MATH 2 part 1

The following examples will illustrate two special cases you may encounter whileapplying the elimination method.Example 4: Solve by the elimination method: a. 2x + y = 2 Equation 1 4x + 2y = -5 Equation 2Solution:Multiply equation 1 by –2 to eliminate y. -2(2x + y) = -2(2) Equation 3 -4x – 2y = - 4Add equation 3 and equation 2, and solve for x. -4x – 2y = - 4 Equation 3 4x + 2y = - 5 Equation 2 0x + 0y = - 9 0=-9 The result 0 = - 9 is a false statement, which means that the system of linearequation has no solution and is said to be inconsistent.b. Solve the system by elimination: 5x – 7y = 9 15x – 21y = 27Solution:Multiply Equation 1 by –3. -3(5x – 7y) = -3(9) Equation 3 -15x + 21y = -27Add Equation 3 and Equation 2, and solve for x. -15x + 21y = - 27 Equation 3 15x – 21y = 27 Equation 2 0 =0 The result 0 = 0 is true. When equation 2 is multiplied by –1, you will notice thatyou will arrive at equation 3. Such a system of linear equation is said to be dependent. 6

Try this outA. Solve the following systems by elimination: 1. x – y = 3 x+y=5 2. –x + 2y = 6 x–y=3 3. 2x – y = 4 x+y=5 4. 3x – y = 3 2x + y = 2 5. x + y = 4 x–y=6 6. 5x – 2y = 12 3x + 2y = 12 7. 2x – y = 1 -2x + 3y = 5 8. x + y = 6 x–y=2 9. x + y = 2 2x – y = 4 10. x + y = 7 x–y=-3B. Name Game Do you know the name of the first British woman mathematician on record who wasa Benedictine nun? She lived from 935 to 1000 A.D. (Source: Math Journal) To find the answer, simplify the following systems of linear equations. Match thesolutions with the choices in the box. Copy the letter corresponding to the answer on the blank with the same exercisenumber. Then discover the answer to the trivia question above. 7

1. 5x + 7y = 10 A x=5 y = -2 3x – 14y = 6 B x=0 y=02. 3x – 2y = 0 6x + 5y = 03. 2x – 3y = 16 W x=2 y=0 3x + 4y = 7 I x = 97 y = - 194. 5x + 3y = 7 28 28 2x + 5y = 1 T x = -5.8 y = - 4.5 H x=3 y=45. 7x – 2y = 13 5x + 3y = 276. 8x – 3y = 11 O p = 1.4 q = 20.4 6x – 5y = 11 C x = 32 y =- 97. 3x + 5y = 7 19 19 2x – 6y = 11 R a = -0.5 b=6 S x=1 y = -18. 4x = 6y + 3.8 y – x = 1.39. –0.7p + 0.2q = 3.1 -1.7p + 0.2q = 1.710. 0.4 a – 0.2b = -1.4 0.4a – 0.5b = -3.2Answer: ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ 5 10 9 8 6 1 7 8 5 3C. Solve the following systems of equations by elimination:1. 2x – y = 7 3x + 2y = 02. x + y = 7 -3x + 3y = -93. x + 3y = -4 x+y=0 8

4. p + 5q = 5 p + q = -35. x + 4y = -18 3x + 5y = -196. 2x – y = -8 5x + 2y = -207. 2x – y = 1 x + 3y = 48. 4x – 5y = 22 x + 2y = -19. 2x – y = 1 4x – 2y = 210. 4x + 3y = 15 2x – 5y = 1 Lesson 2 Solution of Systems of Linear Equations by Substitution Another method of finding the solution of a system of linear equations in twovariables is by substitution. The following example will illustrate how it is done.Example 1:Solve the system by substitution:2x – 3y = -3 Equation 1 y = 2x – 1 Equation 2Since y is already solved in Equation 2, substitute 2x – 1 for y in Equation 1.2x – 3y = - 3 Equation 12x – 3(2x – 1) = - 3 2x – 1 is substituted to y.Solving for x gives2x – 6x + 3 = - 3 - 4x +3 = - 3 - 4x = - 6 9

x=-6 -4x=3 2Now, solve for y by substituting the value 3 to x in Equation 2. 2y = 2x – 1 Equation 2 3 is substituted to x.y = 2( 3 ) – 1 2 2y= 6–1 2y =3–1y=2The solution is ( 3 , 2) 2To check for the result, you substitute these values in equation 1 and you have:2x – 3y = -3 Substitute 3 to x and 2 to y2( 3 ) – 3(2) = -3 2 2 3–6=-3 -3=-3It checks since both sides of the equation are equal.Example 2: Solve the system by substitution:2x + 3y = 16 Equation13x – y = 2 Equation 2You start by solving Equation 2 because this is the simpler equation.3x – y = 2 Equation 2 - y = -3x + 2 Equation 3. y = 3x – 2Substitute the value of y in Equation 3, which is 3x – 2 in Equation 1.2x + 3y = 162x + 3(3x – 2) = 162x + 9x – 6 = 16 11x – 6 = 16 10

11x = 16 + 6 11x = 22 x=2Now solve for y by substituting the value of x in Equation 3. Y = 3x – 2 Y = 3(2) – 2 Y=6–2 Y=4The solution for the system is (2, 4).Try this outA. Solve the following systems by substitution: 1. 2x + 3y = 7 x=2 2. y = 3 3x – 2y = 6 3. y = x – 3 x+y=5 4. y = x + 2 x+y=6 5. x = y – 2 x + 3y = 2 6. 3x + 5y = -6 x = 5y + 3 7. y = 2x + 3 4x – 3y = 1 8. 3x + y = 4 4x – 3y = 1 9. x – 4y = 9 2x – 3y = 11 10. x = y + 1 x + 2y = 7 11

