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Module 6 - Linear Equations

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Mapúa Institute of Technology**We aren't endorsed by this school
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MATH 0
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Mathematics
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Jan 7, 2025
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Page 1 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.phCOLLEGE OF ARTS AND SCIENCES DEPARTMENT OF MATHEMATICS Vision: “The Home of the Learned committed to WISDOM, INTEGRITY and TRUTH, and the legacy of providing equal opportunities for the underprivileged”Institutional Outcomes: Within 5 years after graduation, the Ideal WIT Graduate can: •Be employed, compete, and lead locally and globally in his/her field of specialization in any Professional, Governmental, Non-Governmental, Civic or Academic Institution •Engage in community service and conduct researches—especially scientific, technological, business, and social researches—that will benefit the community •Be a role model in his/her workplace in terms of the WITtian Identity and Core Values____________________________________________________________________________ Modules for ITE Math 0 (Mathematics for Engineering Science and Technology) Module 6 (Linear Equations) Presentation of the Lesson 6.1Concept of Equations An equationis a statement that two expressions are equal, the expression being called the members, or sides, of the equation. These members will be called leftor rightaccordingly as they precede or follow the sign =in the equation. An equation ordinarily contains one or more literal quantities, which are called unknowns; and the object of the processes of algebra is to find the values of these quantities. Thus, 3? – 5 = 7is an equation in which the unknown is ?. 6.2Definition of Terms Transposition–is the process of transferring a term from one side of an equation to the other side and changing its sign. Conditional Equation–is an equation which is satisfied only by particular values of the unknown. Identical Equation–is an equation which is satisfied by all values of the unknown. Equivalent Equations–are equations that have identical solution sets. Linear Equation–is one having the unknown value raised to the first power only or, if it contains fractions, the unknown does not appear in the denominator. Fractional Equation–is one which contains the unknown in one or more of the denominators. Literal equation–is one in which letters other than the unknown quantity are present. Radical Equation–is an equation in which one or more of the terms contain radicals. Solution of an Equation–a set of values of the unknown which satisfies the equation. When only one unknown is involved in the equation, a solution is called a rootof the equation. 6.3Identical or Conditional EquationsAn equation whose members are equal for all admissible values of the unknown or unknowns which it contains is an identical equation, or an identity. An equation whose members are equal for certain values (or possibly for no values), of the unknown or unknowns which it contains, but not for all admissible values, is a conditional equation. The word “equation”, when used without a modifier, will ordinarily be interpreted to mean “conditional equation.” If an equation contains only one unknown, any number which, when substituted for the unknown makes the members of the equation equal to each other, is called a root, or solution, of the equation, and is said to satisfy the equation. If an equation contains more than one unknown, any set of numbers which, when substituted for the unknowns makes the members equal to each other, is a solution. The equations 2(? – 3) = 2? – 6 (? + ?)2= ?2+ 2?? + ?2?2+2?