Chapter 2 - Equations and Inequalities PDF

Chapter 2 – Equations and Inequalities. Equations and inequalities are fundamental tools used to model and solve a wide range of real-world problems across various disciplines. The study of these mathematical concepts allows us to describe relationships, make predictions, and find solutions in field...

Chapter 2 – Equations and Inequalities. Equations and inequalities are fundamental tools used to model and solve a wide range of real-world problems across various disciplines. The study of these mathematical concepts allows us to describe relationships, make predictions, and find solutions in fields such as sciences, economics, engineering. In economics and finance, equations and inequalities play a crucial role in modeling and analyzing market behavior, optimizing production processes, and making informed decisions. Mathematical models based on equations and inequalities allow economists to examine supply and demand relationships, predict market equilibrium, and determine the impact of policy interventions. Moreover, these tools aid in financial planning, risk assessment, and investment analysis by formulating optimization problems that maximize returns or minimize risks under given constraints. This chapter aims to consider various methods used to solve equations and inequalities and explore their practical applications in business problems and dimension problems that can arise in business. Learning outcomes covered: By the end of this chapter, students will learn how to: Solve linear equations, quadratic equations, absolute-value equations, and equations involving fractional expressions, radicals. Translate word problems to mathematical models, and solve real life problems algebraically. Solve linear and nonlinear inequalities, absolute value inequalities; represent solution set of inequalities graphically. 2.1 Linear Equations and Equations leading to Linear Equations Objectives In Section 2.1 you will learn how to solve Linear equations Literal equations Absolute value equations Fractional equations leading to linear 34 An equation is a statement that two mathematical expressions are equal. Here are some examples of equations: 5x − 2 x − 6 = 3x ; 23 − 4 x = 9 ; =7 4− x The values of the variables (unknowns) that make the equation true are called solutions or roots of the equation. The process of finding the solutions is called solving the equation.  Equivalent Equations Two equations with exactly the same solutions are called equivalent equations. To get equivalent equations, we use the following properties: Properties of Equality Property Description A= B  A+C = B +C Adding same quantity to both sides of an equation gives an equivalent equation A= B  AC = BC Multiplying both sides of an equation by same nonzero quantity gives an equivalent equation  Linear Equations (First Degree Equations) A linear equation in one variable is an equation of the form ax + b = 0 where a and b are real numbers and x is the variable. 1 Linear Equations: 2x + 5 = 7 , 3x = x−5 2 2 Nonlinear Equations: x2 + 5x − 24 = 0 , x − 2x + 4 = 0 , − x =1 x Example 1 Solve the equation 2 ( 3x − 1) + x = 14 − x. Solution 2 ( 3 x − 1) + x = 14 − x 6 x − 2 + x = 14 − x 7 x + x = 14 + 2 Colecting x − terms on one side 8 x = 16 8 x 16 1 = Multiplying by on both sides 8 8 8 x=2 Answer: The solution is x = 2. 35  9x + 8 7x − 3 Example 2 Solve the equation − = 6. 4 2 Solution 1 9x + 8 7x − 3 − =6 LCD = 4 4 2 ( 9 x + 8 ) − ( 7 x − 3)  2 = 6 Use LCD to make the denominators equal 4 2 2 9 x + 8 14 x − 6 − =6 4 4 9 x + 8 − (14 x − 6 ) =6 4 9 x + 8 − 14 x + 6 =6 4 −5 x + 14 =6 4 −5 x + 14 = 24 Cross multiplying −5 x = 24 − 14 −5 x = 10 −5 x 10 =  x = −2 −5 −5 Solution 2 9x + 8 7 x − 3 Multiply both sides by LCD = 4 − =6 4 2  9x + 8 7 x − 3  Perform the operations 4 −  = 6 4  4 2  Simplify the left side ( 9 x + 8 ) − 2 ( 7 x − 3) = 24 9 x + 8 − 14 x + 6 = 24 −5 x + 14 = 24 Divide both sides by −5 −5 x = 10 x = −2 Answer: The solution is x = −2. 36  Literal Equations A literal equation is an equation that involves two or more variables. These variables are represented by letters and can vary independently. To solve a literal equation for a particular variable, we express this variable in terms of other variables which we consider as constants. Example 3 Solve the equation S = P + Prt for r. S−P Solution S = P + Prt  S − P = Prt  r= Pt   n Example 4 Express n in terms of other literals in the equation V = C 1 − . N   Solution  n V = C 1 −  Divide both sides by C ( C  0 )  N V n = 1− C N n V = 1− Multiply both sides by N ( N  0 ) N C  V n = N 1 −   C  Absolute Value Equations  x if x0 The absolute value of a real number x, denoted by x , is x =  − x if x0 An absolute value equation is an equation that involves the absolute value of algebraic expressions. Example 5 Solve the equation 2 x − 1 = 5. Solution Using the definition of absolute value, we replace the absolute value equation 2 x − 1 = 5 by equivalent equations 2x −1 = 5 and 2x −1 = −5 , then solve them separately. 2x −1 = 5 or 2 x − 1 = −5 2x = 6 or 2 x = −4 x=3 or x = −2 Answer: The solutions are x = 3 and x = −2. 37  Fractional equations leading to Linear equations A fractional equation is an equation that contains a variable in the denominator of one or more fractions. In order to solve a fractional equation, we eliminate the fractions by multiplying every term in the equation with the least common denominator (LCD) or by cross multiplying fractions (when it is possible). After solving the equation, we need to check the solutions to ensure they satisfy the equation. If any of the solutions make the denominator(s) zero, they should be excluded from the solution set since division by zero is undefined. 