1 Linear and Absolute Value Equations

Transcription

1 1 Linear and Absolute Value Equations 1. Solve the equation 11x + 6 = 7x Solution: Use properties of equality to bring the x s to one side and the numbers to the other: 11x (7x) + 6 = 7x (7x) x = 15 6 x = Solve the equation 7(5s 2) 8(6s 8) = 141 Solution: First distribute, and then combine like terms: 35s 14 48s + 64 = s + 50 = s = 91 and so s = = Solve the equation 5x 2 4x = 3 6 Solution: First 6 is the common denominator, so multiply both sides of the equation by 6 to obtain ( 5x (6) 2 4x + 5 ) ( ) ( ) ( ) = (6) (5x) (4x + 5) = that is (3)(5x) (4x + 5) = 3 and so 15x 4x 5 = 3 11x 5 = 3 11x = x = Solve the following equation 9 4x 5 = 4x + 5 4x 5 + 1

2 Solution: First observe x 5/4, and simplify the equation by multiplying both sides of the equation by 4x 5 to obtain 9 (4x + 5)(4x 5) (4x 5) = + 1(4x 5) 4x 5 4x 5 9 = 4x (4x 5) 9 = 4x x = 8x 9 = 8x x = 9 8 Thus the proposed solution is x = 9 8. Because 9 8 5, this is a valid solution, that 4 is x = 9 is the solution to the problem; the reader should check that it works by 8 plugging it into the original equation (to make sure no algebra mistakes were made). 5. Solve the absolute value equation, or indicate that there is no solution x = 16 Solution: We first isolate the absolute value expression x = x = x = 7 Now use the fact that Expression = 7 means Expression = 7 or Expression = 7. That is, 1 5x = 7 or 1 5x = 7 In the first case, and in the second case 1 5x = 7 5x = 6 x = x = 7 5x = 8 x = 8 5 Thus the solutions are x = 6 5 and x = Solve the absolute value equation, or indicate that there is no solution x = 49 Page 2

3 Solution: We first isolate the absolute value expression x = x = x = 9 There is no solution because the absolute value of any expression must be positive. 7. Solve the absolute value equation, or indicate that there is no solution. 4 x 4 12 = 8 Solution: First 4 x 4 = , that is 4 x 4 = 20 and so x 4 = 5. The absolute value property now implies x 4 = 5 or x 4 = 5 Therefore x = 4 5 = 1 or x = = A local nursery charges a flat rate of $65 to deliver top soil anywhere in the city of Riverside and charges $38 per cubic yard for their premium top soil, and they will sell this top soil in increments of one-tenth of a cubic yard. How many cubic yards of premium top soil can be delivered to a home in Riverside for $274? Express answer to the nearest tenth of a cubic yard. Solution: The cost of having the nursery x cubic yards of premium topsoil is 38x+65. So we solve 274 = 38x = 38x x = Thus, approximately 5.5 cubic yards can be delivered for 274 dollars. 9. The formulas for exercise heart rates given in beats per minute where a is age in years are: Maximum exercise heart rate =.85(220 a) Minimum exercise heart rate =.65(220 a) (a) Find the minimum exercise heart rate of a 54 year-old. Express answer to nearest whole number. (b) Find the maximum exercise heart rate of a 48 year-old. Express answer to nearest whole number. (c) Find the approximate age of a person whose maximum exercise heart rate is 133 beats per minute. Express answer in years to nearest 0.1 year. Page 3

4 Solution: (a) Min rate = (.65)(220 54) = (.65)(166) 108 beats per minute. (b) Max rate = (.65)(220 48) = (.65)(172) 146 beats per minute. (c) Solve 133 =.85(220 a) which implies 133 = 220 a which implies.85 a = years 10. Suppose the final test in a class is worth 25 percent of the overall grade, and assignments and quizzes are worth 20 percent of the overall grade and term tests are worth the remaining 55 percent of the grade. If a student has an average of 98% on assignments and quizzes and 92% on term tests going into the final. What percentage is needed on the final test for the student to finish with an overall average of 90%. Express answer to nearest whole number. Solution: Note the overall percentage of any student in the class will be (0.25)(F T ) + (0.20)(A) + (0.55)(T ) where F T, A, and T are the student s percentages on the Final Test, Assignments & Quizzes and Term Tests respectively. In this case we know A = 98 and T = 92, and F T is the unknown variable we wish to find. So, let F T represent the percentage needed on the final test. Then we solve (0.25)(F T ) + (0.20)(98) + (0.55)(92) = 90 (0.25)(F T ) = 90 (0.25)(F T ) = F T = 19.80/0.25 = Thus the student will need 79.20% on the final exam, or to the nearest whole number, 79% on the final exam in order to end the term with an overall average of 90%. 11. Ruben is driving along a highway that passes through Barstow. His distance d, in miles, from Barstow is given by the equation d = t, where t is the time, in hours, since the start of his trip and 0 t 16. When will Ruben be exactly 60 miles from Barstow? Express your answers in hours, accurate to two decimal places. Solution: Solve t = 60. This implies t = 60 or t = 60. The first equation implies = 60t and so t = 150/ hours. The second equation implies = 60t and so t = 270/ hours. Page 4