−1= ? + 1 +3?−1are identities, being satisfied by all admissible values of the unknowns. Note: In the last identity, 1is not an admissible value for ?; the members are not defined for ? = 1, since this value would involve division by zero. The equation 3? – 5 = 7, on the other hand, is a conditional equation, since it is satisfied only by ? = 4. An identity is often written with the sign≡instead of =to emphasize that it is an identity rather than a conditional equation. Thus 3(? – 5) 3? – 156.4Equivalent EquationsTwo equations are equivalent if they have exactly the same solutions. Thus, the equations 2? − 10 = 0 and ? − 5 = 0
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Page 2 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.phare equivalent; each has the solution ? = 5and no other. 6.5Operations Yielding Equivalent EquationsThe following operations on an equation will always lead to an equivalent equation: 1.Adding the same number or expression to both sides or subtracting the same number or expression from both sides. 2.Multiplying or dividing both sides by the same number or expression, provided that this number or expression is not zero and does not contain an unknown. Note: Dividing by 𝒑is just multiplying by ?𝒑, if 𝒑 ≠?and also Subtracting 1is just adding –1In equation form, these statements are as follows: ? = ?is equivalent to ? + ? = ? + ?for every ?? = ?is equivalent to ?? = ??for every ? 0Example 1a.Solve the equation ? – 3 = 5Solution ? – 3 = 5` Given ? = 8Adding 3 b.Solve the equation 3? – 5 = 7Solution 3? – 5 = 7Given 3? = 12Adding 5 ? = 4Dividing by 3 c.Solve the equation 3? = 5? – 7Solution 3? = 5? – 7Given 3? – 5? = – 7Adding –5x – 2? = – 7Combining similar terms ? = 27Dividing by –2 d.Solve the equation 3? + 5 = 7? – 3 Solution 3? + 5 = 7? – 3Given 3? – 7? = – 3 – 5Adding (– 7? – 5) – 4? = – 8Combining similar terms ? = 2Dividing by – 4e.Solve the equation ?2−23=3?4+112Solution ?2−23=3?4+112Given 12 (?2−23) = 12 (3?4+112)Multiplying by the LCD = 12 6? – 8 = 9? + 1Distributive axiom 6? – 9? = 1 + 8Adding 8 –9x – 3? = 9Combining similar terms ? = – 3Dividing by –3 f.Solve the equation ??+1+58=5?2(?+1)+34Solution ??+1+58=5?2(?+1)+34Given 8(? + 1)(??+1+58) = 8(? + 1) (5?2(?+1)+34)Multiplying by the LCD = 8(? + 1)8? + 5(? + 1) = 4(5?) + 6(? + 1)Removing common factors from each term 8? + 5? + 5 = 20? + 6? + 6Adding (– 26? – 5)
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Page 3 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.ph– 13? = 1Combining terms ? =131−Dividing by –13 g.Solve the equation 2??−3= 1 +6?−3Solution 2??−3= 1 +6?−3Given 2? = 1(? – 3) + 6Multiplying by the LCD = ? – 3, ? 32? = ? – 3 + 6Distributive axiom 2? – ? = – 3 + 6Subtracting ?? = 3Combining terms The number 3 in this example is called an extraneous rootof the given equation because it is not a root of the given equation, although it is a root of a subsequent nonequivalent equation. Remember: Do not multiply each side of the equation by 0 in any form. h.Solve 2?−5+5?−2=7(?−2)(?−5Where we assume ?5and ?2Solution 2?−5+5?−2=7(?−2)(?−5Given 2(? – 2) + 5(? – 5) = 7Multiplying by the LCD = (? – 5)(? – 2) 2? – 4 + 5? – 25 = 7Distributive axiom 7? – 29 = 7Combining terms 7? = 36Adding 29 ? = 736Dividing by 7 6.6Linear Equations in One Unknown?? + ? = 0→General Form of linear equation in the unknown ?(or linear equation in ?) if ? 0“to solve an equation” means to find all of its solutions. ?? + ? = 0Given ??+ ?–?= 0 –?Adding –?to both sides ?? = – ?Combining terms ???