B. Who Was He? This mathematician born in 600 BC was said to be the first mathematician to provea theorem. He was also considered the first true mathematician. Who was he? To find out, solve the following systems of linear equations using the substitutionmethod. Then match the letter that corresponds to the answer. The letters will spell outthe name of this famous mathematician.1. x + 2y = 4 A (-3, -1) x – 3y = -1 B (2, 4) E (2,1)2. x + 2y = 7 2x – y = 4 353. x – 4y = 1 F (2, - 1 ) x – 2y = -1 34. x + 2y = 8 H (3, 2) 2x – y = 6 L (4, 2) S (2, - 1)5. 3x + 5y = 3 T (2, 1) 9x – 10y = 46. x + 3y = -1 3x – 6y = 12Answer: ___ ___ ___ ___ ___ ___ 12 3 4 5 6C. Math Game: How High Is It? The Statue of Liberty in New York, USA was designed by Frederic AugusteBartholdi, a French. The statue was unveiled and dedicated on Bedloe’s Island in UpperNew York Bay in 1886. It was declared a national monument in 1924. (Math Journal)What do you think is the approximate height of the Statue of Liberty in centimeters? To find out, match the following systems of linear equations with the correspondingsolutions.The digits will reveal the height of the Statue of Liberty. 12

a. 3x – 2y = 13 1. x = -2, y = 2 x+y=6 4. x = 5, y = 1 6. x = 4, y = 1b. 2x + 3y = 11 4y – x = 0c. 2x – y = 7 3. x = 2, y = -3 3x + 2y = 0 5. x = -5, y = 3 2. x = 6, y = 4d. 4x + y = 5 x–2=0Answer: The height of the Statue of Liberty is approximately:____ ____ ____ ____ cmab cdD. Solve the following systems by substitution method:1. x + y = 3 3x + 2y = 92. x + 3y = -4 2y + 3x = 33. 2y – 3x = 4 3x = 64. y = 3 2x = 5y + 75. y – x = 4 x+3=y6. 5b – 2 = 2a 3a + 6 = 25b7. 12x + 18y = 12 2x = 2 – 3y8. –5r + 14t = 13 -9r = -72t9. 2x – 5y = -1 2x – y = 1 13

10. x + 3y = 4 y+x=0 Let’s Summarize The following steps are followed in solving a system of linear equations in two variables:A. Solving by Elimination: Steps: 1. Multiply one or both of the equations by a constant so that one of the variables can be eliminated by addition. 2. Add the equations of the equivalent system formed in step 1. 3. Solve the equation found in step 2. 4. Substitute the value found in step 3 in either of the given equations to find the corresponding value of the remaining variable. The ordered pair formed is the solution of the system. 5. Check the solution by substituting the pair of values found in step 4 in the other given equation.B. Solving by Substitution: Steps: 1. Solve one of the given equations for one of the variables. 2. Substitute the expression obtained in step 1 in the other equation of the system to get an equation in a one variable. 3. Solve the equation found in step 2. 4. Substitute the value found in step 3 in the equation derived in step 1 to find the corresponding value of the remaining variable. The ordered pair formed is the solution of the system. 5. Check the solution by substituting the pair of values found in step 4 in both the given equations. 14

What have you learnedA. Tell whether the given ordered pairs is the solution of the given system: 1. (-2, 6) 3x + 2y = 6 -3x – y = 0 2. (-6, 4) x + y = -2 y = x + 10 3. (3, 0) 5x – y = 15 7x + y = 21 4. (5, 7) 7x + y = 42 -3x + y = -8 5. (-1, 2 ) 3 x – 4y = 13 -x + 6y = -18B. Solve the following systems by elimination: 1. 5x + 3y = - 9 7x + y = -3 2. x + 2y = 1 3x - 5y = -8 3. 3x + 2y = 40 x – 7y = - 2 4. 2x + 3y = 8 x–y=2 15

5. 2x + 2y = 3 4y – 5x = 15C. Solve the following by substitution: 1. x – y = 7 y = 2x - 11 2. 2x + 3y = 3 4x = 12y -3 3. 3x + 4y = -3 x + 4y = 1 4. 2x + 3y = 1 x -y=3 5. x + 2y = 4 2x – y = 3 16

Answer KeyHow much do you knowA.1. not a solution2. solution3. solution4. solution5. solutionB. 1. x = 6, y = 2 2. x = 2, y = 0 3. x = 2, y = 5 4. m = -1, n = -1 5. a = 0, b = 1C. 1. x = 2, y = -2 2. x = 5, y = 1 3. x = -3, y = 1 4. s = 1 , t = -2 2 5. x = 3, y = - 5Try this outLesson 1A. Solution is: 1. (4, 1) 2. (6, 3) 3. (3, 2) 4. (1, 0) 5. (5, -1) 6. (3, 3 ) 2 7. (2, 3) 8. (4, 2) 9. (2, 0) 10. (2, 5) 17

B. Name Game:1. x = 2, y = 0 W2. x = 0, y = 0 B3. x = 5, y = -2 A4. x = 32 , y = - 9 C 19 19 H5. x = 3, y = 4 S6. x = 1, y = -1 I7. x = 97 , y = - 19 28 28 T O8. x = -5.8, y = -4.5 R9. p = 1.4, q = 20.410. a = -0.5, b = 6Answer: H R O T S W I T H A 5 10 9 8 6 1 7 8 5 3C. The Solution is:1. (2, -3)2. (5, 2)3. (2, -2)4. (-5, 2)5. (2, -5)6. (-4, 0)7. (1, 1)8. (3, -2)9. infinite number of solutions10. (3, 1)Lesson 2A. The solution is:1. (2, 1)2. (4, 3)3. (4, 1)4. (2, 4)5. (-1, 1)6. (- 3 , - 3 ) 447. (-5, -7) 18

