MATH 2 part 1

3x² - 3x + 2x + 2 = 4(x² - 1) combine similar terms 3x² - 3x + 2x + 2 = 4x² - 43x² - 3x + 2x + 2 – 4x² + 4 = 0 multiply both sides by –1 factor -x² - x + 6 = 0 by zero product property x² + x – 6 = 0 (x + 3)(x-2) = 0 x + 3 = 0 and x – 2 = 0 x = -3 and x = 2Check your answers when dealing with rational equations.Check: When x = 2 3x + 2 = 4 x +1 x −1 3(2) + 2 = 4 2+1 2−1 6+2=4 31 2+2=4 4= 4 When x = -3 3x + 2 = 4 x +1 x −1 3(−3) + 2 = 4 −3+1 −3−1 −9 + 2 =4 −2 −4 18 − 2 = 4 4 16 = 4 4 4 =4In this example both solutions check. 10

Resulting answers may not actually be solutions of the given equation.Solutions that do not satisfy the original equation are called EXTRANEOUS andthey must be discarded.Example 2. Solve: 2 = 1 − 6x 3x +1 x 3x +1 Solution: The LCD is x(3x + 1) with the restriction that x should not be equal to – 1 and 0. Thus, 3 x(3x + 1)  2 = 1 − 6x  multiply by the LCD  3x +1 x 3x +1 x(3x + 1)  2  = x(3x + 1) 1  − x(3x + 1) 6x   3x +1  x   3x +1 2x = (3x + 1) – 6x(x) by distributive property 2x = 3x + 1 – 6x²Since the equation is quadratic, equate the right-hand side to 0. 6x² - 3x – 1 + 2x = 0 6x² - x – 1 = 0 combine similar terms factor (3x + 1)(2x – 1) = 0 by zero product property 3x + 1 = 0 or 2x – 1 = 0 x = - 1 or x= 1 3 2Check: If x = 1 2 = 1 − 6x 2 3x +1 x 3x +1 6 1  2 = 1 − 2 3 1  + 1 1 3 1  + 1 2 2 2 2 =2− 3 3 +1 3 +1 22 2 =2− 3 3+2 3+2 22 22 11

2 =2− 3 55 22 2( 2 ) = 2 – 3( 2 ) 55 4 =2– 6 55 4 = 10 − 6 55 4=4 55 If x = - 1  2 = 1 − 6x  3  3x +1 x 3x +1  6 − 1    2 1  3   1  + 1 = −1 − 3 −   3 3 1   − 3  + 1  3  2 = −3 − − 2 −1+1 −1+1 2 = −3 + 2 00STOP! Division by 0 is undefined. - 1 does not solve the equation. Notice 3that at the beginning of this example, - 1 is already a restriction so that 3you do not need to check this value, because it is an extraneous solution.Therefore, the only solution is 1 . 2B. Radical Equations that Lead to Quadratic EquationsExample 3. Solve for x: x = 6 –xSolution: ( x )² = (6 – x)² square both sides and expand write in standard form x = 36 – 12x + x² x² - 13x + 36 = 0 factor (x - 4)(x – 9) = 0 by the zero product property x - 4 = 0 or x – 9 = 0 x = 4 or x = 9 12

Check: If x = 4 x =6–x 4 = 6 –4 2=2 If x = 9 x =6–x 9 =6–9 3 = -3 9 is an extraneous solution. Therefore, 4 is the only solution. Note that solving equations involving rational expressions, you need tocheck the solutions obtained. This is because the solution that is obtained maybe true to the raised equation but not to the original equation.Example 4. Solve for x: 4x = 6x + 1Solution: Square both sides of the equation to get rid of the radical sign, thus: 4x = 6x + 1 (4x)² = ( 6x + 1 )² 16x² = 6x +1 simplify write in standard form 16x² - 6x –1 = 0 factoring by zero product property (8x + 1)(2x –1) = 0 8x + 1 = 0 or 2x – 1 = 0 x = -1 or x = 1 8 2Check: If x = - 1 4x = 6x + 1 8 4(- 1 ) = 6 1  + 1 8 8 - 1 = −6 +1 28 13

-1 = 1 24 -1 ≠ 1 22 - 1 is an extraneous solution. 8 If x = 1 4x = 6x + 1 2 4( 1 ) = 6 1  + 1 2 2 2 = 3+1 2= 4 2=2 The solution or root is 2.C. Solving Equations that are Quadratic in Form When an equation can be written in the form au² + bu + c = 0 with u as analgebraic expression, it is said to be quadratic in form and is solved by any of themethods you have already learned.Example 5. Solve for x: (x + 3)² + 5 (x + 3) + 6 = 0Solution: Because of the repeated quantity (x + 3), this equation is quadratic in form with u = x + 3. Write (x + 3)² + 5 (x + 3) + 6 = 0 as u² + 5u + 6 = 0 let u = x + 3 (u + 3)(u + 2) = 0 factoring u + 3 = 0 or u + 2 = 0 zero product property u = -3 or u = -2 Note that the equation is not yet solved since x is not yet known. Tofind x, substitute x + 3 for u: x+3 = -3 or x + 3 = -2 x = -6 or x = -5 14

Check: If x = -6 (x + 3)² + 5(x + 3) + 6 = 0 (-6 + 3)² + 5(-6 + 3) + 6 = 0 (-3)² + 5(-3) + 6 = 0 9 – 15 + 6 = 0 0=0 If x = -5 (x + 3)² + 5(x + 3) + 6 = 0 (-5 + 3)² + 5(-5 + 3) + 6 = 0 (-2)² + 5(-2) + 6 = 0 4 – 10 + 6 = 0 0=0 The solutions check.Example 6. Solve for y: y4 – 8y² + 16 = 0Solution: Since (y²)² = y4 the equation can be written as (y²)² - 8y² + 16 = 0 So that u = y² and thus, u² - 8u + 16 = 0 let u = y² (u – 4)² = 0since u² - 8u + 16 is a perfect square u–4=0 by the square root property u=4To solve for y, substitute y² for u: y² = 4 by the square root property y=± 4 y = ±2Check: If y = 2 y4 – 8y² + 16 = 0 (2)4 – 8(2)² + 16 = 0 15

16 – 8(4) + 16 = 0 16 – 32 + 16 = 0 0=0If y = -2 y4 – 8y² + 16 = 0 (-2)4 – 8(-2)² + 16 = 0 16 – 8(4) + 16 = 0 16 – 32 + 16 = 0 0 =0The solutions check.Try this outA. Solve the equations by first clearing the equation of fractions. (See examples 1 and 2.) 1. w = 3 2 w+2 2. y = y + 2 y +1 3y 3. z + 3 = z 2z 4. t = 3 6 t+4 5. 1 = x + 2 x +1 x +3 6. x = 4 x +1 x + 4 7. x + 2 = 5 x x−4 8. x + 2 = x + 3 x+4 x+6 9. 3x + 2 = 4 x +1 x −1 16