=−??Dividing by ?since ?0? =−??6.7Solution of a Linear Equation in One Unknown1.If fractions appear, multiply both sides of the equation by the least common denominator and combine like terms. 2.Transpose all terms having an unknown on one side and all other terms on the other side of the equation. 3.Reduce the resulting equation to the form ?? = ?. 4.Divide both sides of the equation by the coefficient of the unknown (i.e., by an equation ?? = ?) resulting to ? =??. 5.To check, substitute the result in the original equation. Example 2Solve the following equationsa.3? – 4 = ? + 4Solution 3? – 4 = ? + 4Given 2? = 8Adding (−? + 4)? = 4Dividing by 2b.2?3− 1 = ? + 5Solution 2?3− 1 = ? + 5Given 3 [2?3− 1] = (? + 5)3Multiply by the LCD = 32? – 3 = 3? + 15Distributive law – ? = 18Combining similar terms ? = – 18Multiply by (–1)
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Page 4 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.phc.?3?+3= ?+33?+6Solution ?3?+3= ?+33?+6Given ?3(?+1)= ?+33(?+2)Factor the denominator to determine the LCD ?(? + 2) = (? + 3) (? + 1)Multiply by the LCD = 3(? + 1)(? + 2)?2+ 2? = ?2+ 4? + 3Distributive law ?2− ?2+ 2? − 4? = 3Transpose and combine similar terms – 2? = 3Add (−?2− 4?)? = 23−Multiply by −126.8Translating Word Statements into Algebraic Expressions 1.Five more than ?5 + ? 2.Eight decreased by ?8 – ?3.The product of 3and ?3?4.The quotient of ?and ???5.Three more than twice ?2? + 36.The sum of 3and ?3 + ?7.The difference of ?and four ? − 48.The square of the hypotenuse ℎℎ29.The cube of the side ??310.The square root of 5√511.The quantity 3less than ?(? − 3)12.Eleven multiplied by the quantity ? + 2711(? + 27)13.Ella was ?years old four years ago ? + 10Represent her age six years from now 14.If John can walk 𝑘kilometers in ℎhours, How long will it take him to walk ?kilometers? ℎ𝑑𝑘15.?increased by seven equals thirteen Equation: ? + 7 = 1316.?is equal to the difference of ?and five Equation: ? = ? – 517.A number ?increased by five gives ten Equation: ? + 5 = 1018.Two thirds of a number ?is 20 Equation: 23? = 2019.?is 45 more than ?Equation: ? = 45 + ? ?is greater than ?by 45 ? – ? = 4521. the value of ?decreased by 45 is ?Equation: ? – 45 = ??is 45 less than ?? = ? – 4522.The sum of ?and ?is 357 Equation: ? + ? = 357?is 357 diminished by ?? = 357 – ? The values of ?and ?total 357 ? + ? = 35723.The sum of four consecutive integers is 178Equation: ? + (? + 1) + (? + 2) + (? + 3) = 178,where ?is the smallest of the four integers. 24.?is twice ?Equation: ? = 2??is half of ?? = ?225.Chum’s age is 3 more than twice Equation: ? = 3 + 2(? – 10)what it was 10 years ago 6.9Applications of Linear Equations in One Unknown A stated problem, or word problem, is a description of a situation that involves both known and unknown quantities and relationships between them. A problem involving only one unknown can be solved by the use of a single equation. Procedure 1.Read the problem carefully and make sure the situation is thoroughly understood. 2.Identify the quantities, both known and unknown, that are involved in the problem. 3.Select one of the unknown quantities and represent it by a variable (letter), and then express any other unknown in terms of this variable, if possible. 4.Search the problem for information that tells what quantities or combinations of them are equal. 5.Often, making a sketch helps to carry out step 4. 6.Write an equation using an algebraic expression found in step 5. Carrying through the calculations with an initial guess sometimes help to clarify the relationship between variables. 7.Solve the equation obtained in step 6. Check the solution in the original problem. This step is critical since we want a solution of the stated problem rather than of the equation that we write.
