MATH 2 part 2

Solution: Add coefficients and annex their common radical factor= (4 + 5 + 6) 2= 15 22. 6 2 + 3 5 + 2 2 + 5 5Solution:= (6 2 + 2 2 ) + (3 5 + 6 5 ) Group similar radicals.=8 2 +9 5 Add the coefficients of similar radicals.You cannot add the coefficients of different radicand.3. 6 27 + 2 75 Solution:= 6 9 • 3 + 2 25 • 3 Factor each radicand= 6(3 3 ) + 2(5 3 ) Simplify= 18 3 + 10 3 Add the coefficients and annex= 28 3 their common radical factor4. 3 32 + 3 108 Factor each radicand such that oneSolution: factor is a perfect cube. = 3 8 • 4 + 3 27 • 4 = 2 3 4 + 33 4 = (2 + 3) 3 4 = 53 4 3

5. 4 3 + 3 3 25 25 Solution:=4 3 + 3 3 Split the numerator and denominator 25 25 Simplify and add the coefficients=4 3 + 3 Split the numerator and denominator 3 and rationalize the denominator.55 Simplify Get the LCD. LCD = 6=7 3 Add 56. 2 + 3 32 Solution:2+ 332= 2⋅ 3 + 3⋅ 2 33 22= 6+ 6 94= 6+ 6 32= 2 6+3 6 6=5 6 6 4

7. 9a7 + a3 Split a7 such that the exponent is exactly Solution: divisible by the index, 2. = 9a6 • a + a2 • a Simplify = 3a2 a + a a = (3a2 + a) a Keep in mind that our basic approach to these problems has been to firstput each term into simplest radical form before adding and /or subtracting.Try this outPerform the indicated operations.A. 1. 4 2 + 5 2 2. 5 3 + 2 3 + 10 3 3. 16 7 + 8 7 + 7 4. 15 11 + 3 11 + 5 11 + 6 11 5. 1 13 + 2 13 23B. 6. 2 27 + 3 16 7. 180 + 125 + 10 45 8. 3 36a + 100a 9. 75 + 2 27 + 12 10. 5 b + 4b 5

C. 11. 80 + 3 45 + 2012. 12x + 27x + 48x13. 3 + 2 3 4414. 5x2 y2 + 5x2 y2 415. 2 2 + 5 3 32D. What’s the Message? Simplify the expression and find the correct answer in the box, then fill inthe small box with the corresponding letter. Once you have filled in all the smallboxes with the appropriate letter, darken all the blank spaces to separate thewords in the secret message. What’s the message? Have fun!1. E : 7 + 3 + 4 7 9. U: 2 12 + 482. N : 18 + 2 32 + 50 10. H: 2 28y3 + 63y33. T: 3 81 + 2 9 + 5 16 11. O: 3 5 + 5 54. S :6 10 + 10 16 16 12. L: 4 320a3 + 2 180a35. W: 3 25x + 2 4x + 3 9x 13. R: 44x5 + 99x56. I: 100 + 144 + 25 14. C: 2 27 + 3 167. Q: 8 12 + 2 3 15. F: 648. Y: 2 + 12 49 16. V: 3 200 + 2 147 + 338 6

__ __ __ _ _ _ __ 8√3 12 + 6√3 12+6√3 5√7+√3√13 √14 + 3 7√10 __ __ __ _ _ _ __ 27 __7√10 7√10 5 √10 11√3+√5 √7 +√11 7√10 __ __ _ _ __ _ _ __ _√12 8√2 +√5 2√5 16√2 44a√5a √2 +2 √3 4√10+√ 3 58_ ___ _ __ _7 2√5 5x2√11x 4√3 53 7y√7y 2√5 __ _ _ __ _ __ __7√10 5√7 + √3 6√6+5√2 28√x 7y√7y 2√5 2/3√5 __ _ _ _ __ _ _ __ ___15√3+2√5 √2 + √3 16√2 5√7+√3 43√2+14√3 5√7+√3 5x2√11x _ _ _ _ _100+√5 18√3 8√3 27 53 2/3√5 15√3 LESSON 2 Subtraction of Radicals Subtraction of radicals follows the rule in addition of real numbers. Theoperation just like in addition is only possible for similar radical expressions.Examples: Simplify the following: 1. 5 3 − 2 3 7

Solution: Group their coefficients and subtract. 5 3−2 3 = (5 – 2) 3 Factor each radical in such a way that =3 3 one factor is a perfect square.2. 54 − 3 96 Simplify Solution: 54 − 3 96 = 9 ⋅ 6 − 3 16 ⋅ 6 ( )= 3 6 − 3 4 6= 3 6 −12 6 Subtract= −9 63. 25x2 − 9x2 Solution: 25x2 − 9x2 = 5x – 3x = 2x4. 27x3 y2 − 12x3 y2 Solution: 27x3 y2 − 12x3 y2= 9 ⋅ 3 ⋅ x2 ⋅ x ⋅ y2 − 4 ⋅ 3 ⋅ x2 ⋅ x ⋅ y2 Factor the radicals in such a way that one factor is a perfect square. 8

= 3xy 3x − 2xy 3x 9, 4, x2 and y2 are perfect squares.= (3xy − 2xy) 3x Group the coefficients and subtract= 1xy 3x or xy 3x5. 2 3 16 − 5 3 54 Factor the radicand in such a way that Solution: one factor is a perfect cube. 2 3 16 − 5 3 54 8 and 27 are perfect cubes hence, their = 2 3 8 ⋅ 2 − 5 3 27 ⋅ 2 cube root is 2 and 3 respectively. = 2(23 2 ) − 5(33 2 ) Simplify= 4 3 2 − 153 2= -11 3 26. 2 3 x2 y − 3 8x5 y4 Rename the expression such that the radicand can be expressed as a perfect cube and the exponents are exactly divisible by the index if possible then simplify. Solution:2 3 x2 y − 3 8x5 y4= 2 3 x2 y − 3 8x3x2 y3y 8, x3 and y3 are perfect cubes= 2 3 x2 y − 2xy 3 x2 y = (2 - 2xy) 3 x2 ySubtraction of radical expressions with fraction. 9

