MATH 2 part 1

m B. Use the definition of a n to evaluate each expression. 2 2 1) 273 6)  8 3  27  3 7) − (81)34 2) 16 2 3 3) (− )4 8)  9 2 83 4 2 3 4) 1253 9) 812 2 10) (− 243)35 5) 325 C. Math Integration The First Man to Orbit the Earth In 1961, this Russian cosmonaut orbited the earth in a spaceship. Who was he? To find out, evaluate the following. Then encircle the letter that corresponds tothe correct answer. These letters will spell out the name of this Russian cosmonaut.Have fun! 1 Y. 12 Z. 14 O. 9 U. 131) 144 2 Q 25 R. 5 E. 16 I. 6 1 G. 5 H. 25 A. 27 E. 92) 1692 F. –4 G. 4 A. 81 E. -81 13) 1253 14) 2163 15) 6254 36) 9 27) (− )2 838) (− 27 )4 3

9) (− 343 )1 R. –7 S. 7 3 E. –16 I. 16 N. − 8 P. 810) (− 32)54 27 27 311) −  16 4  81Answer: 1 2 3 4 5 6 7 8 9 10 11Source: Math Journal – Volume XI-Number 2 SY 2003-2004 ISSN 0118-1211 Now you can extend your knowledge on rational exponent notation. Using thedefinition of negative exponents, you can write -m 1 an= m anExamples:Simplify each expression. −1 1 = 1 =1 Change the negative exponent to positive exponent. Then simplify.1) 16 2 = 1 16 4 16 2 ( )−2 1 = 1 Follow the same procedure as in example 1.2) 27 3 = 2 3 27 2 27 3 1 = (3)2 1 = 9Try this outSimplify each expression.

1) 16 - 1 -1 4 6) 81 4 -3 -32) 81 4 7) 9 2 -1 -53) 49 2 8) 64 6 -3 9)  4 - 1 24) 4 2  25  1 -1 - 10) 121 25) 25 2 Let’s Summarize As you mentioned earlier, you assume that all your previous exponent propertieswill continue to hold for rational exponents. These properties are restated here.A. For any nonzero real numbers a and b and rational numbers m and n, 1) Product Rule am • an = am+n 2) Quotient Rule am = am−n an ( )3) Power Rule am n = amn 4) Product-power rule (ab)m = ambm 5) Quotient-power rule  a m = am b bm 1B. In general, n x = xn if x and n are real numbers and n > 0.C. For any real numbers a and positive integers m and n where n > 1 , an( )m= n m =n am . a What have you learnedA. Express the following in radical form.

1) (5x )1 ( )1 2 6. 6x3y3 5 ( )1 7. (2y )1 22) 3x2y2 2 2 8. (3a )1 63) y3 3 9. (8x )1 34) 254 1 15) x5 10. 6x2B. Express the following in exponential form.1) 3a 6) x2y42) 4 x3y4 7) 5 32x5y53) 3 9xy 8) 6 y 5 ( )3 9)  4 4 254) xy ( )3 ( )2 10) 3 x5) 5 27x5C. Change the indicated roots into radicals and evaluate. 3 31) 362 6) 814 5 32)  25 2 7)  49 2  64   1213) (225 )1 4 2 8)  27 34) (169)32  64 5) (27 )2 2 3 - 9) 8 3 -1 10) 81 2

Answer KeyHow much do you know A. 1) 9 2) 4 81 3) 3 x2 4) 25 81 5) 3 3x 6) 3 27 125 7) 102 8) 4 25 9) − 144 10) 4b10 1 B. 1) 642 2 2) 83 3) (ab)12 5 4) 244 5) (5xy)32 ( )1 6) 10m2 3 5 7) 94 3 8) 16 2

( )1 9) 24b4 4 ( )1 10) x2y2 2 C. 1) 16 = 4 2) − 121 = −11 3) 169 = 13 4) 3 64 = 4 5) 3 82 = 4 6) 4 163 = 8 7) 253 = 125 8) 5 − 323 = −8 9) 365 = 7776 10) 6 645 = 32Try this outLesson 1 A. 1) 4 a3 2) 6 m5 3) 23 x 2 4) 35 x 2 ( )5) 5 3x 2 ( )3 6) 4 2y 7) 93 8) 24 y3