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Page 5 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.ph6.9.1Number ProblemsExample 3a.A number is twice another number and their sum is 15. What is the number? Solution 𝐿?? ? =one number 2? =the other number ? + 2? = 15The required equation 3? = 15Combining terms ? = 153Dividing by 3 ? = 5The required number b.The sum of three numbers is 27. The second number is 2 less than twice the first number, and the third number is 5 greater than 3 times the first. Find the three numbers. Solution Let ? =the first number 2? – 2 =the second number 3? + 5 =the third number ? + 2? – 2 + 3? + 5 = 27The required equation 6? = 24Combining similar terms ? =246Dividing by 6 ? = 4The first number 2? – 2 = 6The second number 3? + 5 = 17The third number c.One number exceeds three times a second number by 3. Half of the larger number exceeds one–fifth of the smaller number by 8. Find these numbers. Solution Let ? =the smaller number 3? + 3 =the larger number 12(3? + 3) −15(?) = 8The required equation 5(3? + 3) – 2(?) = 80Multiply by the LCD = 10 15? + 15 – 2? = 80Distributive Law 3?=65Adding –15 and combining similar terms ? = 6513Multiply by 113? = 5The smaller number 3? + 3 = 18The larger number d.Separate 357 into 2 parts such that if the larger is divided by the smaller, the quotient is 15 and the remainder is 5. Solution Let ? =the smaller number 357 – ? =the larger number 357−??= 15 +5?Adding –15 and combining similar terms quired equation 357 – ? = 15? + 5Multiplying by ?; →?0352 = 16?Combining similar terms 16? = 352Add (– 16? – 352)and multiply by – 1? =35216Multiplying by 116? = 22The smaller number 357 – ? = 335The larger number e.The unit’s digit of a two digits number is 2 greater than the ten’s digit. If the number is divided by the sum of the digits, the quotient is 4 and the remainder is 9. Find the number. Solution Let ? =the ten’s digit ? + 2 =the unit’s digit 10 ? + ? + 2 =the number 11?+22?+2= 4 +92?+2The required equation 11? + 2 = 4(2? + 2) + 9Multiplying by 2? + 211? + 2 = 8? + 8 + 9Distributive law 3? = 15Combining similar terms ? = 5Multiplying by 13? = 5The ten’s digit ? + 2 = 7The unit’s digit 6.9.2Age Problems Example 4a.Mama is four times as old as Isabelle. Twenty years from now, Mama will be only twice as old as Isabelle. What are their ages now? Solution: Let ? =the present age of Isabelle
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Page 6 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.ph4? =the present age of Mama Present Future Isabelle ?? + 20Mama 4?4? + 204? + 20 = 2(? + 20)The required equation 4? + 20 = 2? + 40Distributive law 2? = 20Add (– 2? – 20)? =202Multiply by 12? = 10 ?????The present age of Isabelle 4? = 40 ????? The present age of Mama b.Starla’sage is 3 more than twice what it was 10 years ago. What is her age now? Solution: Let ? = Starla’sage now ? – 10 =her age 10 years ago Past Present Starla ? – 10?? = 3 + 2(? – 10)The required equation ? = 3 + 2? – 20Distributive law ? – 2? = 3 – 20Add – 2?– ? = – 17Combining terms ? = 17Multiply by – 1? = 17 ?????Starla”s age now c.I am 20 years old. My mom was as old as I am now when I was one fourth of her age now. How old is my mom? Solution: Let ? =Mom's age now Past Present Me 14?20Mom 20?20 −14? = ? − 20The required equation 80 – ? = 4? – 80Multiply both sides by 4 −5? = – 160Add (– 4? – 80)? =−160−5Multiply by −1−5∴ ? = 32Mom’s age now 2.Ella's age is four–fifths of that of Marie's. In three years, the former will be as old as the latter is now. How old are they? Solution Let ? =Marie's age Present Future Marie ?? + 3Ella 45??? −45? = (? + 3) − ?The equation 5? − 4? = 5(? + 3) − 5?Multiply both sides by 5 ? = 5? + 15 − 5?Remove symbols of grouping ? = 15Combine similar terms ∴ ? = 15Marie's age 3.Timothy is 23as old as his sister Isabelle. 20years hence he will be 78as old. How old is each now? Solution Let ? =Isabelle's present age 23? = Timothy’s present age Present Future Isabelle ?? + 20Timothy 23?23? + 2023? + 20 = 78(? + 20)The required equation 16? + (24) (20) = 21 (? + 20)Multiply by the 𝐿?? = 2416? + 480 = 21? + 420Distributive law 5? = 60Combining similar terms