5 Thus Ruben will be 60 miles from Barstow in approximately 2.50, and 4.50 hours (where at 2.50 hours he will have not yet reached Barstow, and in 4.50 hours, he will have passed through Barstow and be 60 miles beyond it). 12. Identify each of the equations below as a conditional equation, identity or contradiction. (a) 4(5x + 2) = x (b) 20 11x = 4(5 2x) 3x (c) 4(5x + 2) 20x 7 = 0 Solution: (a) This equation simplifies to 20x x = 8 or 1x = 0; which is true when x = 0 but not true for other values of x, so this equation is a conditional equation. (b) This equation simplifies to 20 11x = 20 11x which is an identity because both sides are the same. (c) This equation simplifies to 20x x 7 = 0 or 1 = 0; because 1 is not equal to 0 this equation is a contradiction, it is never true. Page 5

6 2 Formulas and Applications 1. Solve the following equation of a plane (in three-dimensional space) for y 2x 4y 2z = 3 Solution: Adding 4y to both sides yields 2x 2z = 3 + 4y Subtracting 3 from each side yields 2x 2z 3 = 4y The dividing by 4 on each side results in y = 2x 2z Solve the following equation for x 1 x + 11y = 8 Solution: We would like x alone, so to get it in the numerator, we multiply both sides of the equation by x (it is often easier to get rid of fractions when solving equations) to obtain 1 11xy = 8x Next add 11xy to both sides so all of the x s are on the same side of the equation: 1 = 8x + 11xy Now factor out x on the right side so we can isolate it 1 = x(8 + 11y) Finally, divide both sides by y to solve for x x = y 3. Many physical formulas involve reciprocals of the variables. Here is a random example to give you practice with that type of situation where you are asked to solve the following equation for x 8 x + 15 y = 8 z Page 6

7 Solution: We would like x alone, so to get it in the numerator, we multiply both sides of the equation by xyz (it is often easier to get rid of fractions when solving equations) to obtain 8yz + 15xz = 8xy Next add 15xz to both sides so all of the x s are on the same side of the equation: 8yz = 15xz + 8xy Now factor out x on the right side so we can isolate it 8yz = x(15z + 8y) Finally, divide both sides by 15z + 8y to solve for x x = 8yz 8y + 15z 4. Many physical or mathematical formulas involve combinations of products and sums of the variables. Here is a random example to give you practice with that type of situation where you are to solve the following equation for C A = 1 H(B + 12C) 8 Solution: We wish to isolate C. To start, we can multiply both sides by 8 H to obtain Next, subtract B from each side to obtain 8A H = B + 12C 12C = 8A H B = 8A H B H H Last, we divide both sides by 12 to have = 8A BH H C = 8A BH 12H 5. A retailer determines the price of a dress by first computing 170% of the wholesale price of the dress and then adding a markup of $ What is the wholesale price of a dress that has a retail price of $131.60? Page 7

8 Solution: Let x be the wholesale price of the dress. Then Retail Price = 1.70x and therefore we solve = 1.70x , and so = 1.70x x = Thus the wholesale price of the dress is $ = = A worker can paint a house in 25 hours. With the help of an assistant, the house can be painted in 19 hours. How long would it take the assistant to paint the house alone? Express final answer to nearest 0.1 hour. Solution: Let A be the amount of time in hours that it would take the assistant to build the fence alone. Because the two workers can paint the house in 19 hours time, we get that A = 1. Therefore, 19 A = = A = and so = so to the nearest 0.1 hours, it would take the assistant 79.2 hours to paint the house alone. 7. A standard pump can drain a certain large swimming pool in 60 hours, and a turbo pump can drain the same pool in 20 hours. Suppose the standard pump and turbo pump worked for 5 hours to drain the pool until the turbo pump broke down. How many additional hours will it take the standard pump working alone to finish draining the pool? Solution: The rate of the standard pump is 1 and the rate of the turbo pump is In 5 hours the pumps were able to drain = = (12)(4) = 1 of the pool. 3 Thus it remains to drain = 2 of the pool using only the standard pump. The 3 time t it will take the standard pump alone to finish draining the pool is t 60 = 2 3 t = (2)(60) 3 = 40 hours. Page 8

