) 3x + 12 = 2 2x 3x x = 2 2x + 2x 5x + 12 = 2 7 Linear equations 427

4 7.4 4 Use inverse operations to solve for x. ( 12 first, then 5) 5 Check by substitution that you have found the solution. 5x = x = -10 5x = x = -2 LHS = 3(-2 + 4) RHS = 2(1-2) = 6 = 6 x = -2 is the solution (b) 1 Write the equation. (b) 3x = 2x Multiply both sides by the LCD x + 4 = (LCD = 10), expand and simplify x (3x + 4) = 2(2x 1) 15x + 20 = 4x 2 3 Use inverse operations to collect the variable onto the left-hand side of the equals sign. ( 4x from both sides) 4 Use inverse operations to solve for x. ( 20 first, then 11) 5 Check by substitution that you have found the solution. 15x 4x + 20 = 4x 4x 2 11x + 20 = -2 11x = x = x -22 = x = LHS = RHS = = -1 = -1 x = -2 is the solution Try to organise your equation so the variable is on the left-hand side and its coefficient is positive. If brackets are present in the equation, always expand first. If a fraction is present in the equation, always multiply both sides by the LCD (lowest common denominator) and expand the result. Solving graphically We can use graphs to solve equations when the variable appears on both sides of the equals sign by: equating the expression on each side of the equals sign to y sketching the graph of each linear equation locating the point of intersection. The x-coordinate of the point of intersection of the two graphs gives us the solution of the original equation. It is the only x-value that makes the original equation true. 428 PEARSON mathematics 8

5 7.4 3 Worked Example Solve -x 1 = 2x 4 graphically by finding the point of intersection of relevant graphs. Thinking 1 Because the expressions on each side of the equation equal each other, we can equate both of them to y. Write each expression in the form of y = mx + b. 2 Graph each linear equation on the same Cartesian plane (you can use by-hand techniques or technology). y = -x 1 and y = 2x 4 y x Find the point of intersection and read off the values for x and y. 4 The x-value is the solution of your equation, so write this as the solution. 5 Check by substitution that you have found the solution. Substituting the x-value found will give the value of each expression and will give the y-coordinate of the point of intersection. Point of intersection is at (1, -2). Therefore, x = 1 and y = -2. x = 1 is the solution LHS = -1 1 RHS = = -2 = -2 x = 1 is the solution Finding the point of intersection of two linear graphs allows us to solve equations where the variable appears on both sides of the equation, as it is the only point that satisfies both equations. The y-coordinate of your graph gives the value of the expression on both sides of the equals sign. Some equations have no solutions. For instance, 2x + 4 = 2x 7 has no solution. Graphing y = 2x + 4 and y = 2x 7 will show that they are parallel lines. Lines that are parallel to each other have the same gradient, so they will never intersect. Hence, there is no solution for equations when the coefficients of the variable on both sides of the equation are the same. 7 Linear equations 429

6 7.4 Answers page Navigator Solving equations where the unknown appears on both sides Q1 Column 1, Q2 Column 1, Q3 Column 1, Q4, Q5, Q6, Q8, Q9, Q14 Q1 Column 2, Q2 Column 2, Q3 Column 2, Q4, Q5, Q6, Q8, Q9, Q11, Q13, Q14 Q1 Column 3, Q2 Column 3, Q3 Column 3, Q4, Q6, Q7, Q8, Q9, Q10, Q11, Q12, Q13, Q Equipment required: A graphing program can be used for Question 3 Fluency 1 Solve the following equations using algebra. (a) 7c + 2 = 6c 5 (b) 4b + 1 = 2b 15 (c) 3a 5 = a + 7 (d) 5d 4 = 2d + 17 (e) 5e 3 = 2e + 9 (f) 9f + 5 = 5f 3 (g) 3g + 12 = 2g + 5 (h) 2h 7 = h 13 (i) 4x 3 = 7 x (j) k + 6 = 9 k (k) 10j + 1 = 25 2j (l) 3y 22 = -4 3y (m) 2m + 5 = 4m 3 (n) 5 + 4n = 7n 1 (o) 7x 4 = 9x 6 (p) 2 y = 3 2y (q) p + 16 = 19 2p (r) 6 + 3x = 21 2x (s) 6 5p = 30 p (t) 14 5x = 5 2x (u) 17 15a = 39 4a 2 Solve the following equations using algebra. (a) 3(x 2) = 2x + 1 (b) 4(x 1) = x + 8 (c) 5(x + 3) = 2x + 9 (d) 5(2x + 3) = 2(3x + 1) (e) 7(x + 4) = 3(x 4) (f) 4(3x 2) = 3(2x + 9) (g) 2(3x + 7) = 3(5 x) (h) 3(2x + 1) = 4(3 x) (i) 4(2x + 3) = 3(4 5x) 2x (j) x = x (k) x = x (l) x = x (m) x = x (n) x = x (o) x = Solve the following equations graphically by finding the point of intersection of relevant graphs. (a) x + 6 = -x (b) 5x 2 = 2x + 7 (c) 2x 6 = -2x + 10 (d) -2x + 7 = x 5 (e) 5x 1 = 3x + 7 (f) x 2 = 2x + 7 (g) 7x 2 = 2x + 3 (h) 5x 1 = -x + 11 (i) -2x + 1 = 4x 5 4 (a) 9x + 4 = 2x 3 has the solution: A x = 1 B x = -1 C x = 2 D x = 4 (b) 2y + 7 = -8 3y has the solution: A y = 2 B y = 1 C y = -2 D y = -3 (c) 4(3k 2) = 5(k + 4) has the solution: A k = 3 B k = -2 C k = 1 D k = PEARSON mathematics 8