9) 4 163 10) 3 82B. 1) (7a )1 2 ( )1 2) 27p6 3 ( )1 3) 81x 8 y16 4 ( )1 4) 25w 4 2 ( )1 5) 8m6n9 3 ( )1 6) 32r s10 15 5 ( )1 7) 64x2 2 ( )1 8) 32y 5 5 9) (121)21 10) (64 )1 3C. 1) 2 2) 8 3) –8 4) 32 5) 2 5 6) 9 4 7) 27 8) 16Answer: TURKEYLesson 2A. 1) 6 2) 10

3) 5 Y 4) 8 U 5) 3 R 6) –4 I 7) 3 G 8) –2 A 9) 2 G A 3 R 3 10) 2B. 1) 9 2) 64 3) 16 4) 25 5) 4 6) 4 9 7) –27 8) 27 8 9) 729 10) -27C. 1) 12 2) 13 3) 5 4) 6 5) 5 6) 27 7) 4 8) 81 9) -7

10) 16 I11) − 8 N 27Negative ExponentsTry this out 1) 1 2 2) 1 27 3) 1 7 4) 1 8 5) 1 5 6) 1 3 7) 1 27 8) 1 32 9) 5 2 10) 1 11What have you learnedA. 1) 5x 2) 3x2y2 3) 3 y2 4) 4 253

5) 5 x 6) 5 6x3y3 7) 2y 8) 6 3a 9) 3 8x 10) 6 xB. 1) (3a )1 2 ( )1 2) x3y4 4 3) (9xy )1 3 4) (xy )3 2 ( )2 5) 27x5 5 ( )1 6) x2y4 2 ( )1 7) 32x5y5 5 5 8) y 6 4 9)  4 2  25  3 10) 3x 2( )C. 1) 36 3 = 216 2)  25 5 = 3125 64 32768 3) 225 = 15 ( )3 4) 169 = 2197 ( )5) 3 27 2 = 9

( )6) 4 81 3 = 277)  49 3 = 343 121 13318)  3 27 4 = 81 64 256( )9)3 1 2 = 1 8 410) 1 = 1 81 9

Module 2 Rational Algebraic Expressions What this module is about This module is about addition and subtraction of rational expressions. As you goover the exercises, you will develop skills in finding the LCD of rational expressions, andfinding the sum and difference of the given rational expressions. You will also recallsome concepts on how to find the least common denominator, add and subtractfractions. Treat the lessons with fun and take time to go back and review if you thinkyou are at a loss. What you are expected to learnThis module is designed for you to: 1. Recall LCM, addition and subtraction of rational numbers 2. Recall how to simplify rational expression 3. Find LCD of rational expressions 4. Add rational expressions having the same denominator 5. Add rational expressions having different denominators 6. Subtract rational expressions having the same denominator 7. Subtract rational expressions having different denominatorsHow much do you know1. Give the least common denominator of _3_ and _5_ . 15y2 36y4a. 36y2 b. 36y4 c. 90y2 d. 180y42. Give the least common denominator of __7_ and __2_ . 8–a a–8a. (8 – a) b. (a – 8) c. a and b d. a or b 1

3. Give the least common denominator of ___3____ and _____2_____ . x2 – 9x x(x – 9)(x – 5)a. x2 – 9x b. x(x – 9)(x – 5) c. (x – 9)(x – 5) d. x2 + 14x + 454. Rewrite 17 ____ with indicated denominator. 9r 36r25. Rewrite _12__ __________ with indicated denominator. r2 – 7r r(r – 7)(r + 7)6. Add __a___ and ___b _ .2(a2 – b2) 2(a2 – b2)7. Add _p_ and __q _ .p–q q–p8. Add r – 3s_ and ____r – s__ .r2 – 4s2 r2 – 4rs + 4s29. Find the difference between ____2_____ and ______m_ _ . (m + 3)(2 – m) (m + 3)(2 – m)10. Subtract ____3x___ from _____x_ _ . x2 – 3xy + 2y2 x2 + xy – 2y2 What you will do Review 1 Finding Least Common Multiples Let’s recall LCM (Least Common Multiple) by solving mathematical problems. As marketing strategy, Jack, the owner of JACKASS Snack Bar decided to give free sandwich to every 6th customer and every 8th customer will get a free fruit shake. Suppose 64 customers come into the store. Which ones will win both a free fruit shake and a free sandwich?To solve this problem you need to find the LCM of 6 and 8. The least common multiple (LCM) of two numbers is the smallest nonzeromultiple which the numbers have in common. 2