9 8. An investment adviser invested $14200 in two accounts. One investment earned 5% annual simple interest, and the other investment earned 2% annual simple interest. The amount of interest earned for one year was $419. How much was invested in each account? Solution: Use the formula I = P rt, that is Interest is Principal times rate times time. Let x be the amount invested at 5% and then x is the amount invested at 2%. Therefore 419 = 0.05x + (14200 x)(0.02) Solving for x one obtains 135 = 0.03x, or x = 135 = Thus $4500 was invested 0.03 at 5% and $9700 was invested at 2%. 9. A tea merchant wants to make 150 pounds of blended tea costing $4.90-per-pound. The blend is made using a $4.00-per-pound grade of tea and a $6.50-per-pound grade of tea. How many pounds of each grade of tea should be used? Solution: Let x be the number pounds of tea that costs $4.00 per pound and then there are 150 x pounds of tea that costs $6.50 per pound. Then Tea Price per Pound Quantity Total Cost x 4.00x x (6.50)(150) 6.50x Blend (4.90)(150) Thus we have 4.00x (150 x) = (150)(4.90) and so 2.50x = and so 2.50x = = Then x = /2.50 = 96. Thus the mixture should contain 96 pounds of the tea that costs $4.00 per pound and 54 pounds of the tea that costs $6.50 per ounce. 10. Jogging at an average rate of 5 meters per second, a runner ran to the end of a path. The runner then jogged back to the starting point at an average rate of 2 meters per second. The total time for the jog to the end of the path and back was 1 minutes and 38 seconds (or a total of 98 seconds). Find the distance from the point the runner started to the end of the path. Solution: Let t be the time in seconds it took the runner to run to the end of the path. Then the return trip took 98 t seconds. Then 5t = length of path = 2(98 t). Therefore, 7t = 196 and so t = 196/7 = 28 seconds. Then the distance from where the runner started until the end of the path is (5)(28) = 140 meters. 11. A triangle has a perimeter of 72 centimeters. Each of the longer sides of the triangle is 4-times as long as the shortest side. Find the length of each side of the triangle. Page 9

10 Solution: Let x be the length of the shortest side in centimeters. Then each of the other sides has length 4x. Because the perimeter is 72 centimeters, we obtain x + 4x + 4x = 72 implies 9x = 72 implies x = 8. Therefore, the shortest side has length 8 cm and the other two sides have length 32 cm. 12. The length of a rectangle is 3 feet less than 2 times the width of the rectangle. If the perimeter of the rectangle is 84 feet, find the width and the length of the rectangle. Solution: First, let w be the width of the rectangle and l be the length of the rectangle. Then l = 2w 3. We know that the perimeter, p, satisfies Therefore, p = 2w + 2l and so 84 = 2w + 2l = 2w + 2(2w 3) 84 = 6w 6 90 = 6w w = 90 6 = 15 and then l = 2w 3 = 27. Thus, the width is 15 feet, and the length is 27 feet. 13. A chemist has acid solution. The first solution is 32% acid; the second solution is 72% acid. Set-up and solve a linear equation to determine how many ml of each solution should be mixed to produce 700 ml of an acidic solution that is 54% acid? Solution: Let x be the number of ml of 72% acid, and then 700 x is the number of ml of 32% acid solution. Then 0.32(700 x) x = (0.54)(700) and so x = Then 0.40x = and so x = = 385. So the chemist should use 385 ml of 72% acid solution, and = 315 ml of the 32% acid solution. That is, the chemist should mix 315 ml of 32% with 385 ml of the 72% acid solution. Page 10

11 3 Quadratic Equations 1. Solve the quadratic equation 2x 2 + 7x = 49. Solution: First write the equation in standard form as 2x 2 + 7x 49 = 0 Then factoring the left side we find (2x 7)(x + 7) = 0, and so x = 7 or x = Solve the quadratic equation x 2 4x = 77 by factoring. Solution: First, write as a quadratic equation in standard form, that is x 2 4x 77 = 0 Therefore, (x + 7)(x 11) = 0 and so x = 7 or x = Solve the quadratic equation by using the square root principle 25 (19x + 5) 2 = 0 Solution: The solution is as follows 25 (19x + 5) 2 = 0 (19x + 5) 2 = 25 19x + 5 = ± 25 19x + 5 = ±5 19x = 5 ± 5 x = 5 ± 5 19 So the two real solutions are and 0 4. Solve the quadratic equation x 2 = 4x 23 using the quadratic formula. Solution: First, write as a quadratic equation in standard form, that is x 2 4x + 23 = 0 Page 11

12 Using the quadratic formula with a = 1, b = 4 and c = 23, we find x = b ± b 2 4ac 2a = ( 4) ± ( 4) 2 4(1)(23) 2(1) = 4 ± = 4 ± i (4)(19) 2 = 4 ± 76 2 = 4 ± 2i 19 2 = 4 ± i 76 2 = 2 ± i Solve the following quadratic equation by completing the square 2x 2 16x + 25 = 0 Solution: Following the completion of square steps, we find 2x 2 16x + 25 = 0 2x 2 16x = 25 x 2 8x = 25 divide by 2 2 x 2 8x + ( 4) 2 = ( 4)2 make perfect square x 2 8x + (4) 2 = (4)2 2 (x 4) 2 = x 4 = ± square root principle x = 4 ± 2 = 4 ± x = 4 ± = 8 ± Solve the following quadratic equation by completing the square 3x 2 36x + 91 = 0 Page 12