3 5 Example 6 Solve the equation =. x +1 x − 3 Solution It is important to note that here x  −1, x  3. 3 5 = x +1 x − 3 Cross multiply 3 ( x − 3) = 5 ( x + 1) Expand and simplify 3x − 9 = 5 x + 5 Collect x-terms on one side 3x − 5 x = 5 + 9 −2 x = 14 Dividing both sides by −2 x = −7 Check the solution Substitute x = −7 into the original equation: 3 5 3 5 1 1 =  =  − =−  L.H.S. = R.H.S. ( −7 ) + 1 ( −7 ) − 3 −6 −10 2 2 Answer: The solution is x = −7. In example 7, we applied cross-multiplication to clear the equation of fractions. However, it can be solved multiplying both sides by LCD. We demonstrate this technique in the following example. 3x − 1 x+2 −8 Example 7 Solve the equation − =. 3x − 4 x−2 3 x − 10 x + 8 2 Solution To find the LCD, we factor the denominator of the fraction in the right side. 3x − 1 x+2 −8 − = LCD = ( 3x − 4 )( x − 2 ) 3x − 4 x−2 ( 3x − 4 )( x − 2 ) 38  ( 3x − 1)( x − 2 ) − ( x + 2 )( 3x − 4 ) = −8 Make the denominators equal in ( 3x − 4 )( x − 2 ) ( x − 2 )( 3x − 4 ) ( 3x − 4 )( x − 2 ) the left side ( 3x − 1)( x − 2 ) − ( x + 2 )( 3x − 4 ) = −8 Perform operations and simplify ( 3x − 4 )( x − 2 ) ( 3x − 4 )( x − 2 ) 3x 2 − 7 x + 2 − ( 3x 2 + 2 x − 8 ) −8 = ( 3x − 4 )( x − 2 ) ( 3x − 4 )( x − 2 ) 3x 2 − 7 x + 2 − 3x 2 − 2 x + 8 −8 = ( 3x − 4 )( x − 2 ) ( 3x − 4 )( x − 2 ) −9 x + 10 −8 = Multiply both sides by LCD ( 3x − 4 )( x − 2 ) ( 3x − 4 )( x − 2 ) −9 x + 10 = −8 − 9 x = −18 x=2 Check the solution. The original equation is not defined for x = 2. When we substitute 2 for x in the original equation, the denominator becomes zero and that is not possible. Thus, there is no solution for the equation, or the solution is . Answer: No solution x + 2 x +1 Example 8 Solve the equation − =0. x −1 x − 3 Solution The equation is defined for x  1, x  3. x + 2 x +1 Make the denominators equal by using − =0 x −1 x − 3 LCD ( x + 2 )( x − 3) − ( x + 1)( x − 1) = 0 ( x − 1)( x − 3) ( x − 3)( x − 1) ( x + 2 )( x − 3) − ( x + 1)( x − 1) = 0 ( x − 1)( x − 3) x 2 − x − 6 − ( x 2 − 1) = 0 x2 − x − 6 − x2 + 1 = 0 −x − 5 = 0 x = −5 Check the solution Answer: The solution is x = −5. 39 Exercises 2.1 1 – 10. Solve the linear equations. 1. 2 (1 − x ) = 3(1 + 2 x) − 5 2. 1 (8u + 12 ) = 7 − u 4 3. 0.7 x + 0.4 = 0.6x − 2.4 4. 0.9x −1.25 = 0.75x + 1.75 1 2 1 2 2 1 y +1 5. a+ = a− 6. y + ( y − 3) = 3 5 5 5 3 2 4 x 7x −1 4x + 8 7. = 3x + 5 8. −1 = 7 4 5 1 5 1 1 1 2 9. w+ = w− 10. 2 = x− x+ x 3 4 4 3 2 3 11 – 16. Solve the literal equations for the indicated letters, express in terms of the remaining letters. 11. S = P − PD, for P 1 12. V =  r 2 h, for h 3 ax + b 13. k 1 + p  − r = 0, for p 14. = 2, for x  100  cx + d 1 1 1 2ml 15. + = , for q 16. r = , for n p q f B ( n + 1) 17 – 22. Solve the absolute value equations. 17. 3x + 5 = 1 18. 7 − 5x = 4 2 4 19. x − 5 + 7 = 16 20. − 3x − 8 = −2 3 9 21. 8x − 3 = x 22. 2 x − 7 = −1 23 – 34. The given equations are equivalent to linear equations. Solve the equations. 3x − 9 2x − 3 23. =0 24. =6 x+3 4x − 5 y+2 2− y 3 1 1 25. − = y−2 26. − = 3 6 t + 1 2 3t + 3 40  2 p +1 3 p −1 5 3 35 27. = 28. − = 2 2p + 3 3p + 4 x −1 x +1 x −1 2 + 3x 2 − x 3x − 1 7 x − 2 2 x + 5 29. − = x+2 30. − = 2 6 3 14 21 x + 2 x −1 y −3 y −2 31. − =0 32. = x −1 x − 3 y +1 y + 3 ( 2a − 5) + ( 3a + 1) = 13a 2 + 5a + 7 ( x − 3) = ( x + 3) − 24 2 2 2 2 33. 34. 35. x2 − 7 = 7 − x 36. 2x − 3 + 5 = 0 2.2 Quadratic Equations In section 2.1, we solved linear equations which are first degree polynomial equations. In this section, we will concentrate on quadratic equations, polynomial equations of second degree. Objectives: In section 2.2 you will learn how to Solve quadratic equations by factoring, completing square, and applying quadratic formula Solve fractional equations leading to quadratic Solve radical equations leading to quadratic. A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0 , where a, b and c are constants and a  0.  Solving Quadratic Equation by Factoring When dealing with quadratic equations, it is usually not possible to isolate the variable on one side just by using the properties of equality, because the variable is raised to different powers. Instead, we solve quadratic equations by factoring and using the zero – product property. Zero – Product Property: Consider two real value expressions A and B. If A B = 0 , then A = 0 or B = 0. 41 Example 1 Solve the equation x2 + x = 6. Solution First, rewrite the given equation in standard form (move all terms to one side). x2 + x − 6 = 0 To factor the quadratic form in the left side, we look for two numbers a and b, such that a + b = 1 and a  b = −6. The numbers are 3 and −2. Therefore, we can factor x2 + x − 6 as ( x + 3)( x − 2 ). x2 + x − 6 = 0 ( x + 3)( x − 2 ) = 0 x+3= 0 or ( x − 2) = 0 Zero – Product Property x = −3 or x=2 Check the solutions Both values make the given equation true. Answer: The solutions are x = −3 and x = 2. Example 2 Solve the equation 3 y 2 = 5 y. Solution Rewrite the given equation in standard form, then factor the quadratic form. 3y2 = 5 y 3y2 − 5 y = 0 y ( 3 y − 5) = 0 y=0 or 3 y − 5 = 0 5 y=0 or y= 3 5 Answer: The solutions are y = 0 and y =. 3 Example 3 Solve the equation ( 2 x − 1)( x − 5) = −7. Solution First, we rewrite the quadratic equation in standard form. 2 x 2 − 10 x − x + 5 = −7 2 x 2 − 11x + 12 = 0 ( 2 x − 3)( x − 4 ) = 0 x=3 , x=4 2 Answer: The solutions are x = 1.5 and x = 4. 