8. (1, 1) 9. ( 17 , - 7 ) 55 10. (3, 2)B. Who was he? 1. x = 2, y = 1 T 2. x = 3, y = 2 H 3. x = -3, y = -1 A 4. x = 4, y = 2 L 5. x = 2 , y = 1 E 35 6. x =2, y = - 1 S 3 Answer: T H A L E SC. How high is it? A = no. 4 B = no. 6 C = no. 3 D = no. 3Answer: The height of the Statue of Liberty is approximately 4,633 cm.D. 1. (3, 0) 2. ( 7 , - 11 ) 11 3 3. (2, 5) 4. (11, 3) 5. no solution 6. (- 4 , 6 ) 7 35 7. dependent, infinite number of solutions 8. (-4, - 1 ) 2 9. ( 3 , 1 ) 42 10. (-2, 2) 19

What have you learnedA. 1. solution 2. solution 3. solution 4. solution 5. not a solutionB. The solution is 1. (0, -3) 2. (-1, 1) 3. (12, 2) 4. ( 14 , 4 ) 55 5. (-1, 5 ) 2C. The solution is 1. (4, -3) 2. (-2, 3 ) 4 3. (7, 14) 4. (2, -1) 5. (2, 1) 20

Module 3 VariationWhat this module is about This module deals with the variation of more than two variables. Both direct andinverse variations may occur in the same problem. Joint variation are quantities that aredirectly related. But when joint variation is combined with inverse variation, then it is calledcombined variations.What you are expected to learn 1. identify relationship involving two or more variables. 2. find the relation and constant of variation 3. apply the concept of proportionality.How much do you know A. Using k as the constant of variation, write the equation of variation for each of the following: 1. The area (A) of a parallelogram varies jointly as its base (b) and its altitude (a). 2. The volume (V) of a pyramid varies jointly as its base area (b) and its altitude (a). 3. The area of the circle varies directly as the square of its radius. 4. U varies jointly as the square of m and inversely as n. 5. V varies jointly with l, w and h. 6. The volume (V) of a cube varies directly as the cube of its edge (e). 7. The force (F) needed to push an object along a flat surface varies directly as the weight (w) of the object. 1

8. The altitude (h) of a cylinder is inversely proportional to the square of its radius (r).9. M varies directly as r and inversely as s.10. Q varies jointly as R and T.What you will do Lesson 1 Joint Variation This lesson deals with another concept of variation, the joint variation. Some physical relationships, as in area or volume, may involve three or morevariables simultaneously. Consider the area of a rectangle which is obtained from the formula A = lw wherel is the length w is the width of the rectangle. The table shows the area in squarecentimetres for different values of the length and the base.l2 4 5 6 6 8 8 10w3 3 3 5 7 7 11 13A6 12 15 30 42 56 88 130 Observe that A increases as either l or w increase or both. Then it is said thatthe area of a rectangle varies jointly as the length and the width. Consider the area of a triangle, which is obtained from the formula: A = 1 ab 2where b is the base and a is the altitude of the triangle. The table shows the area insquare centimetres for different values of the base and altitude, both being in centimetres.b 2 4 4 6 6 8 8 10a 3 3 5 5 7 7 11 13A 3 6 10 15 21 28 44 65 Observe that A increases as either b or a increase or both. We say that the area ofa triangle varies jointly as the base and the altitude. 2

Examples: 1. Find an equation of variation where a varies jointly as b and c, and a = 36 when b = 3 and c = 4 .Solution: a = kbc substitute the set of given data to find k 36 = k(3)(4)k = 36 apply the properties of equality 12k =3Therefore, the required equation of variation is: a = 3bc2. z varies jointly as x and y . If z = 16 when x = 4 and y = 6 , find the constant of variation and the equation of the relation.Solution: z = kxy: 16 = k(4)(6) substitute the set of given data to find kk = 16 apply the properties of equality 24 k=2 3The equation of the variation is: z = 2 xy 33. The area A of a triangle varies jointly as the base b and the altitude a of the triangle. If A = 65cm2 when b = 10cm and a = 13cm , find the area of a triangle whose base is 8cm and altitude is 11cm.Solution: A = kab the equation of the relation substitute the set of given data to find k : 65 = k(13)(10) apply the properties of equality k = 65 130 k=1 2 3

The equation of the variation is: A = 1 ab 2 Therefore, when a = 11and b = 8 , the area of the triangle is A = 1 (11)(8) 2 A = 44cm24. The area A of rectangle varies jointly as the length l and the width w and A = 180cm2 when l = 9cm and w = 5cm . Find the area of a rectangle whose length is 20cm and whose width is 5cm .Solution : A = klw the equation of the relation: 180 = k(9)(5) substitute the set of given data to find k k = 180 apply the properties of equality 45 k =4Therefore, when l = 9cm and w = 5cm . A = 4lw A = 4(20)(5) A = 400cm25. The volume (V ) of a prism on a square base varies jointly as the height ( h ) and the square of a side ( s ) of the base of the prism. If the volume is 81cm3 when a side of the base is 4cm and the height is 6cm, write the equation of the relation.Solution: Express the relation as: V = ks2h 81 = k(4)2 (6) substitute the given values for V ,s and h 4

81 = k(16)(6)k = 81 reduce to lowest term 96 k = 27 32The equation of variation is V = 27 s2h 326. Extending the problem on the previous example, find the volume of the prism if a side of the base is 7 cm and the height is 12 cm.Solution: V = 27 s2h from the previous example 32 V = 27 (7)2 (12) substitute the given values for s and h 32 V = 27 (49)(12) 32 V = 27 (588) 32 V = 15876 32 V = 496.125cm37. The volume (V ) of a prism on a square base varies jointly as the height ( h ) and the square of a side ( s ) of the base of the prism.A. If the volume is 72 cm3 when a side of the base is 3 cm and the height is 8 cm, write the equation of the relation.Solution: where k is the constant of variation V = ks2h72 = k(3)2 (8) substitute the given to find k 5