7. 2a 3 − 3a2 9 Solution:2a 3 − 3a2 9= 2a 3 − 3a2 Split the numerator and denominator 9 and simplify.= 2a 3 − a 3 or 3= 2a 3 − a 3 3=  2a − a  3 Group the coefficients and simplify.  3=  6a − a  3 3= 5a 3 3Addition and Subtraction of Radical ExpressionsWith the above knowledge, you can now perform a combination of operations.1. 5 3 + 6 2 − 2 3 + 5 2 Solution:5 3+6 2−2 3+5 2= 5 3−2 3+6 2+5 2 Group similar terms and perform the= 3 3 + 11 2 necessary operations.Addition of dissimilar radicands is not allowed. 10

2. 18x2 y − 8x2 y + 50x2 y Solution: 18x2 y − 8x2 y + 50x2 y= 9 ⋅ 2x2 y − 4 ⋅ 2x2 y + 25 ⋅ 2x2 y Factor in such a way that one factor is a perfect square.= 3x 2y − 2x 2y + 5x 2y Simplify= (3x – 2x + 5x) 2 y Perform the necessary operations.= 6x 2 y3. 9a7 - a3 Solution: 9a7 - a3= 9a6a - a2a= 3a3 a - a a= (3a3 – a) a4. 8a3b3 + 3 ab − 3 8a4b4 − 4 4a2b2 In this problem, no two radicands are identical and the radical expressionshave indices 2, 3 and 4. The radical expression 4 4a2b2 can be converted into aradical expression with index 2.Solution: 4 4a2b2 = 4 (2ab)2 Note that 4 (2ab)2 is the same as (2ab)42 or (2ab)12 = 2ab . You can nowcomplete the solution. 11

8a3b3 + 3 ab − 3 8a4b4 − 4 4a2b2Since 4 4a2b2 = 2ab then,8a3b3 + 3 ab − 3 8a4b4 − 2ab= 4 ⋅ 2 ⋅ a2 ⋅ a ⋅ b2 ⋅ b + 3 ab − 3 8a3 ⋅ a ⋅ b3 ⋅ b − 2ab Factor and simplify.= 2ab 2ab + 3 ab − 2ab 3 ab − 2ab= 2ab 2ab − 2ab + 3 ab − 2ab 3 ab Group similar radicands.= (2ab −1) 2ab + (1 − 2ab) 3 ab By additionTry This OutPerform the indicated operations.A. 1. 3 - 2 3 2. 5 7x - 4 7x 3. 3 6a - 7 6a 4. 4 x - 3 x - 2 x 5. 12 - 5 8 - 7 20B. 6. 36 - 100 7. 6 18 - 32 + 2 50 8. -2 63 + 2 28 + 2 7 9. 5 8 + 3 72 - 3 50 10. - 27 + 2 48 - 75 12

C. 11. 3 72m2 - 5 32m2 + 3 18m2 12. - 3 54 − 2 3 16 13. 2 3 27x − 2 3 8x 14. 5 4 32 − 3 4 162 + 184 2 15. 3 4 x5 y − 2x 4 xyD. Mental MathWhy are Oysters greedy?Perform and Simplify the following radicals. Write the letter in the box above itscorrect answer. Keep working and you will discover the answer the questions.H. 5 - 18 5 + 4 5 - 5 5 E. 4 7 + 7R. 35 - 6 5 I. 7 10 + 90T. 1 27 - 3 3 F. 3 2 - 2 32 Y. 4 27 + 5 7 - 2 48 3S. 2 27 - 3 48 + 1 75 5L. 3 16 + 3 3 54 A. 20 + 5 24 13

-2√3 -18√5 5√7 4√3+5√7 3√5 2 3√35 - 6√5 5√7 _ __ 113√2 113√2 -5√2 10√10 -5√3 -18√5-5√3 -18√5 5√7 Let ‘s Summarize When working with radicals, remember the following: 1. Put each radical into simplest form. Sums and difference of radical expressions can be simplified by applying thebasic properties of real numbers. The sum and difference of two radical expressions cannot be simplified if theradicals have different indices and different radicands. 2. Perform any indicated operations, if possible. 3. Make sure the final answer is also in simplest radical form. What have you learned A. Simplify each radical and perform the indicated operations. 1. 2 3 + 4 3 2. 5 5 - 5 3. 6 3 + 2 3 - 3 3 4. 2 7 - 5 7 -8 7 14

5. x 2 – 3x 2 + 5 26. 5 3a + 5 3a - 9 3a7. 3 2 - 2 328. 12 + 5 8 - 7 209. 6 2 - 5 8 + 2 3210. 7 72 + 3 20 - 4 511. 3 + 2 1 312. 6 1 + 2 2 2313. 10 49 - 50 214. 20 3 - 3a2 915. 2ab + 4 4a2b2 15

Answer KeyHow much do you know 1. 11 5 2. 7 3 3. 5 3 6 4. 14 2a 5. 2 b 6. -3 5 7. 4 2 8. 6 10x - 6 7 y 9. 11 5 10. 5 2xLesson 1Try this outA. 1. 9 2 2. 17 3 3. 25 7 4. 29 11 5. 7 13 6B. 6. 12 + 6 3 16