10. 14 = x − 5 x B. Find all solutions by first squaring. (See examples 3 and 4.) 11. 2x = 11x + 3 12. 4x = 8x + 3 13. –2 p = 8 – p 14. k – 12 = - k 15. r = 20 −19r 6 C. Find all solutions. (See examples 5 and 6.) 16. x4 – 18x² +81 = 0 17. x4 – 10x² + 9 = 0 18. (t + 5)² - 7(t +5) + 6 = 0 19. 3(m + 4)² - 2(m + 4) – 8 = 0 20. 4x4 + 5x² + 1 = 0 Lesson 3 Solving Verbal Problems Involving Quadratic Equations Since you have the ability to solve quadratic equations, you can nowproceed in solving verbal problems. Remember that when you are solving averbal problem, you do not know whether the resulting equation will be linear orquadratic. Be sure to look carefully at the equation before deciding on themethod to use.Example 1. The sum of a number and its reciprocal is 29 . Find the number. 10 Solution: Translating the words in the problem: Let x = the number, then 1 = its reciprocal x 17

The sum of a number and its reciprocal is 29 10 x+ 1 = 29 x 10Thus, the equation is x + 1 = 29 x 1010x (x + 1 ) = 10x ( 29 ) multiply both sides by the LCD x 10 by distributive property10x (x) + 10x ( 1 ) = 10x ( 29 ) write equation in standard form x 10 the factoring method works10x² + 10x ( 1 ) = 10x ( 29 ) x 1010x² + 10 = 29x10x² - 29x + 10 = 0(5x – 2)(2x – 5) = 05x –2 = 0 or 2x – 5 = 0x = 2 or x= 5 5 2Notice, that the two solutions are reciprocals of each other.Check that both solutions satisfy the statement of the problem.Example 2. The length of a rectangle is three less than twice the width and its area is equal to 44 square centimeters. Find the dimensions of the rectangleSolution: Draw a diagram of the Rectangle and label the sides according to the information in the problem x 2x – 3 18

The area of the rectangle is equal to its length times its width.Therefore the equation is:x (2x – 3) = 44 multiply2x² - 3x – 44 = 0 factor(2x – 11)(x + 4) = 02x – 11 = 0 or x + 4 = 0x = 11 or x = -4 2 Since it makes no sense for the width of a rectangle to have anegative dimension, reject the negative answer.Thus, the answer to the problem is that the Width = 11 cm. 2and the length = 2( 11 ) – 3 = 11 – 3 = 8 2Length = 8 cm.Check to see that the area is equal to 44 square centimeters.11 · 8 = 11 · 4 = 442Example 3. Erica and Pauline are both driving to a tennis tournament 200 km away. They both leave at the same time, but Erica arrives 1 hour ahead of Pauline because she was driving 10 km per hour faster than Pauline. Find the rate at which each drove to the tournament.Solution: This is a motion problem that requires the formula, d=rt, where d is the distance, r is the rate and t is the time. How is it used here? The problem asks you to find out the rates at which they drove. Let r = Pauline’s rate Then r + 10 = Erica’s rate (she was 10 kph faster) The problem also tells you that Erica arrived 1 hour ahead of Pauline, which means that (since they left at the same time). Erica’s driving was 1 hour less. Thus we have the following “time relationship”; t Erica = t Pauline –1 19

Use the fact that d = rt in the equivalent form of t = d , by rsubstituting into the time relationship: d Erica = d Pauline - 1 r Erica r Pauline However, you know that both girls traveled the same distance,which is 200 km. Having represented the rates, substitute these quantitiesin the last equation. 200 = 200 −1 multiply both sides of ther + 10 r equation by the LCD: r(r + 10).(r)(r + 10) 200  = (r)(r + 10) 200  −1(r)(r + 10 r + 10  r200r = 200(r + 10) – r(r + 10)200r = 200r + 2000 – r² - 10rr² + 10r – 2000 = 0(r + 50)(r – 40) = 0r + 50 = 0 or r – 40 = 0r = -50 or r = 40 Since it makes no sense for rate to be negative, the answer to theproblem is that r = 40Thus, we have Pauline’s rate is 40 kph and Erica’s rate is 50 kph.The checking is left to you. Sometimes a verbal problem involving quadratic equations involvesnothing more than substituting values into a given formula, as you will see in thenext example. Problems like this may seem worth discussing so that you willknow what to do and how to do them when you encounter them in the future.Example 5 The profit in Pesos (P) on each TV set made daily by the Curray television company is related to the number of TV sets produced (x) at the Curray Factory according to the equation 20

P= - 5x2 = 450x – 16,250 2 What is the company’s profit on each TV set if it produces sixty (60) TV sets per day? Solution: This example requires you to find P when x = 60. Therefore, simply substitute x = 60 into the given equation and compute for P. P = - 5x2 + 450x – 16,250 2 P = - 5(60)2 + 450(60) – 16,250 2 2 P = - 5(3600) + 27,000 – 16,250 2 P = -9000 + 27,000 – 16,250 P = P1750 The profit is P1750 per TV set.Try this outSolve each problem algebraically. 1. The sum of the number and its reciprocal is 13/6. Find the number. 2. The sum of the number and three times its reciprocal is 79/10. Find the number. 3. The sum of two numbers is 20, and their product is 96. find the two numbers. 4. One number is five more than three times a second number. If their product is –2, what are the numbers? 5. The length of a rectangle is three more than twice its width, and its area is 90 square meters. Find its dimensions. 6. The width of a rectangle in one - third its length. If the area of the rectangle is 20 square inches, what are the dimensions of the rectangle? 21