72 = k(9)(8) 72 = 72k k = 72 72` k =1The equation of the relation is V = s2hB. Find the volume when a side of the base is 5 cm and the height is 14 cm.Solution: V = (5)2 (14) substitute the given values for s and hV = (25)(14)V = 350cm3 C. By how many percent is the original volume V1 increased if a side is increased by 10% and the height is 20%.Solution: Denote the new side by S and the new height by H . As a result; S = 1.1s since the side is increased by 10% and H = 1.2h since the height is increased by 20%. Then the new volume V2 is: V2 = S 2H V2 = (1.1s)2 (1.2h) V2 = (1.21s2 )(1.2h) V2 = 1.452s2hSince V1 = s2h , we may substitute V1 into the results, which gives V2 = 1.452V1 .The increase in volume is 0.452 of the original volume V1 . To change 0.452to percentage, multiply it by 100% that will give you: (0.452)(100%) = 45.2% 6

The following illustrations are applications of variation in different fields ofmathematics like Geometry, Engineering, etc.Examples: 1. The volume of a right circular cylinder varies jointly as the height and the square of the radius. The volume of a right circular cylinder, with radius 4 centimetres and height 7 centimetres, is 352 cm3. Find the volume of another cylinder with radius 8 centimetres and height 14 centimetres. r Solution:The equation of the relation is V = khr 2 hFrom the given set of data: r = 4 cm h = 7 cm V = 352 cm3To find k substitute the values above:V = khr 2k = V rearranging the equation above hr 2 simplifying the fractionk = 352 (7)(4)2k = 352 (7)(16)k = 22 7To find the volume of a cylinder with r = 8 cm and h = 14 cm: V = 22 (13)(8)2 7 V = 22 (14)(64) 7 V = 2816 cm3 7

2. The horsepower h required to propel a ship varies directly as the cube of its speed s . Find the ratio of the power required at 14 knots to that required at 7 knots.Solution: The equation of the relation is h = ks3 The ratio of power required at 14 knots to 7 knots is h2 = k(14)3 h1 k(7)3h2 = (14)3 the k ' s cancel outh1 (7)3h2 = 2744h1 343h2 = 8h1 13. The pressure P on the bottom of a swimming pool varies directly as the depth d of the water. If the pressure1 is 125 Pa2 when the water is 2 metres deep, find the pressure when it is 4.5 metres deep.Solution 1: P = kd solving for the constant of variation k=P since P = 125 when d = 2 d k = 125 2k = 62.5P = 62.5dP = (62.5)(4.5)P = 281.25 Pa1 Pressure is defined as the force exerted per unit area2 Pascal (Pa) is the metric unit for pressure 8

Solution 2: In this solution, you do not need to find k. The equation P = kdmaybe written as k = P , meaning that the ratio P is a constant. Therefore: dd P1 = P2 d1 d2125 = P2 2 4.5P2 = (125)(4.5) 2P2 = 281.25 Pa4. The horsepower required to propel a ship varies directly as the cube of its speed. If the horsepower required for a speed of 15 knots is 10 125, find the horsepower required for a speed of 20 knots.Solution: let P = required horsepower s = speed, in knotsSince P varies directly as s3, you haveP = ks3 (1)10,125 = k(15)3k = 10, 125 (15)3k = 10,125 3, 375k =3 substitute k = 3 and s = 20 in (1)P = 3(20)3P = 3(8000)P = 24, 000 hp 9

5. The weight of a rectangular block of metal varies jointly as its length, width and thickness. If the weight of a 13 by 8 by 6 dm block of aluminum is 18.7 kg, find the weight of a 16 by 10 by 4 dm block of aluminum. Solution: Let W = weight in kilograms l = length in decimeters w = width in decimeters t = thickness in decimeters Since the weight of the metal block varies jointly as its length, width and thickness you have: W = klwt k= W lwt k = 18.7 (13)(8)(6) k = 18.7 576 Substitute k = 18.7 , l = 16 , w = 10 and t = 4 in the equation W = klwt to get 576 the weight of the desired block: W = 18.7 (16)(10)(4) 576 W = 20.8 kg weight of the 16 by 10 by 4 dm block6. The amount of coal used by a steamship traveling at uniform speed varies jointly as the distance traveled and the square of the speed. If a steamship uses 45 tons of coal traveling 80 km at 15 knots, how many tons will it use if it travels 102 km at 20 knots?Solution: Let T = number of tons used s = the distance in miles v = the speed in knotsand then T = k(sv2) (1)hence, when T = 45, s = 80 and v = 15, you have 10

45 = k(80)(152) k = ___45___ (80)(225) k = __1__ 400 Substituting this value for k in (1), you have T = __1__ (120)(202) 400 T = 48000 400 T = 120 tonsTry this out A. Translate each statement into mathematical statement. Use k as the constant ofvariation. 1. P varies jointly as q and r . 2. V varies jointly with l , w and h . 3. The area A of a parallelogram varies jointly as the base b and altitude h . 4. The volume of a cylinder V varies jointly as its height h and the square of the radius r . 5. The heat H produced by an electric lamp varies jointly as the resistance R and the square of the current c . 6. The area A of a parallelogram varies jointly as the base b and altitude a 7. The volume V of a pyramid varies jointly as the base area b and the altitude a . 8. The area A of a triangle varies jointly as one-half the base b and the altitude h . 9. The appropriate length (s) of a rectangular beam varies jointly as its width (w) and its depth (d). 10. The area A of a square varies jointly as its diagonals d1 and d2 . 11