42  Quadratic equation of form x2 = d , ( d  0 ) , has two solutions x = d and x = − d. Example 4 Solve the equation ( x − 2 ) = 5. 2 Solution In the given equation, the unknown x is isolated inside of a perfect square. It makes possible to solve this equation by taking square root on both sides. ( x − 2) =5 2 x−2 = 5 x−2= 5 or x−2= − 5 x = 2+ 5 or x = 2− 5 Answer: The solutions are x = 2 + 5 and x = 2 − 5.  Solving Quadratic Equation by Completing the Square Some quadratic equations cannot be factored readily. In these cases, we may use another method called completing the square. Using this method, we add or subtract terms to both sides of the equation to get a perfect square trinomial on one side of the equal sign. Before completing the square, we make sure that the coefficient of x 2 is 1. Completing the square 2 b To make x + bx a perfect square, add   , the square of half of the coefficient of x. 2 2 This gives the perfect square 2 2 b  b x + bx +   =  x +  2 2  2 Example 5 Solve each quadratic equation by completing the square. (a) x2 + 6 x − 40 = 0 (b) 3x 2 − 6 x − 1 = 0 Solution (a) x 2 + 6 x = 40 To isolate x-terms, bring 40 to the right side 2 2 6 6 2 x + 6 x +   = 40 +   2 Complete the square by adding  6  to both sides 2 2 2 x 2 + 6 x + 9 = 49 Simplify ( x + 3) = 49 2 Perfect square 43  x + 3 =  49 Take square root x+3= 7 or x + 3 = −7 x=4 or x = −10 Answer: The solutions are x = 4 and x = −10. (b) 3 x 2 − 6 x − 1 = 0 3x 2 − 6 x = 1 Bring constant 1 to the right side 3( x2 − 2 x ) = 1 Factor out 3 in the left side (x 2 − 2x) = 1 3 Divide both sides by 3 2 2 2 1 2 Complete the square x − 2x +   = +   2 2 3 2 4 Simplify x2 − 2 x + 1 = 3 4 ( x − 1) = 2 Perfect square 3 4 x −1 =  Take square root 3 2 x = 1 3 2 2 Answer: The solutions are x = 1 + and x = 1 −. 3 3  Solving Quadratic Equations Using Quadratic Formula. Applying the technique of completing square to the quadratic equation in general form, we get the quadratic formula, which is universal tool for solving any quadratic equation. The solutions (roots) of the quadratic equation ax2 + bx + c = 0 , where a  0 , are −b  b 2 − 4ac x= 2a The quantity b2 − 4ac is called the discriminant and denoted by D. If D  0 , then the equation has two distinct real roots; If D = 0 , then the equation has exactly one real solution; If D  0 , then the equation has no real solution. 44 Example 6 Use Quadratic Formula to find the solutions of each equation. (a) 3x 2 − 5 x + 2 = 0 (b) 4 x 2 − 20 x + 25 = 0 (c) 3x 2 + 2 x + 1 = 0 Solution (a) 3x 2 − 5 x + 2 = 0 a = 3, b = −5, c = 2 Determine values of a, b, and c − ( −5 )  ( −5 ) − 4 ( 3)( 2 ) 2 x= Apply the quadratic formula 2 ( 3) 5  1 5 1 = = 6 6 2 x = 1, x= 3 Here D = b2 − 4ac = 1  0. Therefore, the quadratic equation has two distinct solutions. 2 Answer: The solutions are x = 1 and x =. 3 (b) 4 x 2 − 20 x + 25 = 0 a = 4, b = −20, c = 22 Determine values of a, b, and c − ( −20 )  ( −20 ) − 4 ( 4 )( 25 ) 2 x= Apply the quadratic formula 2 ( 4) 20  0 5 = = 8 2 Here D = b2 − 4ac = 0 and there is only one solution. 5 Answer: The solution is x =. 2 (c) 3x 2 + 2 x + 1 = 0 a = 3, b = 2, c = 1 − ( 2)  ( 2 ) − 4 ( 3)(1) −2  2 −8 x= = 2 ( 3) 6 Here D = b2 − 4ac = −8  0. Therefore, the quadratic equation has no real solutions. Answer: The equation has no real solution. 45  Fractional Equations Leading to Quadratic Equations 5 2 Example 7 Solve the equation − =4. x −1 x Solution The equation is defined for x  0, x  1. 5 2 − =4 x −1 x  5 2 x ( x − 1)  −  = 4 x ( x − 1) Multiply both sides with LCD: x ( x − 1)  x −1 x  5 x − 2 ( x − 1) = 4 x ( x − 1) Expand expressions and simplify 5x − 2 x + 2 = 4 x2 − 4 x Write the quadratic equation in standard form 4 x2 − 7 x − 2 = 0 Factor the quadratic form ( 4 x + 1)( x − 2 ) = 0 Apply Zero – Product Property 4x + 1 = 0 or x−2=0 1 x=− or x=2 4 These solutions are solutions of the quadratic equation 4 x2 − 7 x − 2 = 0. So, we must check the solutions with the original equation. Check the solutions 1 When x = − , L.H.S. = 5 − 2 = 5 − 2 = 4 = R.H.S. 4 x −1 x  1  1  −  −1  −   4  4 When x = 2 , L.H.S. = 5 − 2 = 5 − 2 = 4 = R.H.S. x −1 x ( 2) − 1 ( 2) 1 Answer: The solutions are x = − and x = 2. 4  Radical Equations Leading to Quadratic Equations A radical equation is an equation that contains one or more radicals (square roots, cube roots, etc.) with variables or expressions within them. Solving radical equations, we isolate a radical term and then eliminate it. For example, if an equation has a square root, we square both sides of the equation. 46 Example 8 Solve the equation 2x = 3 − x − 1. Solution 2x = 3 − x −1 Isolate square root on one side 2x − 3 = − x −1 Square both sides ( 2 x − 3) ( ) 2 2 = − x −1 Expand the left side 4 x 2 − 12 x + 9 = x − 1 Write the equation in standard form 4 x 2 − 13 x + 10 = 0 Determine values of a, b, and c − ( −13)  ( −13) − 4 ( 4 )(10 ) 2 13  3 Use the quadratic formula x= = 2 ( 4) 8 16 10 5 x= = 2, x= = 8 8 4 Solutions for a radical equation must be checked in the original equation. Check the solutions When x = 2 , we get L.H.S. = 2 ( 2 ) = 4 , but R.H.S. = 3 − ( 2 ) − 1 = 2. So, L.H.S.  R.H.S. Therefore, x = 2 is not a solution. 5 When x = , we get 4 5 5 L.H.S. = 2   = and R.H.S. = 3 −  5  − 1 = 5. 4 2 4 2 5 So, L.H.S. = R.H.S. Therefore, x = is the only solution. 4 5 Answer: The solution is x =. 