7. The denominator of a fraction is one more than the numerator. If the numerator is increased by three, the resulting fraction is one more than the original fractional. Find the original fraction.8. The numerator of a fraction is one less than the denominator. If 7/12 is added to the fraction the result is the reciprocal of the original fraction. Find the original fraction.9. A motorist completes a trip covering 150 kilometers in 2 hours. She covers the first 120 kilometers at a certain rate of speed, and then decreases her speed by 20 kilometers per hour for the remaining 30 kilometers. Find her speed for the first 120 kilometers.10. Arnold travels from town A to town B, which are 300 kilometers apart. His rate going is twice as fast as his rate returning. If his total traveling time was 7½ hours, what was his rate of speed going from A to B?11. Suppose the profit (P) made on the sale of theater tickets is related to the price (x) of the tickets according to the following equation: P = 1000(-x² + 15x –35) How much profit is earned if the price per ticket is P5.00? P4.00? Let’s Summarize 1. Quadratic equations may be solved by any of the following methods: Factoring Method Square Root Method Completing the Square Method Quadratic Formula 2. The factoring method is easiest specifically when the left-hand side of the equation ax² + bx + c = 0 is factorable. It is also used when the value of c = 0. 3. The square root method is most convenient to use when b = 0 or when the equation is written in the form a(x – p)² + k = 0. 4. The quadratic formula can be used to any type of quadratic equation although it is a little bit messy. 5. Some equations that can be transformed into quadratic equations are: 22

a. equations involving rational expressions. Equations of this type may be solved by first clearing the fractions by multiplying both sides of the equation by the least common denominator. b. radical equations. Equations of this type may be solved by first squaring both sides of the equation. c. equations quadratic in form. Equations of this type may be solved by first taking a suitable substitution. It is best to always check the solutions of these equations. There may be extraneous solutions that may arise while solving them. What Have You LearnedA. Solve each of the following equations. Choose any method you like. 1. (x + 5)(x – 2) = 18 2. (x + 5)² = 10 3. x – 8 = x – 4 2x 4. (x² - 3)² - 2 = (x² - 3)B. Solve the problem. 5. The length of a rectangle is 7 more than twice its width. If the area of the rectangle is 30 square cm, find its dimensions. 6. Arnold travels from town A to town B, which are 300 km apart. His rate going is twice as fast as his rate returning. If his total traveling time was 7 ½ hours, what was his rate of speed going from A to B? 23

Answer KeyHow much do you know 1. 6, 1 2. 6 ± 13 3. 0, -3 4. 2, 3 2 5. 1, -3 6. L = 12 units, w = 8 unitsTry this outLesson 1A. 1. -1, -5 2. –3 ± 14 3. 1 , 1 2 4. 5, -1 5. 2, -9 6. 1 ± 5 2 7. no real solution 8. 1 ± 5 9. 7 ± 3 5 2 10. ±4 11. 0, 9 12. 9, -5 13. − 3 ± 13 2 14. –1, -3 15. –2 ± 7 16. −1 ± 2 2 17. 0,4 18. 0 19. 0 20. no real solution 24

21. 13 ± 133 2 22. 3 ± 7 23. 4 24. 0, - 7 5 25. –8, 5 26. no real solution 27. 3 28. 6, - 2 3 29. –4 30. – 3 , 4 5B. The factoring method is of course the easiest.Lesson 2 1. -1 - 7 2. – 1 , 2 2 3. ± 6 4. –2 ± 22 5. -1 ± 2 6. ±2 7. 8, -1 8. no solution 9. –3, 2 10. 7, -2 11. 3 12. 3 4 13. 16 14. 9 15. 5 6 16. ±3 17. ±3, ±1 18. 1, -4 19. – 16 , -2 3 20. no real solution 25

Lesson 3 1. 2 , 3 32 2. 15 , 2 35 3. 12, 8 4. 2 ,−1 3 5. − 15 ,6 3 6. l = 2 15, w = 2 15 3 7. 2 3 8. 3 ,4 7 9. 8 km/hr 10. P1 = 15,000 P2 = 9000What have you learned 1. –7, 4 2. –5 ± 10 3. 4 4. ± 5 , ± 2 5. w = 2.5 L 6. 120 km/hr 26

Module 3Searching for Patterns in Sequences, Arithmetic , Geometric and others What this module is about This module is about finding the sum of the n terms of an arithmetic sequence.As you go over the exercises, you will develop skills in deriving formula for thearithmetic series , sum of the n terms of an arithmetic sequence and Fibonacci series.You will gain skills in solving problems involving arithmetic series, sum of arithmeticsequences, harmonic and Fibonacci series. You will also recall some concepts onfinding the common difference and the nth term of an arithmetic sequence. Treat thelessons with fun and take time to go back and review if you think you are at a loss. What you are expected to learnThis module is designed for you to: 1. Recall finding the common difference and nth term of an arithmetic sequence 2. Derive the formula for the arithmetic series 3. Solve problems involving arithmetic series, sum of arithmetic sequences and Fibonacci seriesHow much do you know1. Give the first four terms of the arithmetic sequence for which the first term is 8 and the common difference is 5?a. 8, 15, 18, 25 c. 8, 12, 16, 20b. 8, 13, 19, 24 d. 8, 13, 18, 232. Give the arithmetic sequence whose 7th term is 49 and whose 12th term is 84?a. 5, 10, 15, 20, … c. 7, 14, 21, 28, …b. 6, 12, 18, 24, … d. 8, 16, 24, 32, …

3. Find the sum of the first twenty terms in arithmetic sequence 0,1, 2, 3, …a. 190 c. 170b. 180 d. 1604. Find the sum of the first five positive even numbers.5. Find the sum of all the odd integers from 100 to 200, inclusive.a. 7,000 c. 8,000b. 7,500 d. 8,5006. How many terms are there in #57. How many numbers between 150 and 250 are exactly divisible by 12?a. 10 c. 7b. 8 d. 68. Find their sum.9. Find the sum of the five terms of the harmonic sequence 1, _1_, _1_, _1_, … 6 11 1610. Six student players are participating in a chess game. If each of them plays once with each of the others, how many games will be played in all? 2

What you will do Lesson 1 Recalling Arithmetic SequenceA sequence where each succeeding term is obtained by adding a fixednumber is called an arithmetic sequence. This fixed number is thecommon difference d between any two succeeding terms.Any arithmetic sequence is defined by the equation given as tn = t1 + (n –1)dTry this outA. Find the first three terms of the sequence defined by each equation. 1. tn = 2n – 7 2. tn = (-4)n 3. tn n + 6 4 4. tn = n(n + 12) 5. tn = 4n(n)B. Find the seventh term in each sequence.1. 8, 15, 22, 29, … 6. 19, 17, 15, 13 …2. 100, 200, 300, … 7. -8, -3, 2, 7, …3. -7, -2, 3 , … 8. 90, 98, 106, 114, …4. 25, 75, 125, 175, … 9. 75, 65, 55, 45, …5. 87, 97, 107, 117, … 10. 21, 14, 7, 0, … Lesson 2 Deriving the Formula of Arithmetic Series Eight basketball teams are participating in summer sportsfest. If each of the team plays once with each of the others, how many games will be played in all? Let’s simplify the problem by breaking it down into smaller problems. Start withone team, then with two teams, then with three teams, and so on. Each time, determine 3