13 Solution: Following the completion of square steps, we find 3x 2 36x + 91 = 0 3x 2 36x = 91 x 2 12x = 91 3 divide by 3 x 2 12x + ( 6) 2 = ( 6)2 make perfect square x 2 12x + (6) (6)2 = 2 (x 6) 2 = x 6 = ± square root principle x = 6 ± 3 = 6 ± x = 6 ± = 18 ± Find the discriminant, and use it to determine whether 4 5 x2 + 8 x = 2 has (i) two 5 distinct real solutions; (ii) one real solution; or (iii) two complex (nonreal) solutions. You do not neet to find the solutions. Solution: First write the quadratic equation in standard form, and clear the fractions for convenience. 4 5 x x = x x + 2 = 0 4x2 + 8x + 10 = 0 This quadratic equation is now in standard form with a = 4, b = 8 and c = 10 and the discriminant is b 2 4ac = (8) 2 4(4)(10) = = 96. Because the discriminant is negative there are two complex (nonreal) solutions. 8. A television screen measures 117 cm diagonally and its aspect ratio is 16 to 9. Find the dimensions of the television screen. Round your answers to the nearest tenth of a cm. Solution: Let the screen have dimensions 16x by 9x where x is an unknown positive number. Then (16x) 2 + (9x) 2 = x x 2 = x 2 = Therefore x = 13689/ Page 13

14 The longer side of the screen then measures 16x = 16( ) cm and the shorter side of the screen is then 9x = 9( ) 57.4 cm. That is, the dimensions are cm by 57.4 cm. 9. A square piece of cardboard is formed into a box by cutting out 3-inch squares from each of the corners and folding up the sides, as shown in the following figure (not to scale). If the volume of the box needs to be cubic inches, what size square piece of cardboard is needed? Solution: The volume is 3(x 6) 2 and so we solve 3(x 6) 2 = Then (x 6) 2 = and so x 6 = and thus x 6 = 12.5, or x = Thus a sheet of cardboard that measures 18.5 inches by 18.5 inches is needed. 10. A gardener wishes to use 600 feet of fencing to enclose a rectangular region and subdivide the region into two smaller rectangles. See the figure below. The total enclosed area is square feet. Find the possible width(s) of the enclosed area. Round answer(s) to the nearest 0.1 feet. Solution: Let w be the width, and l be the length. The amount of fence used will be 2l + 3w (the 3w is because of the extra fence in the middle to create the two smaller rectangles). Now 2l + 3w = 600 and so l = (600 3w)/2. Also, A = lw and so we get that w(600 3w)/2 = Thus, 3w 2 600w = 0. According to the quadratic formula w = 600 ± (3)(28800) 2(3) Thus the possible widths are feet or 80.0 feet. = 600 ± Page 14

15 11. A model rocket is launched upward with an initial velocity of 200 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h(t) = 16t t. How many seconds after the launch will the rocket be 300 feet above the ground? Round answer to nearest hundredth of a second. Solution: We solve the equation 16t t = 300 for t. In standard form this quadratic equation becomes 16t t 300 = 0 16t 2 200t = 0 Applying the quadratic formula yields t = 200 ± (200) 2 4(16)(300) 2(16) and so we get that the rocket will be 300 feet above the ground 1.74 seconds after the launch, and then seconds after the launch. Page 15

16 4 Other Types of Equations 1. Find all real solutions to the equation 2x 4 9x 2 = 143. Solution: First write the equation in standard form as 2x 4 9x = 0, and observe it is an equation of quadratic type. We let u = x 2, and then 2u 2 9u 143 = 0 Then factoring the left side we find (2u + 13)(u 11) = 0, and so u = 11 or u = Because u = x 2, this implies x 2 = 11 or x 2 = The equation x2 = 11 implies x = ± 11, but x 2 = 13 2 has no real solutions. Therefore, 11, 11 are the real solutions to the equation. 2. Find all solutions to the equation 2x 2/3 7x 1/3 30 = 0. Solution: Observe this is an equation of quadratic type. Indeed, let u = x 1/3, and then 2u 2 7u 30 = 0 Then factoring the left side we find (2u + 5)(u 6) = 0, and so u = 6 or u = 5 2. Because u = x 1/3, this implies x 1/3 = 6 or x 1/3 = 5 2. The equation x1/3 = 6 implies x = (6) 3 = 216, and x 1/3 = 5 2 implies x = ( 2) 5 3 = Therefore, 216, are 8 8 the solutions to the equation. 3. Solve the equation x x x 2 = 0 by factoring. Solution: This implies x 2 (x 2 +12x+32) = 0 and so x 2 (x+8)(x+4) = 0. Therefore, the solutions are x = 0, x = 8, x = Solve the equation 4x 3 5x 2 40x + 50 = 0 by factoring. Solution: We will try factoring the left-hand side of the equation by grouping: 4x 3 5x 2 40x + 50 = x 2 (4x 5) 10(4x 5) = (x 2 10)(4x 5) = 0 Therefore, x 2 = 10 or 4x 5 = 0. This implies the solutions are 10, 5, Page 16