4 Note: We must check the solutions of radical equations because the process of solving them involves squaring both sides of the equation. This can introduce extraneous solutions, which are solutions that do not satisfy the original equation but satisfy the modified equation obtained after applying the radical operations. 47 Exercises 2.2 1 – 10. Solve the quadratic equations by factoring. 1. x2 − 7 x + 12 = 0 2. x ( 3x − 5) = −2 3. 2 y − 7 y + 3 = 0 4. 3 y + 5 y = 2 2 2 5. 4t 2 − 4t − 15 = 0 6. −r 2 − r + 12 = 0 7. 4 x 2 = 5x 8. 4 x2 = 9 ( x − 5) =4 10. x ( 3x + 7 ) = 6 2 9. 11 – 18. Solve the equations by completing the square. 11. x2 + 8x − 11 = 0 12. x2 − 14 x + 13 = 0 13. y 2 + 10 y + 4 = 15 14. 3t 2 − 12t − 15 = 0 15. 2r 2 − 3r − 20 = 0 16. x2 − 6 x − 13 = 0 7 2 1 17. x + 3 x − =0 18. x + x− =0 2 2 4 3 3 19 – 28. Find the solution of the quadratic equations using quadratic formula. 19. x 2 −3x − 5 = 0 20. x2 − 4 x + 21 = 0 21. 2 x2 + x − 3 = 0 22. 3 y − 5 y − 2 = 0 2 23. 4 y − 12 y + 8 = 0 2 24. 3t 2 + 6t − 5 = 0 25. w2 = 3 ( 2w − 3) 26. x2 + x = 4 27. 3x2 − 5x + 1 = 0 28. 2 x − x2 + 9 = 0 48 29 – 42. Find all real solutions of the equations by any method. 1 1 3 2 29. 4 + − 2 =0 30. − =1 x x x+8 x−2 3 x−3 2 t +1 31. + =2 32. − =0 x−4 x t−2 t+4 x 6 1 2 5 4 33. = − 34. − = 4 x 2 x −1 x x + 2 3x + 2 2 x + 1 3 5 35. + =1 36. − +2=0 x +1 2x x2 x 2 7 a + 1 a + 5 7(2a + 1) 37. + −4 = 0 38. + = ( x − 3) x −3 a + 3 a − 2 a2 + a − 6 2 39. 2x + 5 = x − 5 40. 2 x + x + 1 = 8 41. x + 3x − 6 = 0 42. 2r + 1 + 1 = r 2.3 Applications of Equations Equations play a crucial role in solving real life problems. They allow us to break down complex problems into manageable parts and express them in mathematical form. By solving equations, we can find solutions, make decisions, and optimize processes. Objectives In Section 2.3 you will learn how to Model real life problems Convert Units of measurements Solve dimension problems Solve Profit-Loss problems Solve problems on Investment (Simple interest) Mathematical modeling is the process of creating a mathematical representation of a real-world scenario to provide insight and help make a prediction. 49 Mathematical modeling can be thought of as a process made up of the following steps:  Guidelines to solve word problems Step 1. Understand the word problem Read the word problem carefully, make sure you understand the real-life situation. (Use a dictionary to know the meaning of new words) Step 2. Identify the variable Identify the quantity the problem asks to find, then introduce notation for the variable. (For example, “Let x be the length of the classroom”.) Step 3. Translate from words to Algebra Read each sentence of the problem again, and express all mentioned quantities in terms of the variable in Step 2. (To organize the information, it is helpful to draw a diagram or make a table.) Step 4. Set up the model Find the critical fact in the problem that gives a relationship between the quantities in Step 3. Write an equation. Step 5. Solve the equation and check your answer We must make sure that we understand what a mathematical result means in terms of the given problem. A clear report makes the model understandable to others.  Systems of Units of measurements People measure the same quantities in different ways and use different units. The system units of measurements varies from country to country. The most widely used and internationally accepted is the Metric System /International System of Units (SI). Units of distance are meter (m), kilometer (km), centimeter (cm), millimeter (mm). 1 centimeter ( cm ) = 10 millimeters ( mm ) 1 meter ( m ) = 100 centimeters ( cm ) 1 kilometer ( km ) = 1000 meters ( m ) Units of weight are gram (g), kilogram (kg), milligram (mg), ton (ton), etc. 1 gram ( g ) = 1000 milligrams ( mg ) 1 kilogram ( kg ) = 1000 grams ( g ) 1 ton ( ton ) = 1000 kilograms ( kg ) 50 Units of capacity or volume are liter (l), kiloliter (kl), milliliter (ml), etc. 1 kiloliter ( kl ) = 1000 liters ( l ) 1 milliliter ( ml ) = 0.001 liter ( l ) However, in some countries, for example United States of America, United Kingdom, the system of units used are US customary units and Imperial Units respectively. Units of distance are inch (in), foot (ft), yard (yd), mile (mi). 1 foot ( ft ) = 12 inches ( in ) 1 yard ( yd ) = 3 feet ( ft ) 1 mile ( mi ) = 5280 feet ( ft ) Units of weight are ounce (oz), pound (lb), short ton (t), etc. 1 pound ( lb ) = 16 ounces ( oz ) 1 short ton ( t ) = 2040 pounds ( lb ) Units of capacity or volume are gallon (gal), quart (qt), pint (pt), ounce (oz), etc. 1 gallon ( gal ) = 4 quarts ( qt ) 1 quart ( qt ) = 2 pints ( pt ) 1 pint ( pt ) = 16 ounces ( oz ) In a globalized economy, businesses engage in international trade and collaborations. To communicate in business, there must be a conversion from one system of units to another. It allows businesses to communicate effectively with international partners, understand product specifications, and ensure accurate measurements in transactions. Table 1 includes conversion factors between metric units and their corresponding US units for the most widely used measurements. Table 1. Unit Conversion Table From US system to Metric system From Metric system to US system Distance 1 inch (in) = 2.54 centimeters (cm) 1 centimeter (cm) = 0.394 inches (in) 1 foot (ft) = 0.30 meters (m) 1 meter (m) = 3.281 feet (ft) 1 yard (yd) = 0.91 meters (m) 1 meter (m) = 1.094 yards (yd) 1 mile (mi) = 1.61 kilometers (km) 1 kilometer (km) = 0.621 miles (mi) 51  Weight 1 ounce (oz) = 28.349 grams (g) 1 gram (g) = 0.035 ounce (oz) 1 pound (lb) = 0.454 kilogram (kg) 1 kilogram (kg) = 2.205 pounds (lb) 1 short ton ( t ) = 0.907 metric ton ( ton ) 1 metric ton ( ton ) = 1.102 short ton ( t ) Volume 1 ounce ( oz) = 29.573 milliliters ( ml ) 1 milliliter ( ml ) = 0.034 ounce ( oz) 1 quart (qt) = 0.946 liter ( l ) 1 liter ( l ) = 1.057 quarts (qt) 1 pint (pt) = 0.473 liter ( l ) 1 liter ( l ) = 2.114 pints (pt) 1 gallon ( gal ) = 3.7854 liters ( l ) 1 liter ( l ) = 0.264 gallon ( gal )  Dimension Problems To consider real-life problems related to dimensions, we need to remember some formulas: length, l Rectangle Perimeter: P = 2 ( l + w) width, w Area: A = lw a Square Perimeter: P = 4a a Area: A = a2 Triangle b c Perimeter: P = a + b + c height , h 1 Area: A= ah base, a 2 diameter, d Circle d = 2r radius, r Circumference: C = 2 r Area: A =  r2 Example 1 Dimensions of a parking lot. A square parking lot is surrounded by a 368–foot fence. Determine the dimensions of the parking lot. Express your answer in meters. 52 Solution It is mentioned that the fence around the parking lot has a length of 368 ft. It is the perimeter of the parking lot. We need to find the length of side a. a a P = 4a As the perimeter of the parking lot is 368 ft, we can write an equation 368 = 4a. 368 Therefore, a = = 92 ( ft ). 4 meter Now we convert 92 feet to meters: 92 feet  0.3 = 27.6 ( m ). feet Answer: The parking lot has a dimension of 27.6 meters by 27.6 meters. Example 2 Dimensions of a classroom. A rectangular classroom is 2 meters 50 centimeters longer than it is wide. Find the length and width of the classroom if its area is 58.5 meters squared. Solution In the problem, we need to find the classroom dimensions (its width and length). Let x be the width of the classroom. Then the length is ( x + 2.5). To understand the problem, we draw a sketch and label the sides. l = x + 2.5 w= x A = lw To write an equation, we use the fact that the classroom area is 58.5 meters squared: 58.5 = ( x + 2.5)  x Solve the equation (quadratic equation): x2 + 2.5x − 58.5 = 0 Applying the quadratic formula for a = 1, b = 2.5, c = −58.5, we get: −b  b 2 − 4ac −2.5  2.5 − 4 (1)( −58.5 ) −2.5  15.5 2 x1,2 = = = 2a 2 (1) 2 x1 = 6.5 and x2 = −9 We identify the variable x as the width of the classroom, so we reject the negative value. To find the length, we use fact that the room is 2 meters 50 centimeters longer than it is wide. Answer: The classroom width is 6.5 meters and its length is 9 meters. 53  Profit Profit refers to the financial gain obtained by a business after deducting the total expenses from the total revenue generated. It is a fundamental concept in economics that serves as a key indicator of the effectiveness and success of a business. Any manufacturing company has Total Costs (YTC ) and Total Revenue (YTR ). Total costs consist of Fixed Costs (YFC ) and Variable Costs (YVC ) : YTC = YFC + YVC Fixed costs do not depend on production level, they are constant in the company budget. For instance, office rent is a fixed expense. The company pays the same amount every month (or every year). Variable costs depend on the level of production. In the table below, are some examples of Fixed Costs and Variable Costs. Examples of Fixed Costs Examples of Variable Costs Insurance Raw materials Rent Piece-rate labor Equipment Maintenance Advertising Production supplies Loan payments Packing supplies Property taxes Delivery costs Management’s salaries Commissions Total revenue is calculated as YTR = p  q , where p is price per unit, q is quantity (units). The difference between total revenue and total cost is the manufacturer’s Profit: Profit = YTR − YTC When total revenue remains less than total cost, the manufacturer makes Loss. Example 3 A small company produces gift boxes whose selling price is 1 rial 500 baisa. The company has fixed costs 240 rials and the variable cost per unit is 700 baisa. (a) Determine the number of units that must be sold for the company to earn a profit of 1000 rials. (b) Estimate the Profit (or Loss) of the company for a production level of 150 gift boxes. 54 Solution In the problem, we are given YFC = 240 , YVC = 0.7q, p = 1.500. Therefore, total cost is YTC = 240 + 0.7q and total revenue is YTR = 1.5q. (a) Now we can find the profit: Profit = YTR − YTC = 1.5q − ( 240 + 0.7 q ) = 0.8q − 240 We need to determine the production level required to earn 1000-rial profit. Solve the equation: 1000 = 0.8q − 240 → q = 1550 ( units ) (b) We need to estimate the profit/loss when 150 gift boxes are produced and sold. Substitute the value of q into the profit equation: Profit = 0.8q − 240 = 0.8 (150 ) − 240 = −120 ( rials ) ← Loss As we found that the profit is negative, the company gets a loss of 120 rials when they produce and sell 150 gift boxes. Answer: (a) The company must produce and sell 1550 gift boxes to make 1000-rial profit. (b) The company gets a loss of 120 rials at the production level of 150 units.  Simple Interest When you borrow money from a bank or when a bank uses your money from a saving account, the borrower must pay for the privilege of using that money. The fee paid is called Interest. The most basic type of interest is Simple Interest, which is an annual percentage of the total amount borrowed or deposited (invested). The amount borrowed or invested is called Principal ( P ). The annual percentage paid for use of this money is an interest rate ( r ,% ). Simple Interest formula gives the amount of interest ( I ) earned when a principal P is deposited (invested) for t years at annual interest rate r : I = P  r t Total amount ( S ) at the end of t years is calculated by formula: S = P+I 55 Example 4 Ahmed invests 10,000 rials for three years at 7.25 % per year. How much interest will he earn at the end of three years? What will be the total amount of his investment after three years? Solution 7.25 The given quantities are: P = 10000 (RO), r = 7.25 % = = 0.0725, t = 3 ( years). 