B. Solve for the value of the constant of variation k, then find the missing value.1. z varies jointly as x and y and z = 62 when x = 5 and y = 6. a. find z when x = 7 and y = 8 b. find x when z = 72 and y = 4 c. find y when z = 82 and x = 42. z varies jointly as x and y . If z = 3 when x = 3and y = 15 , find z when x = 6 and y = 9 .3. z varies jointly as the square root of the product x and y . If z = 3 when x = 3 and y = 12 , find x when z = 6 and y = 8 .4. d varies jointly as o and g. If d = 15, when o = 14 and g = 5, find g when o = 21 and d = 8.5. q varies jointly as r and s. If q = 2.4, when r = 0.6 and s = 0.8, find q when r = 1.6 and s = .01.6. d varies jointly as e and l . If d = 2.4, when e = 0.6 and l = 0.8, find d when e = 1.6 and l = .01.7. x varies jointly as w, y and z. If x = 18, when w = 2, y = 6 and z = 5, find x when w = 5, y = 12 and z = 3.8. z varies jointly as x and y. z = 60 when x = 3 and y = 4. Find y when z =80 and x =2 .9. The weight W of a cylindrical metal varies jointly as its length l and the square of its diameter d a. If W = 6 kg when l = 6 cm and d = 3 cm, find the equation of variation. b. Find l when W = 10 kg and d = 2 cm. c. Find W when d = 6 cm and l = 1.4 cm.10. The amount of gasoline used by a car varies jointly as the distance traveled and the square root of the speed. Suppose a car used 25 liters on a 100 kilometer trip at 100 km/hr. About how many liters will it use on a 192 kilometer trip at 64 km/hr? 12

C. What did the pig say when the man grabbed him by the tail? 5 16 8 1250 4 30 3 53 18 22 448 3 3 4 40 192 21Directions: Answer the questions below then transfer the letter associated to each questionto the box which contains the correct answer.I If z varies jointly as x and y, and z = 24, S Z varies jointly as x and y and z = 60when x = 2 and y = 4, find z when x = 2 when x = 3 and y = 4. Find y when z = 80and y = 5 and x = 2.N If z varies jointly as x and y and z = 12, E If w varies jointly as x and y and w = 36when x = 2 and y = 4, find the constant of when x = 3 and y = 4, find the constant ofvariation. variation.S If z varies jointly as x and y and z = 24, T If A varies jointly as l and w and A is 36when x = 3 and y = 4, find z when x = 3 when l = 9 and w = 2, find A if l = 6 andand y = 2. w = 4.H If a varies jointly as c and d, and O If w varies jointly as x and y2 anda = 20, when c = 2 and d = 4. Find d when w = 24 when x = 2 and y = 3, find the valuea = 25 and c = 8. of w when x = 9 and y = 4.E If z varies jointly as x and the square of H If varies jointly as w2 and l and A = 48y and z = 20, when x = 4 and y = 2. Find z when w = 3 and l = 4, find the value of Awhen x = 2 and y = 4. when w = 9 and l = 15.T If z varies jointly as x and the square of M x varies jointly as and y and z = 48,y and z = 40, when x = 5 and y = 4. Find z when x = 4 and y = 3, find the constant ofwhen x = 4 and y = 5. variation.F If y varies directly as x and if y = 15 I p varies jointly as r and s and p = 32when x = 5, find the value of y if x = 7. when r = 3 and s = 2. Find the constant of variation.E If w varies jointly as x and y and if w = D If y varies directly as x and y = 6 when15 when x = 2 and y = 3, find the value ofw if x = 3 and y = 4. x = 8, what is the value of y when x = 24? 13

Lesson 2Combined variation Combined variation is another physical relationship among variables. This is thekind of variation that involves both the direct and inverse variations. This relationship among variables will be well illustrated in the following examples.Examples:A. The following are mathematical statements that show combined variations. 1. k = I Prt 2. k = E IR 3. k = c ar 4. Pv = k t 5. k = ab2 cB. Translate each statement into a mathematical statement. Use k as the constant of variation.1. T varies directly as a and inversely as b. T = ak2. Y varies directly as x and inversely as the square of z. b Y = kx z23. P varies directly as the square of x and inversely as s. P = kx2 s4. The time t required to travel is directly proportional to the t = kT temperature T and inversely proportional to the pressure P. P 14

5. The pressure P of a gas varies directly as its P = kt temperature t and inversely as its volume V. V The following examples are combined variation where some terms are unknown andcan be obtained by the available information.C. If z varies directly as x and inversely as y, and z = 9 when x = 6 and y = 2, find z when x = 8 and y = 12.Solution: The equation is z = kx ySubstituting the given values:9 = k6 2k=9 3k =3z = (3)(8) 12z=2D. x varies directly as y and inversely as z . If x = 15 when y = 20 and z = 40 , find x when y = 12 and z = 20 . Solution: The equation is x = ky z Substituting the given values to find k where x = 15 when y = 20 and z = 40 , 15 = k 20 40 k = (15)(40) 20 k = 30 15

To find x when y = 12 and z = 20Using the equation x = ky z x = (30)(12) 20 x = 18E. t varies directly as m and inversely as the square of n . if t = 16 when m = 8 and n = 2 , find t when m = 13 and n = 3.Solution:The equation of the variation: t = km n2To find k , where t = 16 , m = 8 and n = 2 , substitute the given values 16 = k (8) (2)2 k = 16(2)2 8 k = (16)(4) 8 k = 64 8 k =8To find t when m = 13 and n = 3 t = (8)(13) (3)2 t = 104 or 9 t = 11 5 9 16

F.. r varies jointly as s and t and inversely as u . If r = 3 when s = 10 , t = 3 and 28 u = 56 , find r when s = 6 , t = 7 and u = 84 . Solution: The equation of the variation: r = kst u Substitute the given values to find k : 3 = k(10)(3) 28 56 k = (3)(56) (28)(10)(3) k= 2 10 k=1 5To find r when s = 6 , t = 7 and u = 84 . r = kst u 1 (6)(7) r= 5 84 r =  42   1   5   84  r= 1 10G. Given: w varies directly as the product of x and y and inversely as the square of z . If w = 9 when x = 6 , y = 27 and z = 3 , find w when x = 4 , y = 7 and z = 2 .Solution:The equation: w = kxy z2Substituting the first given set of values to the equation, where w = 9 , x = 6 ,y = 27 and z = 3 17