7 Understanding 5 (a) Write an expression to show 11 subtracted from four times a certain number. (b) Write an expression to show two times the number in part (a) subtracted from 7. (c) Equate the expressions in parts (a) and (b). (d) Solve the equation in part (c) to find the unknown number. 6 (a) Write an expression to show 36 added to a certain number. (b) Write an expression to show the number in part (a) being added to 6 and the result doubled. (c) Equate the expressions in parts (a) and (b). (d) Solve the equation in part (c) to find the unknown number. 7 (a) Write an expression to show five less than three-quarters of a certain number. (b) Write an expression to show twice the result of subtracting ten from the number in part (a). (c) Equate the expressions in parts (a) and (b). (d) Solve the equation in part (c) to find the unknown number. 8 (a) Show that (-3, 2) is the point of intersection of y = 2x + 8 and y = -x 1. (b) Write an equation without y that is equivalent to the two equations in part (a). (c) Without using algebra or doing any calculations, write down the solution to your equation in part (b). (d) What does the y-coordinate of the point (-3, 2) tell you about your equation in part (b). Reasoning 9 Find the value of x in the following magic square. The rule for any magic square is that the sum of each row, column and diagonal are equivalent (or add to the same value). x + 3 x + 4 x 1 x 2 6 x x x Sarah has $20 more than Frank, but $40 less than Bruce. If Bruce has $10 more than twice the amount that Frank has, write an equation to find how much money each person has. Let s = the amount of money Sarah has. x 11 Solve 2(x 5) = by using graphs Three times Theo s age a year ago is the same as double his age in five years time. Write this information as an equation and then find Theo s age using graphs. 7 Linear equations 431

8 7.4 Open-ended 13 Make up three equations similar to those in Question 1, where the solution is a positive whole number. 14 Barry solves the equation 6(x + 6) = 2(x 4) as follows. 6x + 36 = 2x 4 step 1 4x + 36 = -4 step 2 4x = 32 step 3 x = 8 step 4 (a) At what step/s did Barry make a mistake? Describe how you would correct this mistake. (b) Solve the equation, remembering to check your answer. Alphametics Problem solving Alphametics are a type of mathematical puzzle in which digits are replaced by letters that form words and phrases. A famous example is: S E N D + M O R E solution The rules of alphametics are: 1 Each letter represents a unique digit. 2 Numbers must not start with a zero M O N E Y Use the example and rules above to solve the following alphametics. 1 E A T 2 H E R E + T H A T + S H E A P P L E C O M E S 3 M E M O 4 C R O S S + F R O M + R O A D S H O M E R D A N G E R 3 The solution is unique (unless otherwise stated). Strategy options Guess and check. Work backwards. Look for a pattern. 432 PEARSON mathematics 8