17 5. Solve the following rational equations (a) x 3 4x + 1 = 1. (b) x 3 4x + 1 = 1 4. Solution: (a) Multiply both sides by 4x + 1 and solve the equation as follows: x 3 4x + 1 = 1 x 3 = 1(4x + 1) x 3 = 4x + 1 3x = 4 x = 4 3 Thus x = 4 is the solution for (a). 3 (b) Multiply both sides by 4x + 1 and solve the equation as follows: x 3 4x + 1 = 1 4 4x 12 = 4x + 1 0x = 13 However, 0 13 and so there is no solution for (b). 6. Solve the rational equation x 5x + 6 x 1 = 3x 14 x 1. Solution: Multiplying both sides of the equation by x 1 yields and this implies x(x 1) (5x + 6) = 3x 14 x 2 9x + 8 = 0 (x 1)(x 8) = 0. Thus the proposed solutions are 1 and 8. However, 1 cannot be a solution since x 1 is in the denominator of the original equation. Therefore, the only solution is x = 8, and the reader should check that it works. 7. Use algebra to solve the equation x x + 11 = 9. Solution: We rewrite the equation as and square both sides to obtain x 9 = x + 11 (x 9) 2 = ( x + 11) 2 or x 2 18x + 81 = x + 11 and then x 2 19x + 70 = 0 and factoring implies (x 5)(x 14) = 0. Thus the proposed solutions are x = 5 and x = 14, which we now check by plugging them into the original equation. x = 5: x = 14: , and so 5 is not a solution = 9, and so x = 14 is the solution. Page 17

18 8. Use algebra to solve 2x + 23 x + 8 = 2. Check all proposed solutions. Solution: First write the equation as 2x + 23 = x and square both sides (remember to use FOIL on the right hand side). ( 2x + 23) 2 = ( x ) 2 2x + 23 = x x x = 4 x + 8 (x + 11) 2 = (4 x + 8) 2 x x = 16(x + 8) x 2 + 6x 7 = 0 (x 1)(x + 7) = 0. The proposed solutions are x = 1 and x = 7. We now check the proposed solutions. First, if x = 1 the right hand side of the original equation becomes 2(1) = 25 9 = 5 3 = 2 as desired, so x = 1 is a solution. Next, if x = 7 the right hand side of the original equation becomes 2( 7) = 9 1 = 3 1 = 2 as desired, so x = 7 is also a solution. Thus, the solutions are x = 1 and x = Working together, two pumps can drain a fish pond in 40 minutes. Working alone, the faster pump can drain the pond in 18 minutes less than the slower pump. How long would it take the faster pump to drain the pond working alone? Solution: Let x be the number of minutes it takes the faster pump to drain the pond. Then the slower pump can drain the pond in x + 18 minutes. Because the pumps can drain the pond together in 40 minutes, we get 40 x + 40 x + 18 = 1 and multiplying both sides by x(x + 18) implies 40x + (40)(18) + 40x = x(x + 18) and so x 2 62x 720 = 0 (x 72)(x + 10) = 0 Page 18

19 Therefore, the faster pump can drain the pond in 72 minutes. A check of this solution is as follows: = = (5)(8) (9)(8) + (4)(10) (9)(10) = = 9 9 = 1 and so the solution works. 10. A car and a truck are making a long trip on the same route. The car travels at a constant rate 4 km/h faster than the truck that also travels at a constant rate. The truck started the trip half-an-hour before the car. The car overtook the truck after traveling 312 km. What is the rate of each vehicle? Solution: Let x be the rate of the truck; then x + 4 is the rate of the car. Using the equation t = d/r (time is distance over rate), we get 312 x = 312 x (the time the car travelled plus one-half is the time the truck travelled). Multiplying both sides by 2x(x + 4) yields Thus, 624x + x 2 + 4x = 624x x 2 + 4x 2496 = 0 x 2 + 4x 2496 = (x 48)(x + 52) = 0 Therefore, x = 48 (the rate of the truck is positive). The rate of the truck is 48 km/h and the rate of the car is 52 km/h (since it is 4 km/h more than the rate of the truck). Page 19