7. 41 5 9. U: 8 3 8. 28 a 10. H: 7y 7 y 9. 13 3 11. O: 2 5 10. 7 b 12. L: 44a 5a 11. 15 5 13. R: 5x2 11x 12. 11 3x 14. C: 12 + 6 3 13. 3 3 15. F: 8 2 7 14. 3xy 5 16. V: 43 2 + 14 3 2 17 15. 19 6 6C. What’s the Message? 1. E: 5 7 + 3 2. N: 16 2 3. T: 53 4. S: 7 10 5. W: 28 x 6. I: 27 7. Q: 18 8. Y: 2 + 2 3

S UCCE __ _ _ _ __ 7√10 8√3 12 + 6√3 12+6√3 5√7+√3 SS IS __ __ __7√10 7√10 27 7√10 ONL Y __ __ __ 2√5 16√2 44a√5a √2 +2 √3 F OR T HO 8 _ ___ 53 __ _ 7 2√5 5x2√11x WH 7y√7y 2√5 S E _ __ O R 28√x 7y√7y __ __ _7√10 5√7 + √3 2√5 NE V E _ __ __ __ ___ 16√2 5√7+√3 5√7+√3 5x2√11x 43√2+ 14√3 QUI T _ _ 27 53 18√3 8√3The Message: SUCCESS IS ONLY FOR THOSE WHO NEVER QUITLesson 2Try this outA. 1. - 3 2. 7x 3. -4 6a 18

4. - x E. 5 7 5. 2 3 - 10 2 - 14 5 I. 10 10B. F. 15 3 2 6. -4 Y. 4 3 + 5 7 7. 24 2 A. 3 5 8. 0 9. 13 2 2 10. 0C. 19 11. 7m 2 11. -7 3 2 12. 2 3 x 13. 19 4 2 15. x 4 xyD. Mental Health H. -18 5 R. 35 - 6 5 T. -2 3 S. 15 3 L. 11 3 2

Why are Oysters greedy?THE Y ARE _ __ __ 3√5 __ _ _-2√3 -18√5 5√7 4√3+5√7 2 3√35 - 6√5 5√7 SHE L L F I SH _ __ _ _ _ __ _-5√3 -18√5 5√7 113√2 113√2 -5√2 10√10 -5√3 -18√5What have you learned 1. 6 3 2. 4 5 3. 5 3 4. -11 7 5. -2x + 5 2 6. 3a 7. -5 2 8. 2 3 + 10 2 - 14 5 9. 4 2 10. 42 2 + 2 5 20

11. 5 3 312. 11 2 313. 30 214. 20 - a 3 or 60 − a 3 3315. 2 2ab 21

Module 4 Rational Algebraic Expressions What this module is about This module is about simplifying complex fractions and solving rational algebraicexpressions. As you go over the exercises, you will develop skills in simplifyingcomplex fractions and solving rational equations. You will also recall some concepts onthe changing of mixed expressions to rational expressions. Treat the lessons with funand 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. Change mixed expressions to rational expressions. 2. Simplify mixed expressions and complex fractions 3. Solve rational equationsHow much do you know1. To simplify 2x2 we will rewrite the complex fraction in this form. __ y3___ _3x_ 9y2a. 2x2 • _3x_ c. 2x2 • _ y3_y3 9y2 3x 9y2b. 2x2 • _9y2 _ d. 2x2 • _ y3_ y3 3x 9y2 3x2. The simplified form of #1 isa. 2x2 b. 2x c. 6x d. 6x y2 yy y23. To simplify x – 3 we will rewrite the complex fraction in this form. __ x + 5___ x+5 x 1

a. x – 3 • x + 5 c. x – 3 • x + 5 x+5 x x2 x + 5b. x – 3 • __x__ d. __x__ • x + 5 x+5 x+5 x–3 x+54. Simplify the rational expression in #3.a. ___x2 – 3x___ c. _x2 + 2x – 15_x2 + 10x + 25 x2(x + 5)b. _x2 + 2x – 15 _ d. ___x2 + 5x___ x2 + 5 x2 + 2x – 155. To simplify a2 – 13a + 40 we will first factor a2 – 13a + 40 in this form _ a2 – 4a – 32 __ _a – 5_ a+7a. (a + 4)(a +10) b. (a – 4)(a – 10) c. (a + 5)(a + 8) d. (a – 5)(a – 8)6. Then factor a2 – 4a – 32 to have _____ as numerator.a. (a + 4)(a +10) b. (a – 4)(a – 10) c. (a + 5)(a + 8) d. (a – 5)(a – 8) (a – 2)(a +16) (a + 2)(a – 16) (a – 4)(a + 8) (a + 4)(a – 8)7. The simplified form of # 6 is ______.8. Solve for k : _9_ _3_ k49. In solving _a – 4 _ _a + 8_ , we will multiply both sides by the LCD which is ___. 4 1610. The value of a is _____. 2

What you will do Lesson 1 Mixed Expressions and Complex Fractions Algebraic expressions such as a + b and 5 + x – y are called mixed c x+yexpressions.Definition of Mixed Expressions Algebraic expression which contain monomials and rational expressions. Changing mixed expressions to rational expressions is similar to changing mixednumber to improper fractions.Mixed Number to Improper Fraction3 2 or 3 + 2 3(5) + _2_ 5 555 3(5) + 2 5 15 + 2_ 17 55Mixed Expression to Rational Expressiona + _a2 + b_ a(a – b) + _a2 + b_a–b a–b a–b a(a – b) +_(a2 + b)_ a–b _a2 – ab +_a2 + b_ a–b _2a2 – ab + b_ a–bExample 1. Combine: 2 + _1_ x2 + _1_ _2x_ + _1_ _2x + 1_x xx xExample 2. Find 8 + __x2 – y2 _ x2 + y28 + __x2 – y2 _ _8(x2 + y2)_ + _x2 – y2 _ x2 + y2 x2 + y2 x2 + y2 3