1st Method: List the nonzero multiples of 6 and 8 until you reach a common multiple. Multiples of 6: 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66 . . . Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, . . . The LCM of 6 and 8 is 24.2nd Method: List the prime factors of each number. 6=2•3 8=2• 2 •2 2 • 3 • 2 • 2 = 24 The LCM of 6 and 8 is 24.Before you answer the problem, answer these questions first.1. Which three customers were first to get free sandwiches? Free fruit shakes?2. Which customer was first to get both a free fruit shake drink and a free sandwich? In question #1, you’re correct if your answers are: the 6th, 12th, and 18thcustomers were the first three to get free sandwiches and the 8th, 16th, and 24thcustomers were the first three to get free fruit shakes. The 24th customer was the first to get both a free fruit shake drink and a freesandwich. Who will win both a free fruit shake and a free sandwich? The 24th and the 48thcustomers.Try this outFind the LCM of each pair of numbers. Use mental math.1. 4, 8 2. 6, 5 3. 9, 7 4. 12, 60 8. 14, 265. 9, 27 6. 45, 18 7. 35, 28 Review 2 Adding and Subtracting Fractions You already studied that you can use this LCM when you add and subtractfractions with unlike denominators. The least common denominator (LCD) of two ormore fractions is the least common multiple (LCM) of the denominator. A least common denominator (LCD) is the least number that all denominatorsdivide into without a remainder. 3

Example: Add 1 and 2 . 65The denominators are 6 and 5 and they are not the same. You need to find the LCD.1 2 5 + 12 The LCD is 30.6 5 30 30Add the numerators and write the same denominator. 5 + 12 17 30 30Try this outAdd. Express each sum in lowest term.1. 7 8 6. 5 11 66 12 182. _9_ 13 7. 9 3 16 16 11 43. _3_ 7 8. 11 3 16 8 16 44. 4 2 9. 7 13 57 9 545. 5 9 10. 8 10 68 49 14Take a break and solve this magic square.The sum in each row, column, and diagonal must be 1. _1_ 15 _1_ _1_ 53 _2_ 15You can also use LCD in subtracting fractions with unlike denominators. 4

Example: Subtract 1 from 5 . 3 12The denominators are 3 and 12 and they are not the same. You need to find the LCD._5 1 _5 _4_ The LCD is 12.12 3 12 12Subtract the numerators and write the same denominator. 5 – 4 _1_ 12 12Try this outSubtract. Express each difference in lowest term.1. 12 7 6. 3 11 66 12 482. _5_ 2 7. 4 14 10 9 7 493. 7 10 8. 13 1 8 12 20 44. 5 6 9. 3 12 12 15 9 365. 5 9 10. 9 13 68 10 15Take a break and solve.Place the digits 2, 4, 6 and 8 in the boxes at the right to make1. The smallest possible sum.2. The largest possible sum.3. The smallest possible difference.4. The largest possible difference. Lesson 1 Finding LCD of Rational Expressions A least common denominator (LCD) of rational expressions is the leastexpression that all denominators divide into without a remainder. It is found by 5

multiplying together each different factor the greatest number of times it appears in anydenominator.Examples1. Find the least common denominator for _5_ and _3_ . 6p2 4p3In finding LCD of rational expressions, you follow the following steps.Step 1. Completely factors all the denominators. 6p2 = 2 • 3 • p • p 4p3 = 2 • p • p • p • 2Step 2. Take each different factor the greatest number of times that it appears as a factor in any of the denominators. The greatest number of times 2 appears is two, the greatest number of times 3appears is one, and the greatest number of times p appears is three.Step 3. The least common denominator is the product of all factors found in step #2.LCD = 2 • 2 • 3 • p • p • p = 12p3Here is another way of finding LCD. 6p2 = 2 • 3 • p • p 4p3 = 2 • p • p • p • 2 LCD = 2 • 3 • p • p • p • 2 = 12p32. Find the least common denominator for _12_ and ___5__ . 5a a2 – 3aFactor each denominator.5a = 5 • aa2 – 3a = a • (a – 3)LCD = 5 • a • (a – 3) = 5a(a – 3)3. Find the least common denominator for _____1____ and ____4_+__ . b2 – 4b – 5 b2 – b – 20Factor each denominator.b2 – 4b – 5 = ( b – 5) ( b + 1)b2 – b – 20 = ( b – 5 ) (b + 4)LCD = ( b – 5 ) ( b + 1) (b + 4)4. Find the least common denominator for __1__ and ___3___ . m–5 5–m 6