20 5 Inequalities 1. Sets are indicated below on a number lines in (a) and (b) below. The sets are solutions to compound inequalities, find the compound inequalities. (a) (b) Solution: (a) x 2 or x > 1. (b) 0 < x and x 4, or more concisely, 0 < x Sets are indicated below on a number lines in (a) and (b) below. Write the sets in interval notation using unions and intersections as appropriate. (a) (b) Solution: (a) (, 4) (1, 6] (b) [ 1, 3) (3, ). 3. Sketch the solution sets on number lines for the following inequalities. (a) x 6 (b) x > 0 (c) x 6 or x > 0 (d) x 6 and x > 0 Page 20

21 Solution: (a) (b) (c) This is the union of the sets in (a) and (b) (d) This is the intersection of the sets graphed in (a) and (b) which is the empty set so there is nothing to graph Solve the following compound inequality 4x 6 > 2 or 4x (a) Express your solution in interval notation. (b) Graph your solution on a number line. Solution: (a) We solve the inequalities separately 4x 6 > 2 4x > 8 x > 2 or 4x x 16 x 4 Then x 4 or x > 2 means the solution set is (, 4] (2, ) (b) Solve the following compound inequality 3x + 7 > 11 and 3x (a) Express your solution in interval notation. (b) Graph your solution on a number line. Page 21

22 Solution: (a) The inequalities imply 11 < 3x < 3x < 3x < 3x > x > x 6 Thus, the solution set is [ 6, 6) (b) Solve the following inequality 7(x + 5) + 36 < 7 x (a) Express your solution in interval notation. (b) Graph your solution on a number line. Solution: (a) Using properties of inequalities we find 7(x + 5) + 36 < 7 x 7x < 7 x 7x + 1 < 7 x 7x x < 7 6x < 7 1 6x < 6 x > 6 6 x > 1 Thus, the solution set is ( 1, ) (b) Solve the inequality 6x Use interval notation to express the solution set. Solution: The inequality implies 6x or 6x Page 22

23 Solving these inequalities individually: Similarly, 6x x 6 12 x 6 6 x 1 6x x 6 12 x 18 6 Thus the solution set is (, 3] [ 1, ). x 3 8. Solve the inequality 4x + 4 < 12. Use interval notation to express the solution set. Solution: The inequality implies 12 < 4x+4 < < 4x < < x < < x < 2 Thus the solution set is ( 4, 2). 9. Solve the quadratic inequality x 2 2x 48. Solution: First, rearrange the inequality so 0 is on one side, that is x 2 2x 48 0 Then factor to find the critical values: (x + 6)(x 8) = 0 and so x = 6, x = 8 are critical values and they separate the real line into the intervals (, 6), ( 6, 8) and (8, ). For these intervals, we choose the test points x = 7, x = 0 and x = 9. When x = 7, we find ( 7 + 6)( 7 8) > 0, and so the inequality is not true if x < 6. When x = 0, we find (0 + 6)(0 8) < 0, and so the inequality is true if 6 < x < 8. When x = 9, we find (9 + 6)(9 8) > 0 and so the inequality is not true if x > 8. The inequality is true when x = 6 or x = 8, and thus the solution is {x : 6 x 8}. 10. Solve the quadratic inequality 8x > 20 x 2. Solution: First, rearrange the inequality so 0 is on one side, that is x 2 8x 20 > 0 Then factor to find the critical values: (x + 2)(x 10) = 0 and so x = 2, x = 10 are critical values and they separate the real line into the intervals (, 2), ( 2, 10) Page 23

24 and (10, ). For these intervals, we choose the test points x = 3, x = 0 and x = 11. When x = 3, we find ( 3 + 2)( 3 10) > 0, and so the inequality is true if x < 2. When x = 0, we find (0 + 2)(0 10) < 0, and so the inequality is not true if 2 < x < 10. When x = 11, we find (11 + 2)(11 10) > 0 and so the inequality is true if x > 10. Also, the inequality is not true when x = 2 or x = 10, and thus the solution is {x : x < 2 or x > 10}. 11. Solve the inequality 12 x x 4 2 and write the answer in interval notation. Solution: The inequality 12 x x 4 the left hand side, we have 2 is equivalent to 12 x x Simplifying 12 x x (12 x) + 2(x 4) x x + 4 x 4 0 Thus the critical values are x = 4 and x = 4. Next create a sign chart using this information. x < 4 x = 4 4 < x < 4 x = 4 4 < x x + 4 neg 0 pos pos pos x 4 neg neg neg 0 pos x + 4 x 4 pos 0 neg und pos The solution includes all numbers where the fraction in the bottom line of the sign chart is positive or 0, but not where it is undefined. In interval notation the solution is (, 4] (4, ). 12. Solve the inequality x + 1 x 3 > 5 and write the answer in interval notation. Solution: The inequality x + 1 x 3 left hand side, we have x + 1 > 5 is equivalent to 5 > 0. Simplifying the x 3 x + 1 (x + 1) 5(x 3) 5 > 0 x 3 x 3 > 0 4x + 16 x 3 Thus the critical values are x = 3 and x = 4. Next create a sign chart to determine when the final inequality is true. > 0 Page 24