100 Substitute the data into the formula I = P  r  t. Therefore, the interest earned: I = (10000 )  ( 0.0725)  ( 3) = 2175 (RO). Total amount of investment after three years: S = P + I = 10,000 + 2175 = 12,175 (RO). Answer: Ahmed’s interest for 3 years will be 2175 rials, and the total amount will reach 12,175 rials. Example 5 To open business in a new mall, Sara’s Card Shop borrows 5 000 rials from a bank at 6.5 % simple interest. If the shop repaid 5 650 rials, how long did it take? Solution 6.5 In the problem given: P = 5000 (RO), r = 6.5 % = = 0.065, S = 5650 ( RO). 100 We need to find time t required to repay the loan. Substituting the data into the formula S = P + I , we get an equation 5650 = 5000 + I. Solve the equation for the interest paid: I = 650 (RO). Now we can find t from Simple Interest formula: I = P  r  t 650 = 5000  ( 0.065)  t 650 = 325  t 650 t= = 2 ( years ) 325 Answer: It took two years for Sara’s Card Shop to repay the loan. 56 Exercises 2.3 1. A farmer has a rectangular pasture surrounded by a 450-m fence. Find the length and width of the pasture if its area is 12,500 m ². 2. A building lot is twice as long as it is wide. Its area is 882 m ². How wide is the building lot? 3. A rectangular bedroom is 1 m 20 cm longer than it is wide. Find the length of the room if its area is 24 m ². 4. A circular flower bed has an approximate area of 5.5 m ². Find the diameter of the flower bed. 5. You want to paint the walls of a room in your house. The room dimensions are 3 m by 4.5 m, and the ceiling is 3.2 m high. How many cans of paint do you need to buy if one can of the paint covers 8 m ² ? 6. Hamed wants to rebuild the backyard of his house. The space measures 15 meters by 20 meters. How many pavers does he need to buy to cover the entire area, if the dimensions of the paver are 20 centimeters by 35 centimeters? 7. A manufacturer sells a product at 7 rials, selling all that is produced. Fived cost is 2500 rials and variable cost per unit is 5 rials 600 baisa. (a) Find the profit / loss when 1000 units are produced and sold. (b) How many units should be sold to get a profit of 2750 rials? 8. A furniture company invests 20,000 rials in equipment to produce a new model of sofa set. Each sofa set costs 140 rials to produce, and it is sold for 300 rials. (a) How many sofa sets must be sold to get a profit of 25,000 rials? (b) Determine profit / loss when 200 sofa sets are sold. 9. A restaurant spends 5000 rials on food, 2000 rials on rent, 1500 rials on staff salaries, and utilities each month. If the restaurant generates a monthly revenue of 15,000 rials, what is the monthly profit? 10. A car dealership buys a certain brand of car for $ 15,000 and sells it for $ 23,000. The dealership spends $ 1200 on marketing, $ 2400 on staff salaries, and $ 500 on maintenance. If the dealership sold 50 cars in a month, determine the total profit. 57 11. A toy store buys a toy for 3 rials and sells it for 5 rials. The store has to pay 350 rials for rent, 240 rials for store staff salaries, and 85 rials for electricity every month. If the store sold 1000 toys in a month, what is the total profit? 12. An initial deposit of 1400 rials is made into an account paying 6 ⅜ % simple interest. Find the total amount at the end of 36 months. 13. Salim invested 4700 rials that he inherited in a saving account paying 7.8 % simple interest. He hopes to buy a new car in 6 years and needs 6500 rials. Is it possible to use this investment alone? 14. Omar has 860 rials in a saving account that pays 5 ⅗ % simple interest. How long will it take for the account balance to reach 1000 rials at this interest rate? 15. A construction company borrows 72,000 rials for a new project. At the end of 4 years, they have to pay back 87,000 rials. If it is a simple interest loan, what is the annual interest rate? 16. What annual rate of interest would you have to earn on an investment of $ 2000 to ensure receiving $ 113 interest after one year? 17. Ahmed borrowed 20,000 rials from a bank at a simple interest rate of 5.45 % per year. If Ahmed pays off the loan in 4 years, what is his monthly payment? 18. A bank offers a simple interest rate of 5 % per year on deposits. If Maryam deposits 1200 rials for a period of 5 years, what is the total interest she earns? 19. The simple interest on a sum of money for 3 years at 6.5 % per year is 507 rials. What is the principal amount? 20. A business invests $ 50,000 in a bond with an annual simple interest rate of 4.5 %. How much interest will the business earn after two years? 58 2.4 Inequalities When two algebraic expressions are not strictly equal, we get an inequality. There are many real-life situations where the mathematical model leads to inequalities. Objectives In Section 2.4 you will learn how to State the solution set for an inequality Solve linear inequalities Solve absolute-value inequalities Solve nonlinear inequalities Applications of inequalities  Solution set of Inequality An inequality is a statement that one expression is less than ( < ), greater than ( > ), less than or equal to ( ≤ ), or greater than or equal to ( ≥ ) another expression. For instance, 2 7−x 9x − 2  3.5x ; 3 − 4x  ; 2 x2 − 10  3x ; 2. 