The expression 5 – m can be written as – 1(m – 5), because of this, either m – 5or 5 – m can be used as the LCD. Once the least common denominator has been found, you can use thefundamental property to rewrite rational expressions with the LCD.4. Rewrite 9x ___ with the indicated denominator. 25 50x9x 9x(2x) To get a denominator of 50x, multiply both25 25(2x) numerator and denominator by 2x. 18x2 50x5. Rewrite _12y__ ___________ with the indicated denominator. y2 + 8y y(y + 8)(y – 4)_12y__ __12y_ • y – 4 y(y-8) is a factor of y2 + 8y. You stilly2 + 8y y(y + 8) y – 4 need to multiply by (y – 4). __12y(y – 4)__ y(y + 8)(y – 4)Try this outA. Find the least common denominator of the following rational expressions.1. _2_ , _8_ 6. _ 2 , __3__ 15k 4k 18x3 9x – 362. _5_ , _2_ 7. _ 7 , __1___ 5y3 15y5 5y – 30 6y – 363. _ 1__ , __7_ 8. _ 6 , __ 3__ 25m3 10m7 8n 12n – 244. __1__ , _3_ 9. _ 5 , __3___ 5a2b3 15a5 12p + 60 p2 + 5p5. _ 7___ , ___9___ 10. _ 1 , __10__ 6w – 12 9w – 18 d2 + 7d 5d + 35B. Find the least common denominator of the following rational expressions.1. _ 3__ , ____2____ 6. _ _ 2_ , ____7_____ 8y + 16 y2 + 3y + 2 9a – 18 a2 – 7a + 10 7

2. __3_ , __5__ 7. _ 2 , ____7___ p–3 3–p r2 + 6r r2 + 3r –183. _ _2__ , ____5_____ 8. _ ___7__ , _____3____ k2 – 5k k2 – 2k – 15 2y2 + 7y – 4 2y2 – 7y + 34. ____4_____ , ____7_____ 9. _ _1___ , _____3____ x2 + 2x – 35 x2 + 3x – 40 5q2 + 13q – 6 5q2 – 22q + 85. _ __1____ , ____9____ 10. _ 8 ___ , ____10____ z2 + 4z – 12 z2 + z – 30 2d2 – 11d + 14 2d2 – d – 21C. Rewrite each rational expression with the given denominator.1. _–11 _ ____ 6. _ –5p_ _______ k 8k 6p + 18 24p + 722. _12_ ___ 7. _ 6 _____________ 35y 70y3 m2 – 4m m(m – 4)(m + 1)3. 15m2 ____ 8. __15_ ___________ 8k 32k4 a2 – 9a a(a – 9)(a + 8)4. _19z_ ______ 9. _ _36r ______________ 2x – 6 6x – 18 r2 – r – 6 (r – 3)(r + 2)(r + 1)5. _ –2a__ _______ 10. _ 4m ___ _________________ 9a – 18 18a – 36 m2 – 8m + 15 (m – 5)(m – 3)(m + 2) Review 3 Simplifying Rational Expressions The fundamental property of rational expressions permits us to write arational expression in lowest terms, in which numerator and denominator have nocommon factor other than 1.Examples:1. Write the rational expression 3x – 12 in lowest terms. 5x – 203x – 12 3 (x – 4) Begin by factoring both numerator and denominator.5x – 20 5 (x – 4) 8