9 Technology Exploration Excel Equipment required: 1 brain (but you do need someone to try the trick on), 1 computer with an Excel spreadsheet program How tricky was that! Technology Versions of this Exploration for other technologies are available in Pearson Reader. You have probably seen people on television claiming to be able to read minds. Here, we will show you how you can read someone else s mind. Go home and amaze your parents tonight! Step 1 Get your partner to think of a number from 1 to 10. Step 2 Tell them to double it and then add ten. Step 3 Double this answer. Step 4 Divide the new answer by four. Step 5 Subtract the number you started with. You should then wave your hands about a bit before announcing that the answer is five. Watch the amazed look on your partner s face. We will set up a spreadsheet to check that this does, indeed, always give an answer of five. 1 In row 1, enter the numbers 1 to 10 in columns B to K as shown. 2 We need to enter a formula into each of the cells B2, B3, B4 and B5 that will perform the operation required in cells A2, A3, A4 and A5. Because the operation required in A2 is to double and add 10, the formula for B2 is =B1*2+10. Enter an appropriate formula for each of the other cells into column B. To check that you have them correct, you can fill in the rest of the cells using the fill tool. (To do this, highlight cells B2, B3, B4 and B5, then using the little black cross on the bottom right-hand corner of cell B5, drag the mouse across until column K is reached.) If you are correct you should get 5 as the last entry in every column. How did this work? A little algebra will help us here. We will let x stand for the number our partner thought of initially. The starting number = x Double the number and add ten = 2x + 10 Double the result = 2(2x + 10) Factorise by taking out the = 2 2(x + 5) common factor = 4(x + 5) 4 Divide this result by four = ( x + 5) 4 = x + 5 Subtract the number first thought of = x + 5 x from your answer = 5 Aha! 3 Here is another one of these tricks. Think of a number from 1 to 10 and triple it. Then, add nine and double that answer. Then, divide by six and take away three. You should be left with the number you started with. (a) Construct a spreadsheet to demonstrate that this works. (b) Use algebra to explain and prove how this works. 4 You should be getting good at these by now. Again, think of a number from 1 to 10. Add two to the number and then double the result. Add five and then add the number you started with. Divide by three and subtract three more than the number you started with. The answer should be zero. (a) Construct a spreadsheet to demonstrate that this works. (b) Use algebra to explain and prove how this works. 7 Linear equations 433

10 5 (a) Make up your own mind-reading number rule. (b) Construct a spreadsheet to demonstrate that this works. (c) Use algebra to explain and prove how your trick works. Taking it further 6 Can you solve the riddle to find how long Diophantus lived? Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 We are going to use Excel to help solve the riddle. On a new spreadsheet, in cell A1, enter a guess for the number of years Diophantus lived; for example, try 75. Reading the riddle you should be able to see that Diophantus youth was one-sixth of his life. In cell B1, enter =A1/6. Because he grew a beard after half more, enter =A1/12 in cell C1. Because he married after another one-seventh of his life, enter =A1/7 in cell D1. Because he had a son 5 years later, enter 5 in cell E1. Because his son lived exactly half as long as his father, enter =A1/2 into F1. Step 7 Because Diophantus died 4 years later, enter 4 into cell G1. Step 8 Add all the values from B1 to G1 using =SUM(B1:G1) in cell H1. Not much is known about Diophantus, the Greek mathematician who became known as the Father of Algebra. Some scholars believe he lived sometime between 100 AD and 400 AD. One of his admirers wrote the following riddle about Diophantus. Diophantus youth lasted one-sixth of his life. He grew a beard after half more. After another one-seventh of his life he married. Five years later he had a son. His son lived exactly half as long as his father and Diophantus died only four years after his son. All of this adds up to the years Diophantus lived. (Note that the sentence in the riddle He grew a beard after half more refers to half of his youth, not half of his life.) Step 9 If the answer in H1 is the same as the guess in A1, you have guessed correctly. As you can see, 75 is not the correct answer. Try another guess by changing the value in A1. Keep doing this until you guess correctly. Step 10 Write down how old Diophantus was when he died. 7 Write the information as an equation and solve it to find how old Diophantus was when he died. 434 PEARSON mathematics 8

11 Solving problems using equations 7.5 In everyday life, we are faced with problems that can be solved with equations. We need to be able to write equations to represent the problem and then solve them with equation-solving techniques. A simple process for solving problems using algebraic equations is: Step 1: Identify the variable that needs to be found and represent it with a pronumeral. Step 2: Use the information to write an equation, focusing on key phrases, such as is more than, is less than and triple the amount. Turn these phrases into corresponding operations, remembering the correct order of operations and ensuring that you convert all given amounts to the same units. Step 3: Use inverse operations to solve the equation. Step 4: Check that your answer is the solution by substituting it into the original equation and evaluating. Worked Example For the following situation, write an equation to show the situation, define the pronumeral clearly, and then solve the equation. Jared has $6.40 to spend on lollipops that cost 40 cents each. How many lollipops can he buy? Thinking 1 Identify the variable and represent it with a pronumeral. 2 Use the information to write an equation ensuring that you convert all given amounts to the same units. ($6.40 = 640c) 3 Use inverse operations to solve the equation. ( 40) Let n represent the number of lollipops. 40n = n = n = 16 Jared can buy 16 lollipops. 4 Check that your answer is the solution. LHS = 40n = = 640 = RHS Therefore, n = 16 is the solution. 7 Linear equations 435