_8(x2 + y2) + (x2 – y2)_ x2 + y2 _8x2 + 8y2 + x2 – y2_ x2 + y2 _9x2 + 7y2_ x2 + y2Practice Exercises 1Combine. 1. 4 + _1_ 3 2. a + _b_ c 3. 2 + _3_ x 4. 4 – _8_ a 5. 3 – _10_ e 6. 7 – _5_ d 7. 2z – z + 1 z 8. 3c – c + 1 c9. x + 3 – __4__ x–210. ___y___ + 5y y+1 If a fraction has one or more fractions in the numerator or denominator, it iscalled a complex fraction. Some complex fractions are shown below.31 ___8___ a+b 1–1 2__ _ a_ c____ x y__ b 1+142 a–b xy 3 cDefinition of Complex Fractions If a fraction has one or more fractions in the numerator or denominator it is calleda complex fractions. 4

_3_Consider the complex fraction 5___. To simplify this fraction, rewrite it as 7_ 8_3_ ÷ 7_ and proceed as follows.58_3_ ÷ 7_ _3_ • 8_ 24 Recall that to find the quotient you multiply by _8_the58 5 7 35 reciprocal of _7_ . 7 8 _a_Similarly, to simplify b___, rewrite it as _a_ ÷ c_ and proceed as follows. c_ b d d_a_ ÷ c_ _a_ • d_ ad Recall that to find the quotient you multiply by _d_thebd bc bc reciprocal of _c_ . c dSimplifying Complex Fractions _a_ Any complex fraction _ b___ , where b ≠ 0, c ≠ 0, and d ≠ 0, may be c_may be expressed as ad . d bcPractice Exercises 2 4. _12a2_ 9b2 __Simplify. _4a_ 1. _25_ 3b6 4___ _ 5_ 5. _2n_ 20 m2 __ 2. _2k_ _4n_ 3s2_ 3m2 _5k3_ 6. ___3___ 6s 2h + 5___ 3. _3x5_ _4h + 3_ 8___ 2h + 5 _x3_ 4 5

_1_ – _1_Example 3. Simplify ___x y _ _1_ + _1_ xy_1_ – _1_ _y_ – _x__x y _ _xy xy _ The LCD of the numerator and the denominator is xy._1_ + _1_ _y_ + _x_xy xy xy _y – x_ Subtract to simplify the numerator. ____xy____ _y + x _ Add to simplify the denominator. xy _y – x_ • __xy__ The reciprocal of _ y + x _ is __xy__ xy y + x xy y + x _y – x_ • __xy__ _y – x_ xy y + x y+x _1_ + _1_Example 4. Simplify __m n _ _2_+ _2_ 3m 3n_1_ + _1_ _n_ + _m__m n _ ___mn mn ___2_ + _2_ _2n_ + _2m__3m 3n 3mn 3mn _n + m_ Add to simplify the numerator. ____mn____ Add to simplify the denominator. _2n + 2m _ 3mn _n + m_• __3mn__ The reciprocal of _ 2n + 2m _ is __3mn__ mn 2n + 2m 3mn 2n + 2m _n + m_• __3mn__ mn 2(n + m) _ 3_ 2Example 5. Simplify : a – __3 _ a + 2___ 1 + __1__ a+2 6

a – __3 _ a(a + 2) – 3 a + 2___ ____a + 2_____1 + __1__ a+2 (a + 2) + 1_ a+2 (a + 2) – 3 • __a + 2__ a+2 (a + 2) +1 a – 1 • __a + 2__ a+2 a+3 a–1 a+3Example 6. Simplify : x – _x + 4 _ x + 1___ x+2x – _x + 4 _ x(x + 1) – (x + 4) x + 1___ ______x +1_______ x+2 x+2 x2 + x – x – 4 _____x + 1_____ x+2 _x2 – 4_ (x + 2)(x – 2) ___x + 1__ ____x + 1______ x+2 x+2 (x + 2)(x – 2) • ___1___ x+1 x+2 _x – 2_ x+1Try this outA. Find each sum. 6. _5n – 5 + __9__1. 8 + _3_ 3 n–1 a2. 3b + _b + 1_ 7. _ _4_ + 3t2 2b 2t + 13. _5_ + 8 8. 5 + _a2 + 11_ 3c a2 – 1 7

4. 2b + _4 + b_ 9. c + _c – 2_ b 3c – 15. b2 + __2__ 10. _ 2u – 1 + u b-2 u+2B. Combine. Simplify if possible. _1_ + _1_1. ___a b _ _1_ + _1_ 2a 2b _1_ + _1_2. ___g h _ _4_ + _4_ 5g 5h _1_ – _1_3. ___x y _ _3_ – _3_ 2x 2y _m_ + _2_4. ___n n2 _ 2 – _m_ n2 _5_ + _c_5. ___b2 b _ _c_ – 3 b 1 + _4_6. ___ k __ 1 – _16_ k2 1 + _5_7. ___ y __ 1 – _25_ y2 3 – __12__8. ___ s + 4 _ 2 – __8__ s+4 8