9 = k(6)(27) (3)2 9 = k(162) 9 81 = 162k k = 81 or 162 k=1 2 Find the value of w when k = 1 and use the second set of values when x = 4 , 2y = 7 and z = 2 , you havew = kxy z2 1 (4)(7)w = 2 22w = (2)(7) 4w = 7 or 2w = 3.5H. The current I varies directly as the electromotive force E and inversely as the resistance R . If in a system a current of 20 A flows through a resistance of 20 Ω with an electromotive force of 100 V, find the current that 150 V will send through the system.Solution: Let I = the current in A (ampere) E = electromotive force in V (volts) R = Ω ( ohms)The equation: I = kE RSubstitute the first set of given data: 18

I = 20 A E = 100 V R = 20 Ω By substitution, find k: 20 = k100 20 k = (20)(20) 100 k = 400 100 k =4 To find how much (I) current that 150 V will send through the system I = (4)(150) 20 I = 30 Notice, the system offers a resistance of 20 Ω.Try this outA. Using k as the constant of variation, write the equation of variation for each of thefollowing. 1. W varies jointly as the square of a and c and inversely as b. 2. The electrical resistance (R) of a wire varies directly as its length ( l ) and inversely as the square of its diameter (d). 3. The acceleration A of a moving object varies directly as the distance d it travels and varies inversely as the square of the time t it travels. 4. The heat H produced by an electric lamp varies jointly as the resistance R and the square of the current C. 5. The kinetic energy E of a moving object varies jointly as the mass m of the object and the square of the velocity v.B. Solve the following 1. If r varies directly as s and inversely as the square of u, then r = 2 when 19

s = 18 and u = 2. Find: a. r when u = 3 and s = 27. b. s when u = 2 and r = 4 c. u when r = 1 and s = 36 2. p varies directly as q and the square of r and inversely as s. a. write the equation of the relation b. find k when p = 40, q = 5, r = 4 and s = 6 c. find p when q = 8, r = 6 and s = 9 d. find s when p = 10, q = 5 and r = 2. 3. w varies directly as xy and inversely as v2 and w = 1200 when x = 4, y = 9 and v = 6. Find w when x = 3, y = 12 and v = 9. 4. Suppose p varies directly as b2 and inversely as s3 . If p = 3 when b = 6 and 4 s = 2 , find b when p = 6 and s = 4 .Let ‘s summarize Definition: Joint Variation: The statement a varies jointly as b and c means a = kbc , or k = a , where k is the constant of variation. bc Combined Variation: The statement t varies directly as x and inversely as y means t = kx , or k = ty , where k is the constant of variation. yx 20

What have you learnedA. The following are formulas and equations which are frequently used in mathematics andin science. State whether the relationship is considered direct, inverse, joint or combinedvariation.____________________1. C = 2Π____________________2. A = lw____________________3. D = rt____________________4. I = prt____________________5. V = lwh____________________6. A = Πr2____________________7. V = Πr2____________________8. E = mc2____________________9. F = ma___________________10. V = 2Πr T_____________________11. K = 1 mv2 2_____________________12. P = F A_____________________13. W = Fd_____________________14. V = IR_____________________15. Q = mcδtB. For each given equation with k as the constant of variation and solve for the unknownvalue. Choose the letter of the correct value of the unknown.1. w varies inversely as z and w is 4 when z is 6. what is z when w is 14?a. 12/7 b. 11/7 c. 10/7 d. none of these2. x varies jointly as y and z. x is 4 when y is 3 and z is 2. What is z if x is 8 and y is10?a. 5/6 b. 6/5 c. 6 d. 53. m varies directly as n but inversely as p. What is n if m is 16 and p is 18?a. 284 b. 286 c. 288 d. 2904. p varies inversely as q and r, and p = 2/3 when q is 4 and r is 14. What is q when pis 6 and r is 10?a. 45/28 b. 40/25 c. 42/25 d. 28/45 21

5. F varies directly as g and inversely as the square root of the product of I and h, andF = 5 when g = 7.5, I = 2 and h = 18. What is F when g = 4, I = ¼ and h = 16?a. 8 b. 16 c. 18 d. none of these6. W varies directly as u2 and v, and W is 75 when u is 5 and v is 9. What is v when Wis 150 and u is 10?a. 4 b. 4.5 c. 5 d. none of these7. S varies directly as t and inversely as u2, and S is 9 when t is 4 and u is 12. What is twhen u is 8 and S is 16?a. 3.1 b. 1.16 c. 3.16 d. none of these8. y varies directly as w2 and inversely as the cube of x. What is y when w is 3 andx is 2?a. 8/9 b. 9 c. 8 d. 9/89. P varies as the product r and inversely as the square of t. What is t when P = 2,r = 33?a. 33 b. 33/2 c. 11/2 d. none of these 210. y varies directly as x, and y = 6 when x = ½. What is x if y = 12? a. 3 b. 1 c. 2 d. 0 22

Answer Key V = ke3 7. F = kwHow much do you know 8. h = kA. r21. A = kba or A = kab 9. m = kr2. V = kba or V = kab3. A = kr2 s4. U = km2 10. Q = kRT n 6. A = kab4. V = klwh 7. V = kab 8. A = 1 bhTry this out 2Lesson 1 9. s = kwd 10. A = kd1d2 A. 1. P = kqr 2. V = klwh 6. k = 5 , d = 0.08 3. A = kbh 7. x = kwy , k = 3 or 0.3, x = 54 4. V = khr2 5. H = krc2 10 8. z = kxy , k = 5 , y = 8B. 1. Z = kxy , k = 31 9. a. k = 1 , W = 1 ld 2 15 99 a. 115.7 b. l = 22.5 cm b. 8.71 c. 5.6 kg c. 9.92 10. k = 0.025 , 38.4 liters 2. k = 1 , z = 3.6 23 15 3. z = k xy , k = 1 , x = 18 2 4. k = 3 , g = 1.8 14 5. k = 50 , q = 0.8

C. What did the pig say when the man grabbed him by the tail? T HI S IS 8 12 50 5/4 30 16/3 T HE E ND 48 4/3 3 5/2 3/2 18 OF ME 192 21 4 40Lesson 2 A. 1. W = ka2c b 2. R= kl d2 3. A = kd t2 4. H = kRC2 5. E = kmv2B.1. r = ks , k = 4 3. k = 1200 u2 9 w = 133 1 3 a. r = 4/3 4. k = 1 b. s = 36 6 d. u = 4 b = 482. a. P = kqr2 s b. k = 3 c. P = 96 d. s = 6 24