12 7.5 Worked Example A repair company charges a fixed travel fee of $50 and an hourly rate of $30 for labour on each job completed in a client s home. If h is the number of hours a job takes: (a) write an expression for the total cost of the job. (b) Use an equation to find the number of hours the job took if the total cost of the job was $200. Thinking (a) 1 (b) 1 Identify the unknown and represent it with a pronumeral. 2 Form an expression around the pronumeral. Form an equation using the information given. 2 Use inverse operations to solve the equation. ( 50 first, then 30) 3 Check by substitution that you have found the solution. (a) Let h represent the number of hours. 30h + 50 (b) 30h + 50 = h = h = h = h = 5 The job takes 5 hours. Substitute h = 5 into the original equation. LHS = 30h + 50 = = = 200 = RHS Therefore, h = 5 is the solution. Answers page Navigator Fluency Solving problems using equations Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q9, Q13, Q15, Q16, Q17, Q22, Q24 Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8, Q10, Q14, Q15, Q18, Q19, Q20, Q23 1 For each of the following situations, write an equation to show the situation, define the pronumeral clearly, and then solve the equation. (a) Elena buys 3 apples for $1.17. What is the cost of each apple? Q1, Q2, Q3, Q4, Q5, Q6, Q7, Q8, Q11, Q12, Q15, Q16, Q17, Q19, Q21, Q PEARSON mathematics 8

13 7.5 (b) The mass of a shark pup is 935 kg less than that of its mother. If the pup has a mass of 9 kg, what is its mother s mass? (c) A bus has the capacity to seat 49 passengers. If 21 passengers have boarded the bus, how many seats are still available? (d) After a cool change, the outside temperature fell 9 C to a temperature of 27 C. What was the outside temperature before the cool change arrived? 2 Darren walked to the park while Ann rode her bicycle along the bike path to the park. If Darren walked x km while Ann rode 7 km: (a) write an expression for the total distance travelled by both. (b) Use an equation to find the distance walked by Darren if together they travelled a total distance of 11 km. 3 Alan bought a salad roll and a container of orange juice for his lunch. (a) If the salad roll cost $x and the orange juice cost $2, write an expression for the total cost of Alan s lunch. (b) Use an equation to find the cost of the salad roll if Alan spent $4.50 on his lunch Olivia spent -- of her weekly pay from a part-time job on a new shirt. (a) If Olivia s weekly pay is $w, write an expression for the cost of the shirt. (b) If the shirt cost $24, solve an equation to find how much Olivia is paid each week. 5 The Kittens basketball team won 14 games more than it lost during the last season. (a) If the team lost x games during the season, write an expression for the total number of games the team played. (b) Use an equation to find the number of games won and lost by the Kittens if they played a total of 20 games in the season. 6 (a) Daniel is travelling at a speed of 28 km/h on his bicycle. How far does he travel in 2.5 hours if he maintains this speed? If d represents the distance (in km) Daniel travels in 2.5 hours, the equation to solve is: A d = 28 B d 2.5 = 28 C d = D 2.5d = (b) Melita buys 5 lamingtons from the local bakery and receives $5.50 change from $10. Find the cost of each lamington. If c represents the cost of a lamington (in cents), the equation to solve is: A 5c + 10 = 5.5 B 5c = 1000 C 5c = 550 D 5c = Linear equations 437