b – 2 – _25_9. ___ b–2 _b + 2 – __5__ b–2x–1– _ 3 _10. ___ x+1 _x + 5 + __3 __ x+1C. Combine. Simplify if possible._2n + 6_ – _n – 2 _1. ___n + 3 n+2 _ __n2 – 9 _ n2 + 5n + 6_2t + 2_ – _t – 5_2. ___ t t–3 _ __t2 – 4 _ t2 – 3t_x – 1_ – _x + 1 _3. __ x + 1 x–1 __x – 1_ + _x + 1 _x+1 x–1_b + 2_ + _b – 2 _4. __ b – 2 b+2 __b + 2_ – _b – 2 _b–2 b+2_x + 2_ – _ 3 _5. __ x x+1 __x – 1_ + _ 3 _x2 + x x+1D. Solve and simplify when needed. __a2 + 2a _1. __ a2 + 9a + 18 _ __a2 – 5a _ a2 + a – 30 __x2 + 4x – 21 _2. __ x2 – 9x + 18 _ __x2 + 3x – 28 _ x2 – 10x + 24 9

_y2 + 8y + 15_3. __ y2 + y – 6 _ _y2 + 2y – 15_ y2 – 2y – 3 __b2 – 6b + 5 _4. __ b2 + 13b + 42 _ __b2 – 4b + 3 _ b2 + 3b – 18 __h2 + 2h – 15 _5. __ h2 + h – 12 _ __h2 + 3h – 10 _ h2 + 2h – 8We can now summarize the two methods for simplifying a complex fraction.Method #1 Simplify the numerator and denominator of the complex fractions separately.Then divide the simplified numerator by the simplified denominator.Method #2 Multiply numerator and denominator of the complex fraction by the least commondenominator of all the denominators appearing in the complex fraction. Lesson 2 Solving Rational EquationsRational equations are equations containing rational expressions.Example 1. Solve _11_ – _2_ _1_ 6 2x 3x The LCD is 6x_11_ – _2_ _1_2x 3x 66x _11_ – _2_ _1_ 6x 2x 3x 63(11) – 2(2) = x33 – 4 = x29 = x 10

Check: _1_ _11_ – _2_ 6 2x 3x__11 _ – __2 _  _1_2(29) 3(29) 6__11 _ – __2 _  _1_ 58 87 6_11(87) – 2(58)_  _1_(58)(87) 6_957 – 116_  _1_(58)(87) 6_841_  _1_ Check using calculator.5 046 6  _1_ _1_ 66Example 2. Solve _x_ + _x_ 10 + x 34_x_ + _x_ 10 + x 3412 _x_ + _x_ 10 + x 12 34_12x_ + _12x_ 120 + 12x 344x + 3x 120 + 12x 7x 120 + 12x7x – 12x 120– 5x 120 x – 24Check: _–24_ + _–24_  10 + (–24) 34 11

–8+–6 – 14  –14 – 14Example 3. Solve _y_ – _y – 1 _ 1 23_y_ – _y – 1 _ 1 236 _y_ – y – 1 16 236 _y_ – y – 1 16 23_6y_ – _6(y – 1) 6 23 Be careful to put parentheses around y – 1 ; otherwise an incorrect solution maybe found._6y_ – _6y + 6_ 6 233y – 2y + 2 6 y4Check:_4_ – _4 – 1 _  1 23_4_ – _3 _  1 2 3 1 2–1Example 4. Solve __x__ __2_ _ + 2 x–2 x–2__x__ __2_ _ + 2x–2 x–2Multiply both sides by x – 2, the LCDx – 2 __x__ _ _2__ + 2 x – 2 x–2 x–2 12

x – 2 __x__ _ _2__ (x – 2) + 2 (x – 2) x–2 x–2 x = 2 + 2x – 4 –x =–2 x= 2Check: __2__ __2_ _ + 2 2–2 2–2 The proposed solution 2 cannot be a solution because 2 makes bothdenominators equal to 0. The equation is meaningless and has no solution.Example 5. Solve __2b__ + __1_ _ __2___ b2 – 2 b – 2 b+2 Since b2 – 2 = (b – 2)(b + 2), use (b – 2)(b + 2) as the LCD. Multiply both sidesby (b – 2)(b + 2). (b – 2)(b + 2) __2b__ + __1_ _ __2___ (b – 2)(b + 2) b+2 b2 – 4 b–2(b – 2)(b + 2) __2b__ + (b – 2)(b + 2) __1_ _ __2___ (b – 2)(b + 2) b+2 b2 – 4 b–2 2b + (b + 2)(1) 2 (b – 2) 2b + b + 2 2b – 4 b –6Check:__2(–6)__ + __1_ _  __2___(–6)2 – 4 (–6)– 2 (–6)+ 2__–12 __ + __1__  __2_36 – 4 –8 –4– 3 _ + _1__  _1_–8 –8 2 _1_  _1_ 2 2 13

Try this outA. Solve each equation and check your answers.1. _1_ _x_ 6. _2 _11_ 42 z 52. _2_ _5_ 7. _6 _4_ 5 b 12 e e3. _9_ _3_ 8. _x _x_ 6 k4 244. _15_ _30_ 9. _3 + _2_ 5 g8 yy5. _7_ _8_ 10. _4 + _2_ 1 h3 c3B. Solve each equation and check your answers.1. _9 5 _1_ a a2. _3m_ + 2 _1_ 5 43. _2d_ 5 d 74. _1_ + _2_ 1 2q5. _b + 1_ _b + 2_ 2 36. _n – 4 _ n+2 37. _3p_ + p 5 28. _ 9 _ 3 x+1 14