5 x +1 To solve an inequality that contains a variable means to find all values of the variable that make the inequality true. The set of numbers that satisfy an inequality is called the solution set. To write a solution set of an inequality, we will use forms of set notation, real-number line graph, and interval notation. Inequality Set notation Real-number line Interval notation xa x xa  a x  ( −, a ) xa x xa  a x   a,  ) a xb x a xb  a b x  ( a, b ) a xb x a xb  a b x   a, b  a xb x a xb  a b x   a, b ) Note: An open dot ( ) on a real-number line graph marks the location of an endpoint that is not included in a solution set, whereas a closed dot ( ) marks the location of an endpoint that is included in a solution set. 59  Rules for Inequalities To solve inequalities, we use the following rules to get simpler equivalent inequalities. In these rules A, B, and C stand for real numbers or algebraic expressions. Here we state the rules for inequalities involving the symbol ≤, but they apply to all four inequality symbols. 1. Adding the same value to each side of an A B  A+C  B +C inequality produces an equivalent inequality. Subtracting the same value to each side of 2. A B  A−C  B −C an inequality produces an equivalent inequality. 3. If C  0, Multiplying each side of an inequality by the then A  B  AC  B C same positive value produces an equivalent inequality. 4. If C  0, Multiplying each side of an inequality by the then A  B  AC  B C same negative value reserves the direction of the inequality. 5. If A  0 and B  0, Taking reciprocals of each side of an 1 1 inequality involving positive quantities then A  B  reserves the direction of the inequality. A B 6. If A  B and C  D, Inequalities which have the same inequality then A + C  B + D symbol can be added.  Solving Linear Inequalities A linear inequality in one variable is one that can be written in the form ax + b  c , where a, b, and c are real numbers and a  0. 2 1 Example 1 Solve the inequality x +  4 , then graph the solution set and write it 3 2 in interval notation. Solution ( 6 )   2 1 Multiply both sides of the inequality by x +   4  (6) 3 2 LCD (Here LCD = 6, and 6 > 0) 4x + 3  24 Isolate x-term on one side (subtract 3) 4x  21 Divide both sides by 4 (4 > 0) x  21 4 21 4 ( Answer: Interval notation: x  −, 21 4 ) OR x  ( −, 5.25). 60 Example 2 Solve the inequality, then graph the solution set and write it in interval notation. 9 − 2 ( 5 x + 3)  3 ( 7 − x ) + 2 x Solution 9 −10x − 6  21 − 3x + 2x Simplify both sides 3 −10x  21 − x Isolate x-term on one side −9x  18 Divide both sides by −9 ( −9  0 ) x  −2 −2 Answer: Interval notation: x   −2,  ). Example 3 Solve the inequality, and write the solution set in interval notation. 3( 2 x − 1) − 5 x  2 ( 9 + x ) − x Solution 6x − 3 − 5x  18 + 2x − x Simplify both sides x − 3  18 + x Isolate x-term on one side 0  21 Since the resulting statement is always true, the original inequality is true for all real numbers. Answer: Interval notation: x  ( −,  ). Example 4 Solve the inequality, and write the solution set in interval notation. 4 ( 5 − 2 x ) + 7 x  8 ( x + 1) − 9 x Solution 20 − 8x + 7 x  8x + 8 − 9x Simplify both sides 20 − x  8 − x Isolate x-term on one side 0  −12 Since the resulting statement is always false, the original inequality is false for all real numbers. The inequality has no solution. Answer: x    or ϕ or No solution. 61  Solving Absolute-value Inequalities The absolute value of a number c is the distance from c to 0 on the real number line. We use the following properties to solve inequalities that involve absolute value. Absolute-value Inequality Equivalent inequality Graph x c −c  x  c −c 0 c x c −c  x  c −c 0 c x c x  −c or xc −c 0 c x c x  −c or xc −c 0 c Example 5 Solve the inequality, then graph the solution set and write it in interval notation. 3 x−2  4 5 Solution 3 Replace absolute-value inequality by −4  x−2 < 4 5 equivalent simultaneous inequality −20  3x −10  20 Multiply each part by LCD −10  3x  30 Isolate x-term in middle part (add 10) 10 −  x  10 Divide each part by 3 (3 > 0) 3 −10 −10 3  1  Answer: Interval notation: x   − , 10  10 or x   −3 , 10 .  3   3  Example 6 Solve the inequality, then graph the solution set and write it in interval notation. 7 − 2 x + 1  12 Solution 7 − 2x  11 Subtract 1 to isolate absolute value 62  7 − 2x  −11 or 7 − 2x  11 Replace absolute-value inequality by equivalent inequalities −2x  −18 or − 2x  4 Solve each inequality for x (subtract 7) x  9 or x  −2 Divide by –2 (–2 < 0), and reverse inequalities Show solution sets on the real-number −2 9 line Answer: Interval notation: x  ( −, −2 9,  ).  Solving Nonlinear Inequalities To solve inequalities involving squares or fractional expressions, we use factoring together with the following principle. The sign of a Product or Quotient If a Product or a Quotient has an even number of negative factors, then its value is positive. If a Product or a Quotient has an odd number of negative factors, then its value is negative. For example, the product ( x − 5)(1 − x ) is positive if both factors are positive or both factors are negative. If one of the factors is negative while the another is positive, the value of the product is negative. We take the following steps to solve nonlinear inequalities. Steps to solve nonlinear inequalities 1. Move all terms to one side. Rewrite the inequality so that all nonzero terms appear on one side of the inequality. If the nonzero side involves fractional expressions, bring them to a common denominator. 