the number of games added to the previous number of games when there is one teamless. Also, determine each time the total number of games played for a given numberof teams. From these data we could be able to get the total number of games playedgiven any number of teams.Let’s make a diagram to illustrate the problem.Let Tn denote the team, n = 1, 2, 3, 4, 5, 6, 7, 8Let an arc joining two teams denote a game played between these team.T1 T2 T1 T2 T3 T1 T2 T3 T41 game 2 games 3 games1 game added 2 games added 3 games addedSummarize your observation using a table.Number of Teams Number of Additional Total Number of Games Games 1 0 0 2 1 0+1=1 3 2 0+1+2=3 4 3 0+1+2+3=6 5 4 0 + 1 + 2 + 3 + 4 = 10 6 5 0 + 1 + 2 + 3 + 4 + 5 = 15 7 6 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21 8 7 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28Observe that the 2nd column of a number form an arithmetic sequence. If there are eight teams, then the total number of games played will be equal tothe sum of the first eight terms of arithmetic sequence 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28Naming the sum of the first n terms of the sequence by Sn, Sn = 0 + 1 + 2 + 3 + 4 + . . . + (n – 1)The indicated sum 0 + 1 + 2 + 3 + 4 + . . . + (n – 1) is an arithmetic series Sn . 4

We can express the sum of the eight games as Sn = t1 + t2+ t3 + t4 + t5 + t6 + t7 + t8The total number of games played by eight teams will be Sn = 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 For smaller number n and with the aid of a calculator, computing Sn will not be aproblem. For 20 teams of big basketball league, however, we need to look for a moreefficient way.Sn = 0 + 1 + 2 + 3 + . . . + (n – 4) + (n – 3) + (n – 2) + (n – 1) Equation 1Sn = (n – 1) + (n – 2) + (n – 3) + (n – 4)+ … + 3 + 2 + 1 + 0 Equation 2, the reversed order of Equation 1Adding Equation 1 and 2 gives2Sn = [0 + (n – 1)] + [1 + (n – 2)] + [2 + (n – 3)] + [3 + (n – 4)] + … + [(n – 4) + 3] + [(n – 3) + 2] + [(n – 2) + 1] + [(n – 1) + 0]2Sn = (n – 1) + (n – 1) + (n – 1) + (n – 1) + … + (n – 1) + (n – 1) + (n – 1) + (n – 1)Note that (n – 1) is taken as addend n times2Sn = n(n – 1) Sn n(n – 1) 2For 20 teams, the total number of games played will beSn 20(20 – 1) 10(19) = 190 games 2 Arithmetic series is an indicated sum of the first n terms of an arithmeticsequence. The sum of n terms is denoted by Sn.The formula in finding arithmetic series is Sn n(n – 1) 2 5

Lesson 3 Solving Problems Involving Arithmetic Series and Sum of Arithmetic SequenceGauss, Carl Friedrich (1777-1855), a German mathematician, is noted for his wide-ranging contributions to the field of mathematics and physics, particularly the study ofelectromagnetism. Do you know that this formula is the trick he used, as a schoolboy, to solve theproblem of summing up the integers from 1 to 100 given as busy-work by his teacher?While his classmates toiled away doing the addition longhand, Gauss wrote a singlenumber, the correct answerUsually problems present themselves in either of two ways. Either the first number inthe sequence and the number of terms are known or the first number and the lastnumber of the sequence are known.Examples:1. Find the sum of the first 30 terms of the arithmetic sequence -15, -13, -11,…Since t1 = -15, d = 2, and n = 30 then we can use the formula Sn n {t1 + [t1 + d(n – 1)]} 2S30 _30_ {-15 + [ -15 + 2(30 – 1)]} 2 = 15 [ -30 + 2(29) ] = 15 [ -30 + 58 ] = 15 (28)S30 = 420 6

2. Find the sum of the first 50 terms of arithmetic sequence 26, 31, 36, 41, …We have t1 = 26, d = 5, and n = 50 so we can use the formula Sn n {t1 + [t1 + d(n –1)]} 2S50 _50_ {26 + [ 26 + 5(50 – 1)]} 2 = 25 [ 52 + 5(49) ] = 25 [ 52 + 245 ] = 25 (297)S50 = 7,425We can also use the formula Sn n (t1 + tn) if we have tn. 2t50 = 26 + (50 -1)5 = 26 + (49)(5)t50 = 271S50 _50_ ( 26 + 271) = 25(297) = 7,425 23. Find the sum of the first 42 terms of arithmetic sequence 5, 8, 11, 14,17, ... ,231t1 = 5 and t42 = 231S42 _42_ ( 5 + 231) = 21(236) = 4,956 24. What is the sum of all positive integers less than 300 that are multiples of 7? The multiples of 7 form an arithmetic sequence such that the first term is 7, andd = 7. To get the sum, lets first find the number of multiples of 7 that are less than 300. 300 is not a multiple of 7. Try 299 ÷ 7 = 42.71, 299 is not a multiple of 7. Tryanother number297 ÷ 7 = 42.43 … until you reach 294 ÷ 7 = 42So, 294 = tn294 = 7 + (n – 1)7n = 42Therefore, S42 _42_ (7 + 294) = 21 (301) = 6,321 2Another solution to get S42 is as follows: 7

Although guess and check is one of the strategies in solving mathematicsproblems, you can have easier way of solving for the largest multiple of 7 less than 300. The largest multiple of 7 less than 200 is tn . Write tn< 300. Then tn = 7 + (n – 1)7 7 + (n – 1)7 < 300 7 + 7n – 7 < 300 7n < 300 n < 42.85 Since n is an integer, n = 42. Therefore there are 42 positive integers that areless than 300 which are multiples of 7. The largest of these multiples is t42 andt42 = 7 + (42 – 1)7 = 7 + (41)7 = 7 + 287 = 294The series formed is 7, 14, 21, 28, … 294 and the sum is7+14 + 21+ 28 + … + 294Using the formula, we haveS42 _42_ (7 + 294) = 21 (301) = 6,321 25. A military unit purchases 10 spare parts during the first month of a contract,15 spare parts in the second month, 20 spare parts in the third month, 25 spare parts inthe fourth month, and so on. The acquisition officer wants to know the total number ofspare parts the unit will have acquired after 50 months.Note that the sequence is 10, 15, 20, 25,…Since the given are d = 5, t1 = 10 and n = 50, use the formula Sn n {t1 + [t1 + d(n – 1)]} 2S50 50 [ 10 + 10 + 5(50 – 1)] = 25 [ 20 + 5(49)] = 25 (265) = 6,625 2After 50 months the military unit will acquire 6,625 spare parts 8