What have you learned B. 1. aA. 2. b 1. direct 3.c 2. joint 4. d 3. joint 5. a 4. joint 6. b 5. joint 7. c 6. direct 8. d 7. direct 9. a 8. joint 10. b 9. joint 10. combined 11. joint 12. combined 13. joint 14. joint 15. jointExample 5. If the volume of a mass of gas at given temperature is 56 in3 when the pressure is 18 lb/in2, use Boyle’s law to find the volume when the pressure is 16 lb/in2 Solution: Boyle’s law states that V = k/P or PV = k, meaning that PV is a constant. a) without finding k, you may write P1V1 = P2V2 where P1 = 18 lb/in2, V1 =, 56 in3, P2= 16lb/in2, V2 = ? Substituting (56)(18) = (16) V2 V2 = 56(18) 26 V2 = 63 in 3Example 8. The load which can be safely put on a beam with a rectangular cross section that is supported at each end varies jointly as the product of the width and the square of the depth and inversely as the length of the beam between supports. If the safe load of a beam 3 in wide and 6 in deep with supports 8 ft apart is 2700 lb, find the safe load of a beam of the same material that is 4 in wide and 10 deep with supports 12 ft apart. 25

Solution: let w = width of beam, in inches d = depth of beam, in inches l = length between supports, in feet L = safe load, in poundsthen L = kwd2 lAccording to the first set of data ,when w = 3, d= 6 and l = 8, thenL = 2700, therefore 2700 = k (3) (62) 8 k = 8(2700) = __8(2700)__ 3(62) 108 k = 200Consequently, if w = 4, d = 10, l= 12 and k = 200, you have L = 200(4)(102) 12 L = 6666 2/3 26

Module 3 Quadratic Equations What this module is about This module is the last of the 3 modules for quadratic equations. In thismodule you will be able to apply all the skills you have learned in solvingquadratic equations. What are you expected to learn This module is designed for you to: 1. Use the best method (factoring, quadratic formula and completing the square) in solving a particular quadratic equation. 2. Solve equations reducible to quadratic equations. 3. Solve verbal problems involving quadratic equations. How much do you know A. Solve each equation. 1. x² - 7x – 6 = 0 2. (u – 6)² = 13 3. 2x + 3 = x + 2 x+6 x+4 4. 2 (m – 1)² - 3 (m –1) + 1 = 0 B. Solve the problems. 5. The product of a number and 2 more than itself is 3. Find the number. 6. A field is in the shape of a rectangle. Its length is 4 units longer than its width. If its area is 96 sq. units, what are the dimensions of the field?

What you will do Lesson 1Choosing A Method At this point, you have already learned all the methods used in solving aquadratic equation. These are:1. Factoring2. Completing the square3. Quadratic formula You have learned that completing the square and the quadratic formulawork in all types of quadratic equations. But you also know that the factoringmethod, when it works is often the easiest to use. The question is, how shouldyou choose a method for a particular equation? Unless an equation is given in the form a(x + p)² = d it is easier to solvequadratic equations by factoring than by using the quadratic formula orcompleting the square. It is important to remember that the first step in solving quadraticequations is to transform the equation in standard form. In the following examples, find why a particular method is chosen insolving a given quadratic equation.Example 1. Solve for x: 3x² = 5xSolution: Begin by writing the equation in standard form.3x² = 5x3x² - 5x = 0 Notice that there is a common factor, the factoring method is mostconvenient to use in this case.x (3x – 5) = 0x = 0 or 3x – 5 = 0x = 0 or x= 5 3 2

Notice that 3x² = 5x is an incomplete form of a quadratic equationwhere c = 0.Example 2. Solve for x: x² = 3x + 28Solution: Again, write the equation in standard form. x² - 3x – 28 = 0 by factoring (x – 7) (x + 4) = 0 x – 7 = 0 or x+4=0 x = -4 x = 7 or This example shows that as long as the left-hand side is factorable,factoring is the easiest method to use.The checking is left for you to do. Examples 1 and 2 showed to you where the factoring is most useful.These are:a. when c = 0 in the equation ax² + bx + c = 0; andb. when the right –hand side of the equation is factorable.Example 3. Solve for x: 10x² + 39x – 100 = 20 – 5x – 14x²Solution: Begin by writing the equation into standard form by combining similar terms. 10x² + 39x – 100 = 20 – 5x – 14x² 24x² + 44x – 120 = 0 Both sides of the equation is divisible by 4 so, 24x2 + 44x −120 = 0 44 6x² + 11x – 30 = 0 You may find it difficult to factor or it may not factor at all. You maythen use the quadratic formula. So that, a = 6, b = 11, and c = -30 x = − b ± b2 − 4ac 2a 3

−11 ± (11)2 − 4(6)(−30)x= 2(6)x = −11 ± 121 + 720 12x = −11 ± 841 12x = −11 ± 29 12x = −11 + 29 = 18 = 3 and x = −11 − 29 = − 40 = −10 12 12 2 12 12 3x= 3 and x = - 10 2 3 Note that your arithmetic skills can get messy and may cause a lotof errors due to carelessness.Factoring can also be used in this case. 6x² + 11x – 30 = 0 (2x – 3) ( 3x + 10) = 0 2x –3 = 0 or 3x + 10 = 0 2x = 3 3x = -10 x = 3/2 or x = - 10 3 Lastly, note that since you have already used two methods insolving the equation and got the same answer, then you don’t have tocheck by substitution. By this time you should know that “checking” mayalso be done by using another method in solving the equation.Example 4. Solve for x: x (x – 2) = 5x –2Solution: Begin by simplifying the left-hand side and write the equation into standard form. x (x – 2) = 5x – 2 use the distributive property x² - 2x = 5x –2 x² - 7x + 2 = 0Not factorable, so use the quadratic formula with a = 1, b = -7, c = 2 4