9. _ 9 _ 3 x–210. _2a + 3_ _3_ a 2C. Solve each equation and check your answers.1. _2p + 8 _ _10p + 4_ 9 272. _5y – 3 _ _15y – 2 7 283. _2b – 1 + 2 _1_ b24. _d _ _17 + d 2d 255. _n – 2 + _n + 1 _10_ 4 336. _r + 5 _r – 1 _7_ 3 4 47. _4 _ _2__ 3 q q+18. _6 + _ _1__ 3 s s–29. __2_ + _1__ _5_ c–1 c+1 410. __5_ + _10_ 7 z–2 z+2D. Try Harder!Solve each equation and check your answers.1. _x + 1_ __4__ + 6 x–3 x–32. _5x + 1_ _5x – 5 + _3x – 1_ 3x + 3 5x + 5 x+1 15

3. __2_ _ 3__ __12__ x–3 x+3 x2 – 94. __1_ _ 3__ __–10_ _ y+5 y–5 y2 – 255. ____3y___ _ __5y____ ___2___ _ y2 + 5y + 6 y2 + 2y – 3 y2 + y – 2 16

Let’s Summarize Rational expression is the quotient of two polynomials with denominator notequal to zero.Definition of Mixed Expressions Algebraic expression which contain monomials and rational expressions.Definition of Complex Fractions If a fraction has one or more fractions in the numerator or denominator it is calleda complex fractions.Simplifying Complex Fractions _a_ Any complex fraction _ b___ , where b ≠ 0, c ≠ 0, and d ≠ 0, may be c_may be expressed as ad . d bcTwo methods for simplifying a complex fraction.Method #1 Simplify the numerator and denominator of the complex fractions separately.Then divide the simplified numerator by the simplified denominator.Method #2 Multiply numerator and denominator of the complex fraction by the least commondenominator of all the denominators appearing in the complex fraction.Rational equations are equations containing rational expressions. 17

What have you learnedPerform the given operation. Write your answer in lowest term. Assume that nodenominator is equal to zero.1. To simplify xp we will rewrite the complex fraction in this form. __ q3___ _p2_ qx2a. xp • _ p2 c. _p2 • _ q3_q3 qx2 xp qx2b. xp • _qx2 d. _p2 • _qx2_ q3 p2 xp q32. The simplified form of #1 isa. _x3 b. q3_ c._ x3 d. _ q3_ q2 p q2pqx2 qx23. To simplify _6x + 3_ we will rewrite the complex fraction in this form. __ x ___ _2x + 1_ 8a. _6x + 3_• _2x + 1_ c. _6x + 3_• _ _8__ x8 x 2x + 1b. _6x + 3_• _x_ d. _6x + 3_• _ _x__ 2x + 1 8 8 2x + 14. Simplify the rational expression in #3.5. Simplify __x2 – x – 6_ we will first factor x2 – x – 16 in this form _ x2 + 2x – 15 __ _x2 – 2x – 8_ x2 + x – 20a. (x + 3)(x + 2) b. (x – 3)(x + 2) c. (x + 6)(x + 1) d. (x – 6)(a + 1)6. Then factor x2 + 2x – 15 to have _____ as numerator.a. (x + 3)(x + 2) b. (x – 3)(x + 2) c. (x + 6)(x + 1) d. (x – 6)(a + 1) (x – 3)(x + 5) (x + 5)(a – 3) (x – 3)(x + 5) (x + 5)(a – 3) 18

7. The denominator of #5 is _______.a. (x – 4) (x + 2) b. (x – 2)(x + 4) c. _(x + 8)(x + 1)_ d. _(x –1)(a + 8)_ (x – 4)(x + 5) (x + 4)(a – 5) (x –10)(x + 2) (x + 10)(a – 2)8. The simplified form of #5 is ______.9. Solve _3_ + __1__ _1_ . The LCD is ______. x x–5 2x10. The value of x is _____. 19

Answer KeyHow much do you know1. b 6. d 10. a = 82. d 7. _a + 7_3. b a+44. a 8. k = 125. d 9. 16Lesson 1 4. ab4Practice Exercises 1. 25 162. __4__ 5. _3_ 5sk2 2 3. _3x2 _ 6. ___3___ 2 4h + 3Lesson 1 6. _5n – 5 (n – 1) + 9(3)_ __32__Try this out 3(n – 1) 3n – 31. _8a + _3_ a2. _6b2 +_b + 1_ 7. _4 + 3t2 ( 2t + 1)_ 6t3 + 3t2 + 4_ 2b 2t + 1 2t + 13. _5_+ 24c_ 8. 5(a + 1) 2 + a2 + 11_ _6a2 + 6_ 3c a2 – 1 a2 – 14. _2b2 + b + 4_ 9. c(3c – 1) + c – 2_ _3c2 – 2_ b 3c – 1 3c – 15. _b3 – 2b2 + 2__ 10. 2u – 1 + u(u + 2)_ _u2 + 4u – 1_ b–2 u+2 u+2B. _b + a_ • __2ab__ 1 _b + a_ ab 2(b + a)1. ___ab__ _ _2b + 2a_ 2ab 20