Try this outA. Find the sum of the terms in the arithmetic sequence for the number of termsindicated.1. 4 + 1 + -2 + -5 + … 40 terms2. 6 +12 +18 +24 + … 15 terms3. 10 + 7 + 4 + 1 + … 35 terms4. 13 + 12 +11 + … 50 terms5. 2 + 9 +16 + 23+ … 25 termsB. Find the term asked by using the given values.1. tn = 45, t1 = 27, d = 9, n = ____2. tn = 79, t1 = 7, d = 3, Sn = ____3. t1 _-3_, d _3_, n = 8, S8 = ____ 444. d = -5, t7 = -11, n = 27, Sn = ____5. t10 = 88, t1 = -8, S10 = ____B. Solve the following problems.1. Find the sum of the first 150 counting numbers.2. Find the sum of the first 50 odd natural numbers.3. Find the sum of all the even integers from 12 to 864, inclusive.4. Find the sum of all the odd integers from 27 to 495, inclusive.5. How many numbers between 10 and 200 are exactly divisible by 7? Find their sum.6. How many numbers between 25 and 400 are exactly divisible by 11? Find their sum.7. Find the sum of all positive integers between 29 and 210 that are divisible by 4?8. If a clock strikes the appropriate number of times on each hour, how many times will it strike in one day? In one week?9. A group of hikers has a trek of 6 days to reach Mt. Apo. They traveled 15 km on the 1st day, 13 km on the 2nd day, 11 on the 3rd day, and so on. How many kilometers did they travel to reach Mt. Apo? 9

10. Luis applied for scholarship and was given battery of test. He made a score of 68 on his first test. The passing average score is 75. Would he make it after four test if he did 6 points better on each succeeding test? What was his score on the fourth test? What was his average score after four tests? Lesson 4 Fibonacci SeriesLet’s recall Fibonacci sequence using this problem. Imagine a pair of a male and female rabbits whiich produces a new pair at the age of two months and every month thereafter. Each new pair, in turn, produces a pair at the same rate.How many pairs of rabbits are there in 8 months? Use drawings and tables to analyze this problem. Consider the rabbit populationat the beginning of 6 months. How many rabbits in all after 6 months? After 8 months?Beginning of Growth in pairs of rabbits month 123456 10

Summarize the result using a tableMonth 1 2345678Rabbit 1 1 2 3 5 8 13 21Pairs (1+1) (1+2) (2+3) (3+5) (5+8) (8+13) Recall that in a Fibonnaci Sequence the numbers are arranged as 1, 1, 2, 3, 5, 8,13, … where the first term is 1. Generalizing the relationship among the terms of any Fibonacci sequence leadsto a formula for the Fibonacci series.The sum of the first n terms of a Fibonacci sequence is called a Fibonacciseriest1 = t3 – t2 1= 2–1t2 = t4 – t3 1= 3–2t3= t5 – t4 2= 5–3⋅⋅⋅tn – 1 = t n + 1 – tn tn = t n + 2 – tn + 1Then, t1 + t2 + t3 + t4 + … + tn – 1 + tn = - t2 + tn + 2From the above equation, it can be deduced that the Fibonacci series t1 + t2 + t3 + t4 +… + tn with sum Sn can be written as Sn = - t2 + tn + 2 In the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55…, find the sum of thefirst three terms using the formula.In the example, with t2 = 1, S3 = -1 + (3 + 2) = -1 + 5 = 4How about the sum of the first 5 terms?S5 = -1 + t5 + 2 = -1 + t7 = -1 + 13 = 12Let’s verify 1 + 1 + 2 + 3 + 5 = 12Let’s go back to the problem and answer. 11

How many rabbits in all after 6 months? After 8 months?S6 = -1 + t6 + 2 = -1 + t8 = -1 + 21 = 20S8 = -1 + t8 + 2 = -1 + t10 = -1 + 55 = 54Try this outConsider the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55… 1. Find the sum of the first 10 terms. 2. Find the sum of the first 12 terms. 3. Find the sum of the first 14 terms. Let’s SummarizeArithmetic sequence is a sequence where each succeeding term is obtained byadding a fixed number.Common difference is the fixed number between any two succeeding terms.Any arithmetic sequence is defined by the equation given as tn = t1 + (n –1)dArithmetic series is an indicated sum of the first n terms of an arithmetic sequence.The sum of n terms is denoted by Sn.The formula in finding arithmetic series is Sn n(n – 1) 2The formula in finding the sum of terms in an arithmetic sequenceSn n [t1 + d(n – 1)] or Sn = n {t1 + [t1 + d(n – 1)]} 2and Sn n (t1 + tn ) 2The sum of the first n terms of a Fibonacci sequence is called a Fibonacci series Sn = - t2 + tn + 2 12

What have you learned1. Find the sum of the first thirty terms in arithmetic sequence 0,1, 2, 3, …a. 425 c. 435b. 430 d. 4402. Find the sum of the first ten positive odd numbers.3. Find the sum of all the even integers from 150 to 300, inclusive.a. 17,000 c. 18,000b. 17,100 d. 18,1004. How many terms are there in #3?5. How many numbers between 200 and 400 are exactly divisible by 15?a. 10 c. 12b. 11 d. 136. Find the sum of the numbers in #5.7. Find the sum of the five terms of the harmonic sequence 1, _1_, _1_, _1_, … 8 15 228. What type of sequence is _1_, _1_, 1 , _3_, _5_, …? Fibonacci or harmonic? 22 229. Find the 8th term of # 8.10. Find the sum of the first 8th terms of #8. 13

Key to correctionHow much do you know 1. d 6. 50 2. c 7. b 3. a 8. 1,584 4. 30 9. 1685 5. b 1232Try this out 10. 15Lesson 1 t3 = 2(3) – 7 = -1A. 1. tn = 2n – 7 t1 = 2(1) – 7 = -5; t2 = 2(2) – 7 = -3 ; -5, -3, -12. tn = (-4)n ; t2 = (-4)2 =16; t3 = (-4)3 = -64 t1 = (-4)1 = -4 -4, 16, -643. tn n + 6 4_7_ , _8_ or 2, _9_44 44. tn = n(n + 12) 13, 28, 455. tn = 4n(n) 4, 16, 36B. Find the seventh terms in each sequence. 1. By listing we have 8, 15, 22, 29, 36, 43, 50 By formula we have t7 = 8 + (7 – 1) 7 = 8 + 42 = 502. 100, 200, 300, … t7 = 100 + (7 – 1) 100 = 7003. -7, -2, 3 , … t7 = -7 + (7 – 1) 5 = 23 14