14 PEARSON mathematics 8 (c) The perimeter of a rectangular swimming pool is 14 m. w + 3 The pool is 3 m longer than it is wide, as shown in the diagram opposite. If w represents the width of the pool (in m), the equation to w w find the width of the pool is: w + 3 A w + 3 = 14 B 4w + 3 = 14 C 4w + 14 = 6 D 4w + 6 = 14 7 Six friends share a bag of jelly beans equally. After dividing them up, there is one jelly bean left over. (a) If each person received x jelly beans, write an expression for the total number of jelly beans in the bag. (b) Use an equation to work out how many jelly beans each person received if there were 25 jelly beans in the bag. 8 (a) For the rectangle shown, where l is the length of the rectangle and w is the width, write an expression for the perimeter. (b) Use the expression for perimeter to form an equation to solve for each of the following problems. (i) Cameron has a 10 m length of wire fencing he wishes to use to protect a rectangular vegetable garden. If the width must be 2 m, how long can the vegetable garden be? (ii) Cameron finds an extra 5 m of wire fencing, so, instead of having a width of 2 m, the vegetable garden is to have a width of 3 m. Find the new length of the vegetable garden. For each of the following questions, form an equation and solve it to find the answer. 9 A birthday card and wrapping paper cost a total price of $5.65. If the card cost $2.90, how much did the wrapping paper cost? (Write 5.65 on the RHS of your equation.) 10 Samir has $70. At a sale, he buys three CDs (marked at the same price) and receives $11.50 change. How much did each CD cost? (Write 70 on the RHS of your equation.) 11 Megan buys three ice-creams and four drinks for her family at a cost of $12. If each drink costs $1.20, find the cost of each ice-cream. (Write 12 on the RHS of your equation.) Understanding 12 It took Sasha two hours and twelve minutes to get between two towns while travelling to Queensland. The computer in the car told her that her average speed was 97 km/h. Sasha forgot to reset the odometer and she wants to know how far the two towns are apart. (a) Write an equation to show this situation, using d to represent the distance in km. (b) Find the distance that Sasha travelled between the two towns. For each of the following, write an equation and solve it to find the answer. 13 A farmer needs to fence off a rectangular paddock, but must use 4 layers of wire at different heights to keep her piglets in. She uses a 110 m roll of wire and has 2 m of wire left over. Find the perimeter of the piglet pen. 14 Jack spent 7 hours at the beach. If he relaxed in the shade reading for three more hours than he surfed, for how long did he surf? Hence, find the time he spent reading. 15 The perimeter of a rectangular block of land is 74 m. The width is 7 m less than the length. Find the dimensions of the block of land. (First, draw a diagram.) l w

15 Natalie bought 5 cartons of milk and received $2 in change. If she gave $10 to the shopkeeper, find the cost of each carton of milk. Reasoning 17 Yvonne buys a car that costs $ She pays a deposit of $ and then arranges to pay an amount each month for the next three years. Use an equation to find how much each equal monthly repayment should be if she wishes to pay off the car in three years. 18 At the end of the term each student in the class was given their average test score in mathematics. During the term the class sat three tests. For two of her tests Leonie obtained results of 93% and 85%, but she has forgotten her result in the third. Use an equation to find this unknown result for Leonie if she scored an average of 91%. 19 A rectangle has an area of 27 cm 2. Its width is 4 cm, and its length is longer than its width. Write an equation to show this situation, and find its length. 20 A boy is twice as tall as his little sister and 30 cm shorter than his father. The combined height of the three family members is 3.8 m. Find (in cm) the height of each person. 21 The sum of three consecutive even numbers is 102. Find the numbers by using an equation. 22 The angles inside a quadrilateral add up to 360. For the following quadrilateral, write an equation to determine the value of x and then find the value of the four angles. Open-ended 23 Two plumbers quote for a job to renovate a bathroom. Plumber Alan quotes $1050 for material and $50 per hour for labour. Plumber Barry quotes $1200 for material and $45 per hour for labour. Choose different numbers of hours and decide which quote would be cheaper for those hours. For what number of hours will the quotes be the same? 24 The formula for the area of a triangle is: A = 1 -- base height 2 Identify at least three different sets of base length (b) and height (h) to give an area of 64 m 2. (2x + 20) (2x + 40) (x 10) (x + 10) Drawing graphs may help you to answer question 23. The perfect holiday Problem solving The cost of a cruise ship holiday from Totally Oz Tours consists of a booking fee of $150 and a charge of $25 per day. The cost of food is an additional $15 per day and there is a one-off beverage charge of $100 for the duration of the cruise. A rival company, Fun in the Sun Cruises, charges $200 as a booking fee and $45 per day (including food and beverage charge). (a) After how many days does each company charge the same amount? (b) Explain under which circumstances either company would be a cheaper option. Strategy options Make a model. Guess and check. Test all possible combinations. 7 Linear equations 439