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Page 7 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.ph? =605Multiply by 15∴ ? = 12 ?????Isabelle’s present age23? = 8 ????? Timothy’s present age6.9.3Work ProblemsIf a man could finish a job in ?days, then in one day he could finish only 1?of the job. If both ?and ?are the number of days each man work alone, then if they work together, they could finish it in “?” days.1?+1?=1𝑡Example 6 1.Jack can paint a house in 10 days while Erick can paint the same house in 15 days. How many days will be required for the painting? Solution Let ? =number of days required to paint the house if they work together. 1?=part of the house that can be painted in 1 day by the two. 110=part of the house that can be painted in 1 day by Jack. 115=part of the house that can be painted in 1 day by Erick. 110+115=1?The required equation 3? + 2? = 30Multiply by the 𝐿?? = 30? →? 05? = 30Combine similar terms ? = 6Multiply by 15? = 6 ????Number of days required for Jack and Erick to paint the house 2.Ka Peter can plow a field in 4 days by using a tractor. Mang Pedro can plow the same field in 6 days by using a smaller tractor. How many days will be required for the plowing if they work together? Solution Let ? =number of days required to plow the field working together. 1?=the part plowed in 1 day by the two of them 14=the part plowed in 1 day by Ka Peter 16=the part plowed in 1 day by Mang Pedro 14+16=1?The required equation 12? (14+16) = 12? (1?)Multiplying by the 𝐿?? = 12? 3? + 2? = 12By distributive axiom 5? = 12Combine similar terms ? =125Multiply by 15∴ ? =125Number of days required for Ka Peter and Mang Pedro to finish plowing the field 3.If, in example 6.2, Mang Pedro worked 1day with the smaller tractor and then was joined by Ka Peter using the larger tractor, how many days were required for them to finish plowing? Solution Since Mang Pedro plowed one–sixth of the field in 1 day, five–sixths remained unplowed. Let ? =number of days required for the two to finish the job ?4=the part plowed by Ka Peter ?6=the part plowed by Mang Pedro ?4+?6=56The required equation 12? (?4+?6) = 12? (56)Multiplying by the 𝐿?? = 123? + 2? = 10By distributive axiom 5? = 10Combine similar terms ? =105Multiply by 15? = 2????Number of days required for Ka Peter and Mang Pedro to finish plowing the field 4.A tank is attached to an intake pipe that will fill it in 4 hours and an outlet pipe that will empty it in 12hours. If both pipes are left open, how long will it take to fill the empty tank? Solution Let ? =number of hours required to fill the empty tank using two pipes 1?=part of the tank filled in 1 hour if two pipes are left open 14=part of the tank filled in 1 hour using the intake pipe 112=part of the tank emptied in 1 hour using the outlet pipe 14−112=1?The required equation
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Page 8 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.ph12? (14−112) = 12? (1?)Multiplying by the 𝐿?? = 12? 3? − ? = 12By distributive axiom 2? = 12Combine similar terms ? =122Multiply by 12∴ ? = 6Number of hours required to fill an empty tank if two pipes were left open 5.A tank is attached to 2 pipes. The first pipe can fill the tank in 10 hours. But after it has been opened for 313hours, the second pipe is opened and the tank is filled up in 4 hours more. How long would it take the second pipe alone to fill the tank? The two pipes have different diameters. Solution Let ? =number of hours required for the second pipe to fill the tank alone 1?=part of the tank filled in 1 hour using the second pipe 110=part of the tank filled in 1 hour using the first pipe (313) (110) + (110+1?) 4 = 1The required equation 13+410+4?= 1By distributive axiom 10? + 12? + 120 = 30?Multiplying by the 𝐿?? = 30?– 8? = – 120Add (– 30? – 120)? = 15Multiply by – 1? = 15 ℎ????Number of hours required for the second pipe to fill the tank alone Exercise 6.1 1.Two numbers are in the ratio 4:3. Find the numbers if their sum is 84. 2.The numerator of a fraction is 3 less than the denominator. If the numerator and denominator are each increased by 1, the value of the fraction becomes 34. What is the original fraction? 3.Find three consecutive integers whose sum is 27. 4.Find two consecutive odd integers whose sum is 44. 