4. 25, 75, 125, 175, … t7 = 25 + (7 – 1) 50 = 325 5. 87, 97, 107, 117, … t7 = 87 + (7 – 1) 10 = 147 6. 19, 17, 15, 13 … t7 = 19 + (7 – 1) (-2) = 7 7. -8, -3, 2, 7, … t7 = -8 + (7 – 1) 5 = 22 8. 90, 98, 106, 114, … t7 = 90 + (7 – 1) 8 = 138 9. 75, 65, 55, 45, … t7 = 75 + (7 – 1)(-10) = 15 10. 21, 14, 7, 0, … t7 = 21 + (7 – 1)(-7) = -21Lesson 3A. 1. d = -3, t1 = 4, n = 40 t40 = 4 + (40 – 1)(-3) = -113 S40 = 20[4 +(-113)] = -2,180 2. d = 6, t1 = 6, n = 15 t15 = 6 + (15 – 1)6 = 90 S15 = 7.5 (6 + 90) = 720 3. d = 3, t1 = 10, n = 35 t35 = 10 + (35 – 1)(-3) = -92 S35 = 17.5 [10 + (-92)] = -1,435 4. d = -1, t1 = 13, n = 50 t50 = 13 + (50 – 1)(-1) = -36 S50 = 25 [13 + (-36)] = -575 5. d = 7, t1 = 2, n = 25 t25 = 2 + (25 – 1)(7) = 170 S25 = 12.5 (2 + 170) = 2,150 15

B. 1. 45 = 27 + (n – 1)9 45 = 27 + 9n – 9 n=32. 79 = 7 + (n – 1)3 79 = 7 + 3n – 3 n = 25 Sn = 12.5(7 + 79) = 1,0753. t8 _-3_ + (8 – 1)_3_ -3 + 21 18 or _9_ 4 4 4 42Sn 4 ( _-3_ + _18_) = 4( _15_ ) = 1544 44. t7 = t1 + (7 – 1)(-5) -11 = t1 + (7 – 1)(-5) t1= 19t27 = 19 + (27 – 1)(-5) = -111S27= 13.5 [ 19 + (-111)] = -1,2425. S10= 5 ( -8 + 88) = 400C. 1. d = 1, t1 = 1, n = 150 t150 = 1 + (150 – 1)(1) = 150S150 = 75 (1 + 150) = 11,3252. d = 2, t1 = 1, n = 50 t50 = 1 + (50 – 1)(2) = 99S150 = 25 (1 + 99) = 2,5003. t1 = 12, tn = 864, d = 2 864 = 12 + (n – 1)(2) = 427S427 = 213.5 (12 + 864) = 187,0264. t1 = 27, tn = 495, d = 2 495=27 + (n – 1)(2) = 235S235 = 117.5 (27 + 495) = 61,335 16

5. t1 = 14, tn = 196, d = 7196 = 14+ (n – 1)(7) = 27orfind n by 200 ÷ 7 = 28, then subtract 1 (since 7 is not included in the sequence,the first positive number divisible by 7) n = 28 – 1 = 27S27 = 13.5 (14 + 196) = 2,8356. t1 = 33, tn = 396, d = 11396 = 33 + (n – 1)(11) = 34 orfind n by 400 ÷ 11 = 36.36, then subtract 2 (since 117 and 22 are not included inthe sequence) n =36 – 2 = 34S34 = 17 (33 + 396) = 7,2937. t1 = 32, tn = 208, d = 4208 = 32 + (n – 1)(4) = 45 orfind n by 210 ÷ 4 = 52.5, then subtract 7 (since 4, 8, 12, …28 are not included inthe sequence) n = 52 – 7 = 45S45 = 22.5 (32 + 208) = 5,4008. The clock strikes 1, 2, 3, … 12 for 12 hours, then repeat for the next 12 hours to complete one day.S12 = 6 (1 + 12) = 78 strikes in 12 hours78 x 2 = 156 strikes in one day156 x 7 = 1,092 strikes in one week9. t1 = 15, n = 6, d = -2 t6 = 15 + (6 – 1)(-2) = 5 S6 = 3 (15 + 5) = 3(20) = 60 kilometers10. t4 = 68 + (4 – 1)(6) = 86 S4 = 2 (68 + 86) = 2(154) = 308 sum of all the scores 308 ÷ 4 = 77 average scoreLuis passed the battery of testLesson 4Consider the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610,987…1. Find the sum of the first 10 terms. S10 = - 1 + t10 + 2 = -1 + t12 = -1 + 144 = 143 17

2. Find the sum of the first 12 terms. S12 = - 1 + t12 + 2 = -1 + t14 = -1 + 377 = 376 3. Find the sum of the first 14 terms. S14 = - 1 + t14 + 2 = -1 + t16 = -1 + 987 = 986What have you learned 1. c 2. 100 3. b 4. 76 5. d 6. 3,900 7. 1.27 8. Fibonacci 9. 21 2 10. 27 18

Module 3 Systems of Linear Equations and Inequalities What this module is about This module is about using systems of linear equations to solve problems,(e.g. number relations, uniform motion, geometric relations, mixture, investment).As you go over the exercises, you will find the importance of systems of linearequations in solving real situation problems. What you are expected to learn This module is designed for you to: 1. solve problems by translating them to systems of linear equations in two variables. 2. find the solution to the problems using different methods in solving systems of linear equations. How much do you know Express the following verbal statement into algebraic equation. 1. If a number (x) is subtracted from 18, the result is three less than twice the number. 2. A car and a truck travel towards each other at points 400 km apart and meet in 2 hours. What is the average speed of the two vehicles if the car travels 10 km per hour faster than the truck? 3. A 60% salt solution is added to 15% salt solution to produce 450 liters of a 25% salt solution. How much of each solution is required?