25 x < 3 x = 3 3 < x < 4 x = 4 4 < x x 3 neg 0 pos pos pos 4x + 16 pos pos pos 0 neg 4x + 16 x 3 neg und pos 0 neg The solution includes all numbers where the fraction in the bottom row of the sign chart is positive, it does not include where the fraction is undefined or 0. From the sign chart, in interval notation, the solution is (3, 4). (x 3)(x + 10) 13. Solve the inequality > 0 using the method of critical values with test x 8 numbers or a sign chart. Write your answer in interval notation. Solution: The critical values are x = 3, x = 10 and x = 8. We look for when the expression is positive. Note it is 0 when x = 3 or x = 10, and it is undefined when x = 8, thus none of the critical values are in the solution set. For the rest we use the sign chart: x < < x < 3 3 < x < 8 8 < x x + 10 neg pos pos pos x 3 neg neg pos pos x 8 neg neg neg pos (x 3)(x + 10) x 8 neg pos neg pos Thus the answer is: ( 10, 3) (8, ) (x 1)(x + 7) 14. Solve the inequality < 0 using the method of critical values with test x 2 numbers or a sign chart. Write your answer in interval notation. Solution: The critical values are x = 1, x = 7 and x = 2. We look for when the expression is negative. Note it is 0 when x = 1 or x = 7, and it is undefined when x = 2, thus none of the critical values are in the solution set. For the rest we use the sign chart: x < 7 7 < x < 1 1 < x < 2 2 < x x + 7 neg pos pos pos x 1 neg neg pos pos x 2 neg neg neg pos (x 1)(x + 7) x 2 neg pos neg pos Thus the answer is: (, 7) (1, 2) Page 25

26 15. Suppose the final test in a class is worth 30 percent of the overall grade, and assignments are worth 15 percent of the overall grade and term tests are worth the remaining 55 percent of the grade. If a student has an average of 79% on homework and 72% on term tests going into the final. Determine the interval of percentages needed on on the final test for the student to finish with an overall percentage between 68% and 72%. Express endpoints of interval to 1 decimal place.. Solution: Note the overall percentage of any student in the class will be (.3)(F T ) + (.15)(A) + (.55)(T ) where F T, A, and T are the student s percentages on the Final Test, Assignments and Term Tests respectively. Let F T represent the percentage needed on the final test. Then we solve 68 < (.30)(F T ) + (.15)(79) + (.55)(72) < < (.30)(F T ) < < (.30)(F T ) < < F T < Thus the student will need between 55.2% and 68.5% on the final exam in order to end the term with an overall percentage between 68% and 72%. 16. (A physics inequality) The equation s = 16t 2 + v 0 t + s 0 gives the height s, in feet above ground level, at the time t seconds, of a projectile launched directly upward from a height s 0 feet above the ground and with an initial velocity of v 0 feet per second. A catapult launches a rock directly upward from an initial a height of 12 feet above the ground with an initial velocity of 144 feet per second. Find the time interval during which the rock will be more than 140 feet above the ground. Hint. This will reduce to a quadratic inequality where all terms have a factor of 16. Solution: The problem suggest v 0 = 144 and s 0 = 12. Thus s = 16t t We then solve the inequality 16t t + 12 > 140. Thus 16t t 128 > 0 and multiplying both sides by 1 we have 16t 2 144t < 0 and dividing by 16 (noted in the hint) yields t 2 9t + 8 < 0 and factoring gives us (t 1)(t 8) < 0. The critical values are 1 and 8, and the following sign chart helps us solve the inequality. t < 1 t = 1 1 < t < 8 t = 8 8 < t t 1 neg 0 pos pos pos t 8 neg neg neg 0 pos (t 1)(t 8) pos 0 neg 0 pos Thus the inequality is true when 1 < t < 8. That is the rock is more than 140 feet Page 26

27 above the ground between 1 second and 8 seconds after it was launched. Page 27

28 6 Variations and Applications 1. The area, A, of a picture projected on a movie screen varies directly as the square of the distance, d, from the projector to the screen. (a) Write an equation that expresses the relationship between the variables. Use k as the constant of variation. (b) If a distance of 15 feet produces a picture with an area of 47 square feet, what distance produces a picture with an area of 1692 square feet. Solution: (a) The relation is A = kd 2. (b) For the projector in (b), we have 47 = k(15) 2 and so k = 47 (15) 2. Then, to produce an area of 1692 square feet, we need 1692 = 47 (15) 2 d2 d 2 = (15)2 and so d 2 = 36(15) 2 which implies d = 6(15) = 90 feet. 2. The speed of a bicycle gear, in revolutions per minute, is inversely proportional to the number of teeth on the gear. If a gear with 42 teeth has a speed of 9 revolutions per minute, what will be the speed of a gear with 18 teeth? Solution: The relation is ω = k/n where ω is the speed in revolutions per minute, and n is the number of teeth. From this we get 9 = k/42, and thus k = (9)(42) = 378. Therefore, a gear with 18 teeth has a speed of ω = = 21 revolutions per minute. 3. The height, h, of a picture projected on a screen varies directly as the distance, d, of the projector from the screen. (a) Express the relation between d and h using k as the constant of variation. (b) If a picture is 44 inches high when the projector is 12 feet from the screen, how far should the projector be placed from a screen that is 88 inches high so that the picture height is exactly the same as the screen height? Solution: (a) The relation is h = kd since h varies directly as k. (b) In this case h = 44 when d = 12 and so 44 = k(12) or k = 44/12 = 11/3. Therefore, if we want h = 88, we find 88 = kd 88 = 11 3 d d = (3)(88) = That is, the projector should be placed 24 feet from the screen. Page 28