2. Factor. Factor the nonzero side of the inequality. 3. Find the intervals. Determine a zero of each factor. Plot the zeros on the real- number line. They will divide the real-number line into intervals. 4. Make a table or diagram. In the first row write the intervals determined in Step 3. Chose a test value to determine the signs of each factor on each interval. In the last row of the table determine the sign of the product (or quotient). 63  5. Solve. Determine the solution set of the inequality from the last row of the table. Check whether the inequality is satisfied by the endpoints of the intervals (if the inequality involves ≤ or ≥ ). Here we use the fact that a linear factor changes sign only at its zero. For instance, the 5 5 factor ( 3x + 5) is negative for any x  − and positive for any x  −. 3 3 Example 7 Solve the inequality 2x2 + 7 x  15 , and write the solution set in interval notation. Solution 2 x2 + 7 x − 15  0 Move all terms to one side ( 2 x − 3)( x + 5)  0 Factor nonzero side 3 x= , x = −5 Find zeros 2 Plot zeros on real-number line, and determine intervals: −5 3 2 ( −, −5) , ( −5, 1.5) , and (1.5,  ) We determine the signs of the factors on each interval using test values. We choose a number inside each interval, for example, number − 6 in interval ( −, −5) , 0 in interval ( −5, 1.5) , and 2 in interval (1.5,  ). Now we construct a table. −6 0 2 ← Test values signs intervals ( −, −5) ( −5, 1.5) (1.5,  ) sign of ( 2 x − 3) sign of ( x + 5) sign of ( 2 x − 3)( x + 5) Analyzing the last row of the table, we make a conclusion that product ( 2 x − 3)( x + 5) is negative on the interval ( −5, 1.5). The endpoints −5 and 1.5 do not satisfy the inequality ( 2 x − 3)( x + 5)  0 , so they are not included in the solution set. Answer: Solution set: x  ( −5, 1.5). 64  9 − 2x Example 8 Solve the inequality  1 , and write the solution set in interval x+4 notation. Solution 9 − 2x −1  0 Move all terms to one side x+4 5 − 3x Bring fractions to common  0 denominator and simplify x+4 5 Find zeros of numerator and x= , x = −4 3 denominator Plot zeros on real-number line, and determine intervals: −4 5 3 ( ) ( −, −4) , −4, 5 3 , and 5 3 ,  ( ) Now choose a test value in each interval and construct a table: −5 0 2 ← Test values ( −4, 5 3 ) ( 5 3 , ) intervals signs ( −, −4) sign of ( 5 − 3x ) sign of ( x + 4 ) 5 − 3x sign of x+4 5 − 3x We can see from the table that the quotient is negative on the intervals ( −, −4 ) x+4 ( ) and 5 3 , . The endpoint 5 3 satisfies the inequality 5 − 3x x+4 0. However, we do not include the other endpoint −4 because the quotient in the inequality is not defined at −4. Answer: Solution set: x  ( −, −4 ) 5 , .  3 ) Note: Instead of a table, we can use a sign diagram. You can see below the sign diagram for example 8. −4 5 3 5 − 3x (+) (+) ( −) → = (−) = (+) = (−) x+4 ( −) (+) (+) 65  Applications of Inequalities Inequalities play a significant role in solving real life problems. Example 9 A retailer sells two types of products, A and B. The profit on product A is 20 % of the total revenue, and on product B is 14 % of the total revenue. The retailer wants to ensure that the overall profit is at least 18 % of the total revenue. If the retailer sells 100 units of product A, how many units of product B should be sold to meet the profit requirement? Write an inequality and solve it. Solution It is given that the retailer sells 100 units of product A. Let q be the number of units of product B sold and YTR be the total revenue. The total profit is: Profit = Profit on A + Profit on B. = 100 ( 0.20 ) YTR + q ( 0.14 ) YTR So, the inequality representing the given condition is: 100 ( 0.20 ) YTR + q ( 0.14 ) YTR  (100 + q )( 0.18 ) YTR. Solve the obtained inequality. YTR ( 20 + 0.14q )  (18 + 0.18q ) YTR 20 + 0.14q  18 + 0.18q 0.14q − 0.18q  18 − 20 −0.04q  −2 q  50 Answer: The retailer should sell at most 50 units of product B to meet the requirement. Exercises 2.4 1 – 6. Express the inequalities in interval notation, and then graph them. 1. x  −4.2 2. −5  x   3. −4  x  2 4. x  −3 or x7 5 5. x  8.5 or x  1.5 6. x  −3 and x  −5 4 66  7 – 12. Solve the inequalities. Express the solution set using interval notation and graph it. 7. 4 − 3x  −2 ( 6 − 2 x ) 8. 5 + 2 ( 3 − 4 x )  3 ( 2 x − 1) + 7 x 2 x − 11 5 1 5 9.  10. x + 2  1− x 9 3 3 6 1 4 − 5x 1 2 4x − 7 11. −   12.   3 2 3 4 5 2 13 – 18. Solve the absolute-value inequality. Express the solution using interval notation and graph the solution set. 13. 1 − 3x  5 14. 9 x + 3 − 4  2 x+7 5x − 3 15.  2 16.  1 3 2 17. 3 − 2 x + 5  1 18. 9 + 3x − 5  2 19 – 26. Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. x2 − 3x  x + 12 ( x − 2) + x  4 ( x − 3) 2 19. 20. x2 − 6 x + 9  16 ( x + 3)  2x + 5 2 21. 22. 23. 4 x − 12 24. 2x − 6  0  2 x +1 x+3 25. 2x + 5 26. 7 − 2x  1  −1 x +1 x+2 27. A company manufactures and sells a product. The production cost per unit is $ 10, and the selling price per unit is $ 15. The company wants to determine the minimum number of units it needs to sell in order to make a profit of at least $ 500. Write an inequality and solve it to find the answer. 28. A company's monthly expenses include fixed costs of 1000 rials and variable costs of 5 rials per unit produced. The company wants to ensure that the total expenses do not exceed 2500 rials in a given month. Write an inequality and solve it to find the maximum number of units the company can produce without exceeding the expense limit. 67