4. An adult ticket in an amusement park costs P2 more than a child’s ticket. When 378 adult and 214 children’s tickets were sold, the total earnings is P2384. How much is the cost of an adult ticket?5. The length of a rectangle is 3 cm. more than its width. What are the dimensions of the rectangle if its perimeter is 42 cm? What you will do There are varieties of problems in using first-degree equations in onevariable. The statements require you to express the quantities in terms of onevariable. However, another way is to assign different variables to represent differentquantities. In this case, you must construct two or more equations from statedconditions in a problem. When you write systems of equations, you must be careful that theconditions giving rise to one equation are independent of the conditions to theother equation. Lesson 1 Number ProblemsExamples:1. The sum of two numbers is 55. Four times the smaller is 5 less than the larger. Find the numbers.Solution: You want to find two numbers. This time, use two letters x and y. Let x = the smaller number y = the larger numberYou need two equations since you introduced two variables.Equation 1 x + y = 55 the sum of two numbers is 55 2

Equation 2 4x = y-5 four times the smaller is five less than the largerSo, you have the system of linear equations. x + y = 55 or x + y = 55 4x = y – 5 4x – y = -5By elimination, solve for x: x + y = 55 4x – y = -5 5x = 50 x = 10Solving for y, substitute 10 in place of x in any of the two equations:Equation 1 x + y = 55 Equation 2 4x – y = -5 10 + y = 55 4(10) – y = -5 y = 45 40 – y = -5 y = 40 + 5 y = 45Hence, the smaller number x = 10 the larger number. y = 45Now let us check with the problem:The sum of two numbers is 55. 10 + 45 = 55Four times the smaller is 5 less than the larger. 4(10) = 45 – 52. If 1 of an integer is added to 1 the next consecutive integer, the sum is 25 17. Find the integers. Solution: Use x and y to represent the two consecutive integers. Let x = the first integer y = the next consecutive integer 3

Now, form your working equation:(1) y – x = 1 the difference between any two integers(2) 1 y + 1 = 17 1 of an integer added to 1 of the next integer x 52 2 5 2y + 5x = 170 multiplying both sides by the LCD, 10.or 5x + 2y = 170Solving the system by substitution, 5x + 2y = 170 Equation 2 5(y – 1) + 2y = 170 substitute y – 1 in place of x 5y – 5 + 2y = 170 Addition property of equality 7y = 175 y = 25Solving for x, substitute 25 in place of y in equation (1) y – x = 1 or y – 1 = x 25 – 1 = x x = 24The consecutive integers are 24 and 25.Now, check with the problem: The sum of 1 of the integer and the next 2consecutive integer is 17. 1 (25) + 1 (24) = 17 523. The sum of two numbers is 17 and one of the numbers is 4 less than twice the other. Find the numbers.Solution: Represent each number with a different variable Let x = one number y = the other numberForm the equation:Equation 1: x + y = 17 the sum of two numbers is 17 4

Equation 2 x = 2y - 4 one number is less than twice the otherEliminating x, subtract equation (2) from equation (1)x + y = 17 or x + y = 17 x + y = 17x – 2y = -4 -(x – 2y ) = - 4 -x + 2y = 4 3y = 21 y=7To solve for x, substitute the value of y in equation (1) or (2). x + y = 17 x + 7 = 17 x = 10The numbers are 10 and 7.Check with the problem:10 + 7 = 17 The sum of two number is 17.2(7) – 4 = 10 One number is 4 less than twice the other.4. The sum of the digits of a two-digit number is 10. If the digits are reversed, the new number is 18 less than the original number. Find the original number.Solution: Let u = units digit of the original number t = the tens digit of the original number 10t + u = the original number 10u + t = the new number when the original is reversed.Form the equations:(1) t + u = 10 the sum of the digits is 10.(2) 10u + t = 10t + u – 18 the new number is 18 lessor 9u – 9t = -18 than the original numberSolving the system, use the elimination method.(1) 9u + 9 t = 90 multiply equation (1) by 9(2) 9u – 9t = 18 18u = 72 5

u=4 the units digitSolving for t, substitute 4 in place of u in any of the two equations.Using Equation 1: the tens digit t + 4 = 10 t=6Hence, the number is 64 6(10) + 4 = 64Check with the problem:(1) t + u = 10 (2) 9u – 9t = -18 6 + 4 = 10 9(4) – 9(6) = -185. The sum of the digits of a two-digit number is 13. The number is 4 more than five times the unit digit. Find the numbers.Solution: Let u = units digit t = tens digit 10t + u = the original numberThe two equations are:(1) u + t = 13 or u = 13 - t the sum of the digits is 13(2) 10t + u = 5u + 4 or 10t – 4u = 4 the number is 4 more than 5 times the units digit.Substitute (13 – t) in u in Equation 2. 10t – 4u = 4 Equation 2 10t – 4(13 – t) = 4 10t – 52 + 4t = 4 14t = 56 t = 4 The tens digitSolving for u, substitute 4 in place of t in Equation 1 u + 4 = 13 Equation 1 u=9 The units digit 6

Hence, the number is 49 4(10) + 4 = 49Check with the problem9 + 4 = 13 The sum of the digits is 1310(4) – 4(9) = 4 The number is 4 more than five times the unit digit 40 – 36 = 4Try this out Solve the following problems about number relations by using a system ofequations in two variables. 1. The sum of two numbers is 50 and their difference is 20. Find the numbers. (Represent one number as x and the other as y) 2. The sum of twice one number and three times a second number is 84. The sum of thrice the first number and five times the second is 136. 3. The sum of two numbers is 64 and the sum of the first and twice the second is 88. What are the numbers? 4. One number is 18 more than three times another number. The difference of the numbers is 54. Find the numbers. 5. The difference between two numbers is 14. Three times the larger number is two more than 5 times the smaller. What are the numbers? 6. The sum of the digits of a two-digit number is 10. if the digits are reversed, the number is increased by 36. What is the original number? 7. A two two-digit number is three less than seven times the sum of its digits. If the digits are reversed, the new number is 18 less than the original number. What is the original number? 8. The sum of the digits of a two-digit number is 12. If the digits are reversed, the number is 18 more than the original number. Find the original number. 9. If 18 is added to a two-digit number, the digits are reversed. The sum of the digit is 18. What is the original number? 10. A number is 4 times the sum of its digits. The unit digits is two more than the tens digit. Find the numbers. 7

Lesson 2 Uniform Motion Here, you will solve motion problems whose solutions can be found usingsystems of linear equations. The motion formula: Distance = rate x time will behelpful to you in solving these problems. The following tips will be helpful when solving motion problems. a. Draw a diagram using an arrow or arrows to represent distance and the direction of each object in motion. b. Organize the information in a chart or table. c. Look for as many things as you can that are the same so that you can write equations.Examples: 1. A train leaves Tutuban traveling east at 35 kilometer per hour (km/hr). An hour later, another train leaves Tutuban on a parallel track at 40 km/hr. How far from Tutuban will the two trains meet? Solution: a. Familiarize. First make a drawing. Trains meet here 8