29 4. (a) The sound intensity of a jet engine, measured in watts per meter squared (W/m 2 ), is inversely proportional to the square of the distance between the engine and an airport ramp worker. For a certain jet engine, the sound intensity measures 0.8 W/m 2 for a ramp worker at a distance of 6 meters from the engine. What is the sound intensity for a ramp worker 9 meters from the jet engine? (Round your answer to three decimal places.) (b) In general, what would happen to the sound intensity for a certain jet engine if a ramp worker doubles her original distance from a jet engine? Solution: (a) Use the relation I = k d 2 where I is the sound intensity,and d is the distance from the engine to the ramp meter. For the given data 0.8 = k 6 2 k = (0.8)(6 2 ) = Thus, the intensity when d = 9 is I = and so Thus to three decimal places, the sound intensity is W/m 2 at a distance of 9 meters from the jet engine. (b) If a ramp worker doubles her original distance from a jet engine, the sound 1 intensity will be of the intensity at the original distance. We obtain this from 4 I = k, and so replacing the distance with 2d we obtain d2 I = which is 1/4 of the original intensity. k (2d) 2 = k 4d = k d 2 5. The volume V of a right circular cone varies jointly as the square of the radius r and the height h. (a) Write an equation representing the relation between the given variables. Use k as the variation constant. (b) Determine what happens to V when the radius is quintupled. (c) Determine what happens to V when the radius is quintupled and the height is quintupled. Solution: (a) V = kr 2 h (b) If the radius is quintupled, we replace r with 5r, and V = k(5r) 2 h = 25(kr 2 h), and so V becomes 25 times as large. (c) In this case, replace r with 5r and h with 5h to get V = (k(5r) 2 (5h) = 125r 2 h, and so V becomes 125 times as large. 6. (a) The maximum load L that a cylindrical column of circular cross section can support varies directly as the fourth power of diameter d and inversely as the square of its height Page 29

30 h. Write an equation that represents the relation between the given variables. Use k as the variation constant. (b) If a column 4 feet in diameter and 11 feet high supports up to 17 tons, use the relation in (a) to determine how much a column 3 feet in diameter and 14 feet high made of the same material can support. Round your answer to the nearest hundredth of a ton. Solution: (a) Let L represent the load, d the diameter and h the height. Then the relation is L = k d4 h 2. (b) Plug in the data for the first column: 17 = k 44 and solve for k: 112 k = (17)(112 ) Thus the new column can support a load of L tons (a) The load L a horizontal beam can safely support varies jointly as the width w and the square of the depth d and inversely as the length l. Write an equation that represents the relation between the given variables. Use k as the constant of variation. (b) If a 15-foot beam with width 9 inches and depth 8 inches safely supports 700 pounds, how many pounds can a 11-foot beam that has width 9 inches and depth 6 inches be expected to support? Round answer to the nearest pound. Assume the two beams are made of the same material. Solution: (a) The relation is L = kwd2. l (b) We use 700 = k(9)(82 ) to find k = (15)(700) Thus the new beam 15 (9)(8 2 ) can support L (9)(62 ) Thus the new beam can support approximately 537 pounds. 8. The force F needed to keep a car from skidding on a curve varies jointly as the weight w of the car and the square of its speed v and inversely as the radius of the curve r. (a) Write an equation that represents the relationship between the variables (use k as the constant of variation). (b) Suppose it takes 3000 pounds of force to keep a 2400 pound car from skidding on a curve with radius 405 feet at 45 miles per hour. What force is needed to keep the car Page 30

31 from skidding when it takes a similar curve with radius 410 feet at 55 miles per hour? Round to the nearest 10 pounds. Solution: (a) F = kwv2 r (b) In the equation in (a), we use F = 3000, w = 2400, r = 405 and v = 45 to find k, and so 3000 = k(2400)(452 ) 405 k = (3000)(405) (2400)(45 2 ) Now, using the value of k , w = 2400, r = 410 and v = 55 in the equation in (a) we find F = ( )(2400)(552 ) and so to the nearest 10 pounds the force is 4430 pounds. Page 31