3 (x – 4) Then use the fundamental property. 5 (x – 4) 3 52. Write the rational expression x2 + 2x – 8 in lowest terms. 2x2 – x – 6x2 + 2x – 8 _(x + 4)(x – 2)_ Factor both numerator and denominator.2x2 – x – 6 (2x + 3)(x – 2) Then, use the fundamental property. _x + 4_ 2x + 33. Write a – b in lowest terms. b–aThe denominator b – a is also equal to –1(– b + a) or –1(a – b). With these,a – b _1(a – b) _1_ –1b – a –1(a – b) –14. Express 8m2 + 6m – 9 in lowest terms. 16m2 – 98m2 + 6m – 9 (2m + 3)(4m – 3) 2m + 316m2 – 9 (4m + 3)(4m – 3) 4m + 35. Write p3 + q3 in lowest terms. p2 – q2p3 + q3 (p + q)(p2 – pq + q2) p2 – pq + q2p2 – q2 (p + q) (p – q) p–q Lesson 2 Adding Rational Expressions To find the sum of two rational expressions, use a procedure similar to that ofadding two fractions.Adding Rational Expressions If _P_ and _R_ are rational expressions, then QQ _P_ _R_ _P + R_ QQ Q 9

The first example below shows how addition of rational expressions compareswith that of rational numbers.Examples:1. Addition of rational numbers and expressions having common denominatorsRational Numbers Rational Expressionsa. _5 _ _ 8_ b. __3x__ __4x___ 11 11 x–5 x–5 Since the denominators are the same, the sum is found by adding thenumerators and using the common denominator.a. _5 _ _8_ _5 + 8_ b. _3x_ _4x_ 3x + 4x_ 11 11 11 x–5 x–5 x–5 _7x_ _13_ or 1 _2_ x–5 11 112. Addition of rational numbers and expressions having different denominators To add two rational numbers or expressions with different denominators, use thesteps given in examples a and b below. These are the same steps that are used to addfractions with different denominators.a. _1 _ _ 7_ b. _3_ _2_ 12 15 3y 4yFirst find the least common denominator.a. 12 = 2 • 2 • 3 b. 3y = 3 • y 4y = y • 2 • 215 = 3•5LCD =2 • 2 • 3 • 5 = 60 LCD =3 • y • 2 • 2 = 12y Now rewrite each rational number and expression as a fraction with the leastcommon denominator, with 60 and 21y, as the denominators of a and b respectively.a. _1 _ _ 7_ _1(5)_ _7(4)_ b. _3_ _2_ _3(4)_ 2(3)_ 12 15 12(5) 15(4) 3y 4y 3y(4) 4y(3) 5_ 28 _12_ _6_ 60 60 12y 12y 10

Since the rational numbers and expressions are now having commondenominators, add the numerators. Copy the same denominator and write in lowestterms if necessary.a. 5 + 28 b. 12_+ 6 60 12y 33 11 18_ 3_ 60 20 12y 2y3. Add _m_ and ___n .m2 – n2 m2 – n2_m_ ___n m + n_m2 – n2 m2 – n2 m2 – n2 ____m + n___ (m + n)(m – n) __1__ m–n4. Add __ 4___ and ____b__ .b2 + 2b – 8 b2 + 2b – 8__ 4___ ____b____ ___4 + b__b2 + 2b – 8 b2 + 2b – 8 b2 + 2b – 8 ___4 + b___ __1__ (b – 2)(b + 4) b – 25. Add _x_ and _x .x2 – 1 x+1_ x _ __x__ Factor x2 – 1 and find the LCD(x – 1)(x + 1) (x + 1) The LCD is (x + 1)(x – 1). _ x _ __x (x – 1)__ Multiply the numerator and denominator(x – 1)(x + 1) (x + 1)(x – 1) of the second fraction by (x – 1).x + x (x – 1)_ Add the numerators(x – 1)(x + 1)_x + x2 – x_ ____x2 ____(x – 1)(x + 1) (x – 1)(x + 1) 11