5.One number is 6 more than a second number. The sum of the two numbers is sixteen. Find the numbers. 6.Five hundred tickets for a benefit show netted Php 3,355. Balcony tickets are sold at Php 7.50 each and Orchestra tickets at Php 2.50 each. How many tickets of each kind were sold? 7.The difference of the squares of two consecutive even integers is 92. Find the numbers. 8.The denominator of a fraction exceeds its numerator by 2. If the numerator is decreased by 1 and the denominator is increased by 3, the value of the resulting fraction is 12. Find the fraction. Exercise 6.2 1.Five years ago, Mon was twice as old as Ara. In 6 years, the sum of their ages will be 40 years. What is the present age of each? 2.Raph’s age is 710of Vance’s age. In four years, Raph’s age will be811of Vance’s age. How old is Raph now?3.I am 20 years old. My aunt was as old as I am now when I was one-fourth her age now. How old is my aunt? 4.Rosenni’s age is four–fifths of that of Angelie’s. In three years, the former will be as old as the latter isnow. How old are they? 5.Ella is 24 years old. Ella is twice as old as Marie was when Ella was as old as Marie is now. How old is Marie now? 6.A girl is one–third as old as her brother and 8 years younger than her sister. The sum of their ages is 38. How old is the girl? 7.Pauline is now 18 years old and her colleague Maui is 14 years old. How many years ago was Pauline twice as old as Maui? 8.Ramon tells his son, "I was your age now when you were born." If Ramon is now 38 years old, how old was his son 2 years ago? Exercise 6.31.Farmer A can plow a field in 4 days. Farmer B can plow the same field in 6 days. How many days will be required for the plowing if they work together? 2.A motor can pump out water from a tank in 11 hours. Another motor can pump out water from the same tank in 20 hours. How long will it take both motors to pump out the water in the tank? 3.A 400–mm pipe can fill the tank alone in 5 hours and another 600–mm pipe can fill the tank alone in 4 hours. A drain pipe 300–mm can empty the tank in 20 hours. With all the three pipes open, how long will it take to fill the tank? 4.A tank is filled with an intake pipe in 2 hours and emptied by an outlet pipe in 4 hours. If both pipes are opened, how long will it take to fill the empty tank? 5.A tank can be filled in 9 hours by one pipe, 12 hours by a second pipe, and can be drained when full by a third pipe in 15 hours. How long will it take to fill an empty tank with all pipes in operation? 6.Ram, Mon and Jay can mow the lawn in 4, 6, and 7 hours respectively. What fraction of the yard can they mow in 1 hour if they work together? 7.Crew No. 1 can finish installation of an antenna tower in 200 man–hours while Crew No. 2 can finish the same job in 300 man–hours. How long will it take both crews to finish the same job, working together? 8.Carl can paint a house in 9 hours while Louie can paint the same house in 16 hours. They work together for 4 hours. After 4 hours, Louie left and Carl finished the job alone. How many more hours did it take Carl to finish the job? 9.A pipe can fill the tank in 10 hours. A second pipe can fill the tank in 40 hours. If both pipes are left open, determine the time to fill the tank. 10.Aldrin can finish a job in 4 days. Bart can finish the same job in 6 days. If Aldrin and Bart plus Carding can finish the job in 2 days, how long will it take for Carding to finish the job alone?
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Page 9 Prepared by: Prof. Ninfa Sua –Sotomil Cell No: 09171022328 Tel No: 5035955 sotomil_ninfas@wit.edu.phCourse Outline No. of Hours Part I: Algebra 1.Polynomials: Four Fundamental Operations 6 hours 2.Special Products 3 hours 3.Factoring 6 hours 4.Fractions 3 hours 5.Exponents and Radicals 3 hours 6.Linear Equations 4 hours (Midterm Examination) 7.Systems of Linear Equations 3 hours 8.Quadratic Equations 3 hours 9.Logarithmic and Exponential Functions 3 hours Part II: Trigonometry 10.Circular & Trigonometric Functions 4 hours 11.Solution of Right Triangles & Applications 3 hours 12.Solution of Oblique Triangles 3 hours (Pre –Final Examination) 13.Trigonometric Identities 4 hours (Final Examination)
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