From the drawing, the distances they will cover are the same. Call thedistance d. You do not know the time.Let t = the time for the faster train t + 1 = time for the slower train since it left 1 hour earlier.You can recognize the information in a chart. distance = rate x timeSlow train distance speed time d = 35 ( t + 1)Fast train d 35 t+1 d = 40t d 40 t b. Translate. In motion problems, look for things that are the same so thatyou can write equations. From each row of the chart, you get an equation: d = rt.You have two equations taken from the chart:d = 35(t + 1) Equation 1d = 40 t Equation 2 c. Solve. The equations tell that distance of one is the distance of theother or the distance by the slower train is the same distance for the faster train.35(t + 1) = 40t Substitute 35(t + 1) for d in Equation 235t + 35 = 40t removing the parentheses subtracting 35t 35 = 5t dividing by 5. 7=t The problem asks you to find how far from Tutuban the trains meet. Thusyou need to find d. Do this by substituting 7 for t in the equation d = 40t;d = 40td = 40(7)d = 280 kmd. Check if the time is 7 hours, then the distance the slower train travels is:d = 35( t + 1)d = 35( 7 + 1)d = 35(8)d = 280 km 9

e. State: The trains meet 280 km from Tutuban.2. A motorboat took 3 hours to make a downstream trip with a 6-kph current. The return trip against the same current took 5 hours. Find the speed of the boat in still water. Solution: a. Familiarize and draw. From the figure, the distance in going upstream and returning downstream is the same. Call this distance d.Let r = the speed of the boat in still water.r + 6 = the speed of the boat traveling downstream. (the boat goes with the current)r – 6 = the speed of the boat traveling upstream (The boat goes against the current) Organize the information in a chart. In this case, going upstream is thesame distance going downstream. Use the formula: d = rt distance = rate x timedownstream distance Speed time d = (r + 6) 3upstream d r+6 3 d = (r – 6) 5 d r-6 5 10

b. Translate: From each row of the chart, use the equation d = rt.d = (r + 6) 3 Equation 1d = (r – 6) 5 Equation 2c. Solve: solve the system by substitution.Substitute (r + 6) 3 for d in Equation 2. d = (r – 6)5 Equation 2(r + 6) 3 = (r – 6) 5 substitute (r + 6) 33r + 18 = 5r – 30-2r + 18 = -30 subtracting 5r subtracting 18 -2r = -48 dividing by –2 r = 24d. Check:The distance going downstream = distance going upstream. (r + 6) (3) = (r - 6 )5 90 = 90e. State: The speed in still water is 24 kph.Try this out 1. A truck and a car leave the same service station at the same time and travel in the same direction. The truck travels at 110 km per hour (kph) and the car 80 kph. They can maintain CB radio contact with a range of 20 kilometer (km). When will they lose contact? Complete the following table to aid the translation. distance = rate x timeTruck distance speed timeCar d 110 t d 11

2. A train leaves a station and travels east at 72 kph. Three hours later, a second train leaves on a parallel track and travels east at 120 kph. When will it overtake the first train?Complete the table for your guide. Distance = rate x time Distance Speed TimeSlow train d 72 d = 72 ( )Fast train d t d=( )t3. Two cars leave town at the same time going in the same direction. One travels at 30mph and the other travels at 40 mph. In how many hours will they be at 72 miles apart? (Hint, formulate a chart)4. A private airplane leaves an air[port and flies due south at 192 kph. Two hours later, a jet leaves the same airport and flies due south at 960 kph. When will the jet overtake the plane? Distance = rate x time Distance Speed TimeAirplane d 192 t + 2Jet plane d 960 t5. A canoeist paddled for 4 hours with a 6-kph current to reach a campsite. The return trip against the same current took 10 hours. Find the speed of the canoe in still water.Complete the following table to aid the translation. distance = rate x time Distance Speed Time r+6 10downstream d d = (r + 6)( ) d = ( )10upstream d 12

Lesson 3 Mixture Problem Mixture problems usually deal with the treatment of solutions, either youdilute or concentrate it. The following examples will help you in solving this typeof problems. In some aspect in business, the concept of mixture is utilized toyield a cheaper variety of items.Examples: 1. A chemist prepares a solution that is 62% acid by mixing two solutions. A chemical has one solution that is 80% acid (the rest is water) and another solution is 30% acid. How many liters of each solution should be used? Solution: a. Familiarize. Draw a picture of the situation. The chemist uses the following representations: Let x = number of liters of the first solution y = number of liters of the second solution 13

You can arrange the information in a table.Type of solutions First Second Mixture y 200 liters x + y = 200Amount of solution x% of acid 80% 30% 62% .8x+3y = 124Amount of acid 0.8x 0.3y 0.62(200)=124b. Translate. The Chemist uses x liters of the first solution and y liters of thesecond. Since the total is to be 200 liters, you have: x + y = 200 Total amount of solution80%x + 30% y = 124 The amount of acid from the two solutions. 0.8x + 0.3y = 124Translate to a system of equations: x + y = 200 Equation 1 .8x + .3y = 124 Equation 2c. Solve. Solve the system. Use elimination method.-.3x – .3y = - 60 multiplying Equation 1 by –3.8x + .3y = 124.5x = 64 by addition dividing by 5. x = 128Go back to Equation 1, substitute 128 to x. x + y = 200 128 + y = 200 y = 72The solution is x = 128 liters and y = 72 litersd. Check. The sum of 128 + 72 = 200 also, 80% of 128 = 102.4 and 30% of 72 = 21.6. Then 102.4 + 21.6 = 124.e. State. The chemist should use 128 liters of the 80% acid solution and 72liters of the 30% acid solution. 14

2. A grocer wishes to mix some nuts worth P45.00 per kio and some worth P80.00 per kilo to make 350 kilo of a mixture worth P65.00 per kilo. How much of each should be used?Solution:a. Familiarize. Arranging the information in a table will help you..Let x = the amount of P45.00 worth of nuts y = the amount pf P80.00 worth of nutsTypes of nuts Inexpensive Expensive Mixture nuts nutsCost of nuts P45.00 P80.00 P65Amount in pounds x y 350Mixture 45x 65(350)= 22750 or 80y 45x + 80y = 22750b. Translate. Translate as follows: x + y = 350 total amount of units 45x + 80y = 22750 total cost of mixtureTranslate to a system of equations: x + y = 350 Equation 1 45x + 80y = 22750 Equation 2c. Solve. Solve the equation by elimination method. -45x – 45y = -15750 multiply Equation 1 by –45 45x + 80y = 22750 35y = 7000 by addition y = 200 dividing by 35Substitute 200 for y in Equation 1: x + y = 350 x + 200 = 350 x = 150d. Check: 150 + 200 = 350 kilo The total weight of the nuts 15