6. Add _1_ and __1 . m+n m–n _1_ __1 _1(m – n)__ ___1(m + n) _m+n m – n (m + n)(m – n) (m + n)(m – n) ___m – n__ ___m + n_ _ (m + n)(m – n) (m + n)(m – n) _m – n_ _m + n_ m2 – n2 m2 – n2 _m – n_+_m + n_ m2 – n2 2m _ m2 – n27. Add _ 2x _ and ___x + 1 . x2 + 5x + 6 x2 + 2x – 3 To begin, factor the denominator completely. _ 2x _ ___x + 1____(x + 2)(x + 3) (x + 3)(x – 1) The LCD is (x + 2)(x + 3)(x – 1). By the fundamental property of rationalexpressions, _ 2x _ ___x + 1____ _ 2x(x – 1) _ ___(x + 1)(x + 2)__(x + 2)(x + 3) (x + 3)(x – 1) (x + 2)(x + 3)(x – 1) (x + 3)(x – 1)(x + 2) Since the two rational expressions have the same denominator, add theirnumerators._ 2x(x – 1) _ ___(x + 1)(x + 2)__(x + 2)(x + 3)(x – 1) (x + 3)(x – 1)(x + 2)2x(x – 1) + (x + 1)(x + 2)_ (x + 2)(x + 3)(x – 1)2x2 – 2x + x2 + 3x + 2_ (x + 2)(x + 3)(x – 1)___ 3x2 + x + 2___ It is convenient to leave the denominator in factored from.(x + 2)(x + 3)(x – 1) 12

Try this outA. Add and express your answer in lowest term if necessary.1. 3 _2_ 6. _ _3s_ __5s__ pp 9s – 8 9s – 82. _8_ _5_ 7. _ r _ ___6_ __ r2 r2 r2 + 3r –18 r2 + 3r –183. _ y__ _2y_ 8. _ 2y__ ____1____ k–5 k–5 2y2 + 7y + 3 2y2 + 7y + 34. __z__ _ 1__ 9. _ _6__ ____b____ z+1 z+1 b2 + 5b – 6 b2 + 5b – 65. _ a – b _a + b_ 10. _ 2d ___ ____– 7 ____ 77 2d2 – 11d + 14 2d2 – 11d + 14B. Add and express your answer in lowest term if necessary.1. 1 _2_ 6. _ _1_ _1_ 2y y m–8 m2. _3_ _5_ 7. _ 2 _ 5_ 3c 4c p–3 p23. _4_ _1_ 8. _ 2x __3x__ 3y y2 x+y 2x + 2y4. _ 2_ _ 3_ 9. _ 5 __4__ 10x 5x2 b–6 b+65. _2xb _a – b _ 10. t – 1 t+1 4x2 2x t+1 t–1C. Add and express your answer in lowest term if necessary.1. _ _ 3 _ __– 2 _ b2 – 5b + 6 b2 – b – 22. 2m + 3_ __m – 4 __ m2 – 4m + 4 m2 + m – 63. _r – 3s_ ____r – s__ _ r2 – 4s2 r2 – 4rs + 4s2 13

4. _ 3p + 2q__ ____3p – 2q___ 3p2 + pq – 2q2 3p2 + 5pq + 2q25. _ 3m ___ _4m_– 1_ m2 + mn – 2n2 m2 – n26. 2x + y_ ___x – 2y_ __ 2x2 + xy – y2 x2 + 3xy + 2y27. _ __x + 3y__ ____x – y__ _ x2 + 2xy + y2 x2 + 4xy + 3y28. _ _r + 1_ ___r – 1__r2 – 3r – 10 r2 + r – 309. _ m ___m_– 1__m2 – 1 m2 + 2m + 110. _ 2w + 1_ ___3w + 5___3w2 +10w – 8 2w2 + 5w – 12 Lesson 3 Subtracting Rational ExpressionsTo find the difference of two rational expressions, use the following rule.Subtracting Rational Expressions If _P_ and _R_ are rational expressions, then QQ _P_ _R_ _P – R_ QQ QExamples:1. Subtract. _ 8_ b. __10c__ __5c___ a. _22 _ 30 c2 – 12 c2 – 12 30 Since the denominators are the same, the difference is found by subtracting thenumerators and using the common denominator. 14