_h_+_g_ _h_+_g_ • __5gh__ _5_ gh 4(h + g) 42. ____gh__ __ _4h +_4g_ 5gh _y_– _x_ _y_– _x_ • __2xy___ _2_ xy 3(y – x) 33. ____xy _ _3y – 3x_ 2xy _mn_+ 2_ _mn_+ 2_4. ___ n2 _ 2n2 – m _2n2 – m_ n2 _5_+ bc_ _5_+ bc_ • __b___ _5_+ bc_5. _____b2 _ b2 c – 3b bc – 3b2 _c – 3b_ b k + 4_ _k + 4_ • ___k2___ _ k___ k k2 – 16 k–46. __ k __ k2 – 16_ k2 y + 5_ y + 5_ • ___y2 ___ _ y___ y y2 – 25 y–57. ___ y __ y2 – 25_ y23(s + 4) – 12__8. ___ s + 4 _ _3_ 22(s + 4) – 8__ s+4_(b – 2)(b – 2) – 25 _9. ___ b–2 _ _b2 – 4b – 21_ _b – 7_ b2 – 9 b–3_(b + 2)(b – 2) – 5_ b–2_(x – 1)(x + 1) – 3 _10. ___ x + 1 _ ___x2 – 4___ _x – 2_ x2 + 6x + 8 x+4(x + 5)(x + 1) +_3 __ x+1 21

C. _ _2(n + 3)(n + 2) – (n – 2)(n + 3) _1. _________(n + 3)(n + 2)_________ __(n – 3)(n + 3) _ (n – 2)(n + 3)_(2n2 + 10n + 12) – (n2 + n + 6)_ _n2 + 11n + 18_ (n + 3)(n – 3) (n + 3)(n – 3)_(2t + 2)(t – 3)_– t(t – 5)_2. ___ t (t – 3) _ _(2t + 2)(t – 3)_– t(t – 5)_ (t – 2)(t + 2) __(t – 2)(t + 2) _ t+3 t(t – 3)_(x – 1)(x – 1)_–_(x + 1)(x + 1) _3. __ (x + 1)(x – 1) _ _ – 4x_ 2x2 + 2_(x – 1)(x – 1)_+_(x + 1)(x + 1) _ (x + 1)(x – 1)_(b + 2)(b + 2)_+_(b – 2)(b – 2) _4. __ (b – 2)(b + 2) _ _2b2 + 8_ 8b_(b + 2)(b + 2) –_(b – 2)(b – 2) _ (b – 2)(b + 2)_(x + 2)(x + 1)_– 3x _5. __ ____x(x + 1) _ _x2 + 2_ 4x – 1 __ x – 1 + 3 _ x(x + 1)D. _a + 6_ ___a(a + 2) _ a+91. __ (a + 2)(a + 9) _ ___a(a – 5) _ (a – 5)(a + 6) _(x + 7)(x – 3) _ 12. __ (x – 3)(x – 6) _ _(x + 7)(x – 4) _ (x – 4)(x – 6) _(y + 5)(y + 3)_ _ _y + 1_3. __ (y – 2)(y + 3) y–2 _(y + 5)(y – 3)_ (y + 1)(y – 3) 22

_(b – 5)(b + 1)_ _b – 5 _4. __ (b + 7)(b + 6)_ _ b+7 _(b – 3)(b – 1) _ 1 (b + 6)(b – 3) _(h + 5)(h – 3) _5. __ (h + 4)(h – 3) __ _(h + 5)(h – 2) _ (h + 4)(h – 2)Lesson 2 6. z _10_ 111. x _1_ 2 7. e _2_ 52. b _24_ 5 8. x 24 9. y _1_3. k 12 54. g 4 10. c 125. h _21_ 6. n –5 8 7. p 2B.1. a 2 8. x 2 9. x 52. m –35 10. a –6 12 4. d 343. d –7 5. h 6 6. r –24. q 45. b 1C.1. p 52. y 23. b _2_ 7 23

with solutions 37. _4(q + 1) – 2q_ q(q + 1)4q + 4 – 2q 3(q2 + q) 3q2 + 3q 2q + 43q2 + q – 4 0(3q + 4)(q – 1) 0q _–4_ q 1 38. _6(s – 2) + s_ 3 s2 – 2s 3(s2 – 2s) 6s – 12 + s3s2 – 13a + 12 0 (3s – 4)(s – 3) 0 s _4_ s 3 39. _2(c + 1) + (c – 1)_ _5_ c2 – 1 4 _5_ _2c + 2 + c – 1_ 4 c2 – 1 5(c2 – 1) 0 4(3c + 1) 0 5c2 – 12c – 9 (5c + 3)(c – 3) c –3_ c 3 7 5 7(z2 – 4)10. _5(z + 2) + 10(z – 2) 0 z2 – 4 0 5z + 10 + 10z – 20 7z2 – 15z – 18 (7z + 6)(z – 3)z –6_ z 3 7D. Try this out1. _x + 1_ _4 + 6 (x – 3)_ x–3 x–3 x+1 4 + 6x – 18 24

x+1 6x – 14 15 5x x 32. _5x + 1__ 5(x – 1) + 5(3x – 1)_ 3(x + 1) 5(x + 1)(5x + 1)(x + 1) 5(x – 1) + 5(3x – 1)_ 3(x + 1) 5 _5(5x + 1)_ 3 5x – 5 + 15x – 5 5(5x + 1) 25x + 5 3(20x – 10) 60x – 30) 35 35x x13. _2(x + 3) – 3(x – 3) __12__x2 – 9 x2 – 92x + 6 – 3x + 9 12 – x + 15 12 x3The equation is meaningless and has no solution because the denominator is 0.4. _(y – 5) – 3(y + 5) _–10_ _y2 – 25 y2 – 25(y – 5) – 3(y + 5) –10y – 5 – 3y – 15 –10 – 2y – 20 –10 y –5The equation is meaningless and has no solution because the denominator is 0.5. _____3y___ _ __5y____ _____2___ _ (y + 3)(y + 2) (y + 3)(y – 1) (y + 2)(y – 1)_____3y___ __5y(y + 2) – 2(y + 3)(y + 3)(y + 2) (y + 3)(y + 2)(y – 1)_3y(y + 3)(y + 2)_ __5y(y + 2) – 2(y + 3)(y + 3)(y + 2) (y – 1)3y(y – 1) 5y2 + 10y – 2y – 62y2 + 11y – 6 0(2y – 1)(y + 6) 0 y –6 y _1_ 2 25