x = − b ± b2 − 4ac 2a − (−7) ± (−7)2 − 4(1)(2) x= 2(1) x = + 7 ± 49 − 8 2 x = 7 ± 41 2 x = 7 + 41 or x = 7 − 41 2 2The checking is left for you to do. Basically, if you are good at factoring, it probably pays to spend a minuteor two in trying to use the factoring method. If it fails, then generally you coulduse the quadratic formula or completing the square.Example 5. Solve for x: (3x - 5)² = 17Solution: Since this equation is given in the form (x – p)² = d then you can readily use the square root method. (3x - 5)² = 17 x = 5 − 17 3x – 5 = ± 17 3 3x = 5 ± 17 x = 5 ± 17 3 x = 5 + 17 and 3 You may use the quadratic formula in solving this equation, bywriting it in standard form first. However, if an equation is given in the formof a perfect square, that is, (x - p)² = d, then it is convenient to use thesquare root method. Just then suppose you use the quadratic formula to see that theanswer is correct. That is, (3x - 5)² = 17 not in standard form 9x² - 30x + 25 = 17 expand the square of a binomial 9x² - 30x + 25 – 17 = 0 write in standard form 5

9x² - 30x + 8 = 0 not factorable x = − b ± b2 − 4ac 2a − (−30) ± (−30)2 − 4(9)(8) x= 2(9) x = 30 ± 900 − 288 18 x = 30 ± 612 18 If you have not yet studied radicals and how to simplify them, youmay find it hard to simplify this one. But using a calculator you should findthat x = 30 + 612 = 3.04 18 x = 30 − 612 = -0.29 18The checking is left for you to do.Example 6. Solve for t: 3t² + 2 = 5t² - 4Solution: Begin by writing the equation in standard form.3t² + 2 = 5t² - 4 Not factorable 0 = 2t² - 62t² - 6 = 0 2t² = 6 t² = 3 t=± 3t= 3 and t = - 3 Note that the square root method worked for this equation. Notice,too, that it is an incomplete form of a quadratic equation where b = 0.The checking is left for you to do. 6

Examples 5 and 6 showed you how the square root method works. Itworks when a. the given equation is written in the form (x – p)² = d b. b = 0 in the equation ax² + bx + c = 0. Unless you are given specific instructions as to which method to use, youare free to choose the method you find most efficient and effective. Based on the examples, here is an outline for you in deciding whichmethod works in solving specific quadratic equations. Outline for Solving Quadratic Equations 1. Simplify both sides of the equation as completely as possible and put the equation into standard form. 2. Factor out any common factors and divide both sides of the equation by any common numerical factor to eliminate it. 3. If an incomplete form of a quadratic equation results and: a. if b = 0, then use the square root method. b. if c = 0, then use the factoring method. 4. If the left-hand side of the equation can be factored, use the factoring method. 5. If it looks too complicated to factor or it does not factor, then use the quadratic formula. 6. Check the equation either by substituting it into the original equation or by solving the equation using a different method.Try this outA. In each of the following exercises, use the method you think is the most appropriate to solve the given equation. Check your answers by using a different method, if possible. 1. x² + 6x + 5 = 0 2. x² + 6x – 5 = 0 3. 2r² + 1 = 3r 4. w² = 4w + 5 7

5. x² + 7x = 18 6. (a + 1)² - (a + 3)(a – 2) = a² + 6 7. 4x² = 16x –28 8. (x – 1)² = 5 9. (x – 1)² = 5x 10. (u + 2)² = 4(u + 5) 11. y² - 4y + 10 = 5(y + 2) 12. (t + 4)( t – 8) = 13 13. (n + 2)(n + 1) = 3 14. y² + 4y + 3 = 0 15. y² + 4y – 3 = 0 16. 2r² + 1 = 4r 17. w² = 4w 18. (z + 3)(z – 2) = (z – 3)(z + 2) 19. 2(3y + 4) = (y + 4)(y + 2) 20. 6x² = 24x - 30 21. (m – 3)² = 7m 22. (m – 3)² = 7 23. ( u + 2)² = u(u + 5) 24. 3y² - 7y – 12 = 4(2y² - 3) 25. (c – 3)(c + 6) = 22 26. (n + 2)(n + 1) = 3n 27. z² - 3z = 3z – 9 28. 16z + 12 = 3z² 29. 4z + 5 = z² + 12z + 21 30. 17z + 12 = 5z²B. Solve the equation 4x² - 4x – 15 = 0 by all three methods. Which is the easiest? the most difficult? why? 8

Lesson 2Solving Equations Reducible to Quadratic Equations So far you have studied how to solve quadratic equations. You have alsolearned to simplify equations that do not seem to be quadratic but with somemanipulations and simplifications, these equations turned out to be quadraticsafter all. A lot more equations can be reduced to quadratic equation and thus besolved by any of the methods you have learned. In this lesson, you will be working on fractional equations. These areequations that included radicals and other equations that can be transformed intoquadratic equations. Before attempting to proceed you must have prior knowledge working withrational expressions and working with radicals. Otherwise, proceed to the nextlesson.A. Rational Expressions whose solution Lead to Quadratic Equations:Example 1. Solve 3x + 2 = 4 x +1 x −1Solution: To solve this equation you should be able to get rid of the denominator. To do this you must first find the LCM of the denominators. Then multiply both sides by the LCM. Note that if -1 is substituted for x in x + 1 and 1 in x – 1 in thedenominator, the result is equal to zero. This is not allowed since divisionby zero is undefined. 3x + 2 = 4 x +1 x −1 3x + 2 = 4 (x+1)(x-1) multiply by the LCD x +1 x −13x(x + 1)(x −1) + 2(x + 1)(x −1) = 4(x + 1)(x −1) x +1 x −1 3x(x -1) + 2(x + 1) = 4(x + 1)(x -1) by distributive property 9