What you have learned1. b 6. b2. c 7. a3. c 8. 14. _24_ 9. 2x (x – 5) 10. x _25_ x5. b 7 26

Module 4 Searching for Patterns in Sequences, Arithmetic, Geometric and others What this module is about This module is about geometric sequence. As you go over the exercises,you will develop skills in identifying a geometric sequence and ability to give thefirst few terms of the sequence. Treat the lesson with fun and take time to goback if you think you are at a loss. What you are expected to learn This module is designed for you to: • demonstrate knowledge and skills related to geometric sequence and apply these in solving problems. • describe a geometric sequence in any of the following ways:  Giving the first few terms of the sequence  Giving the formula for the nth term How much do you know Tell whether or not each number sequence is a geometric sequence. Forthose that are geometric sequence, give the common ratio. 1. 5, 10, 20, 40, 80, . . . 2. 512, 128, 32, 8, 4, . . . 3. 8, 8, 8, 8, 8, 8, . . . 4. 1, 3, 6, 10, 15, . . . 5. 16, 24, 36, 54, 81, . . . 6. 40.5, 13.5, 4.5, 1.5. . . .

7. 1 , 1 , 1 , 1, 2, . . . 842 8. 1 , 1 , 1 , 1 , 1 , . . . 23456 9. 3, 6, 9, 12, 15, . . . 10. 1, 3, 9, 27, 81, . . .B. Review the following sequences, tell which are geometric and arithmetic sequences. State their common ratio or common difference. 1. 100, -50, 25, -12,5, . . . 2. 1, 2, 4, 8, . . . 3. 40, 20, 10 5, . . . 4. -10, -5, 0, 5, 10, . . . 5. 5, 15, 45, 135, . . . 6. 1, -3, 9, -27, 81, . . . 7. 25, 21, 17, 13, . . . 8. 11, 22, 33, 44, 55, . . . 9. 3 , 2, 5 , 3, 7 , . . . 222 10. 1 , 1 , 2 , . . . 239C. Write the next two terms of the following geometric sequences. 1. 2, 14, 98, . . . 2. -2, -6, -18, . . . 3. -5, 10, -20, . . . 4. 1, 4, 16, . . . 5. 7, 3.5, 1.75, . . . 6. 3, 1, 1 , . . . 3 7. 0.8, 0.08, 0.008, . . . 8. 1 , 1 , 1 , . . . 248 2

9. 3, 3, 3 3 10. a, 11ax, 121ax, . . . What you will do Lesson 1 The First Few Terms of a Geometric Sequence Making one fold on a sheet of a paper, you can form two triangles. Nowfold the paper again, and count the rectangles formed (count only the smallesttriangle as shown below). Continue this process until you can no longer fold thepaper.12 4 8 16The number of rectangles formed produce a geometric sequence,1, 2, 4, 8, 16, . . . Notice that each term after the first may be formed by multiplying theprevious term by 2. A geometric sequence or progression is a set of terms in which each termafter the first is obtained by multiplying the preceding term by the same fixednumber called the common ratio which is commonly represented by r. The common ratio may be integral or fractional, negative or positive, and itcan be found by dividing any term by the term that precedes is. 3, 9, 27, . . . is a geometric sequence, 3 is multiplied to any term to get thenext term.Therefore, we can say that 3 is the common ratio.3

Examples:1. Determine whether each sequence is geometric . If so, find the common ratio.a. 2, 4, 6, 8, . . .2 is being added to each term to get the next term. Therefore, thesequence is an arithmetic sequence. The common difference is 2b. 2, 4, 8, 16, . . .2 is being multiplied to each term to get the next term. Therefore, thesequence is a geometric sequence. The common ratio is 2.c. 1 , 1 , 1 , 1 , . . . 3 9 27 811 is being multiplied to each term to get the next term. Therefore, the3sequence is a geometric sequence. The common ratio is 1 . 32. Find the next two terms of the geometric sequence 4, 12, 36, . . .Solution: First, find the common ration by finding the quotient of any twoconsecutive terms from the right to left.36 = 3 The common ratio is 3.1212 = 34To get the next term, a. multiply 36 by 3 to get the fourth term. The fourth term is 108. b. Multiply 108 by 3 to get the fifth term. The fifth term is 324.Therefore, the next two terms of the geometric sequence are 108 and 324. 4

Try this outA. Determine whether each sequence is geometric. If so, find the common ratio. 1. 4, 20, 100, 500, . . . 2. 7, 14, 21, 28, . . . 3. 5, 10, 20, 40, . . . 4. 1, 4, 9, 16, . . . 5. 4, 12, 36, 108, . . . 6. 1, 5, 10, 15, . . . 7. 6, 12, 24, 48, . . . 8. 1 , 1 , 1 , 1 , . . . 2 4 8 16 9. 9, 6, 4, 8 , . . . 3 10. 3 , 9 , 27 , 81, . . . 2 4 8 16B. Find the next two terms of each geometric sequence. 1. 90, 30, 10, . . . 2. 2, 6, 18, . . . 3. 20, 30, 45, . . . 4. 729, 243, 81, . . . 5. 6, 24, 96, . . . 6. 7, 21, 63, . . . 7. 5, -10, 20, . . . 8. 1 , 1, 2, . . . 2 9. 1 , 1 , 1 , . . . 27 9 3 10. - 1 , 1 , -1, . . . 42 5