a. _22 _ _ 8_ b. __10c__ __5c___ 30 30 c2 – 12 c2 – 12 _14_ or _7_ _5c__ 30 15 c2 – 12 Subtraction of rational expressions with different denominators uses the samesteps used to subtract fractions with different denominators.2. Subtract.a. _11 7_ b. _2_ _2_ 12 15 3y 7yFirst find the least common denominator.Least common denominator Least common denominator 3y = 3 • y12 = 2 • 2 • 3 7y = y • 715 = 3 •5LCD =2 • 2 • 3 • 5 = 60 LCD =3 • y • 7 = 21y Now rewrite each rational expression as a fraction with the least commondenominator, with 60 and 12y, as the denominator._11 _ _ 7_ 11(5) _7(4)_ _2_ _2 _2(7) 2(3)_ 12 15 12(5) 15(4) 3y 7y 3y(7) 7y(3) 55 28 _14_ _6_ 60 60 21y 21y Since the fractions are now having common denominator, subtract thenumerators, copy the same denominator and write in lowest terms if necessary. 55 – 28 14_– 6 60 21y 27 _9_ _8_ 60 20 21y3. Find 12 _8_ . m2 m212 _8_ _4_ By the definition of subtraction.m2 m2 m2 15

4. Find __9__ _3 x–2 xThe least common denominator is x(x – 2).__9__ _3_ __9x__ _3(x – 2)_x–2 x x(x – 2) x(x – 2) __9x_– 3x + 6_ x(x – 2) _6x + 6_ x(x – 2)4. Find __1__ __1__ m–1 m+1Here the least common denominator is (m – 1)(m + 1). Hence,__1_ __1__ _ 1__ + _–1__m–1 m+1 m–1 m+1 _(m +1) + (–1)(m – 1)_ (m – 1)(m + 1) _m + 1 + –m + 1_ (m – 1)(m + 1) ___ 1 + 1____ (m – 1)(m + 1) ___ __2______ (m – 1)(m + 1)5. Find ____q_____ ______3_____. q2 – 4q – 5 2q2 – 13q + 15 Start factoring each denominator. _____3______ (q – 5 )(2q – 3)____q____ ______3_____ _____q_____q2 – 4q – 5 2q2 – 13q + 15 (q + 1)(q – 5) Now, rewrite each of the two rational expressions with the least commondenominator, which is (q + 1)(q – 5)(2q – 3). Then subtract numerators. 16

_____q_____ _____3______(q + 1)(q – 5) (q – 5 )(2q – 3)_ ___q(2q – 3)____ ______3(q + 1)_____(q + 1)(q – 5)(2q – 3) (q + 1)(q – 5 )(2q – 3)_ ___2q2 – 3q_____ _____3q + 3______(q + 1)(q – 5 )(2q – 3) (q + 1)(q – 5 )(2q – 3)_ 2q2 – 3q_– 3q – 3_(q + 1)(q – 5 )(2q – 3)_ _2q2 – 6q_– 3_ __(q + 1)(q – 5 )(2q – 3)6. Find ____4_____ ___ m2 – 1__ . m2 – 2m – 3 m2 – 5m + 6Start factoring each denominator.____4_____ ___m2 – 1__ ____4_____ __ m2 – 1____m2 – 2m – 3 m2 – 5m + 6 (m + 1)(m – 3) (m – 2 )(m – 3) Now, rewrite each of the two rational expressions with the least commondenominator, which is (m + 1)(m – 2)(2m – 3). Then subtract numerators.___ 4(m – 2)_____ ___(m2 – 1)(m + 1)___(m + 1)(m – 2)(2m – 3) (m + 1)(m – 2)(2m – 3)___ 4m – 8_______ __m3 + m2 – m – 1___(m + 1)(m – 2)(2m – 3) (m + 1)(m – 2)(2m – 3)4m – 8 – m3 – m2 + m + 1_(m + 1)(m – 2)(2m – 3) – m3 – m2 + 5m – 7 _(m + 1)(m – 2)(2m – 3)The denominator is usually left in factored form in a problem of this type.Try this outA. Subtract and express your answer in lowest term if necessary.1. 3 _2_ 6. _ _3s_ __5s__ pp 9s – 8 9s – 8 17

2. _8_ _5_ 7. _ r _ ___3_ __ r2 r2 r2 + 3r –18 r2 + 3r –183. _ y__ _2y_ 8. _ 2y__ ____1____ k–5 k–5 2y2 – 7y + 3 2y2 – 7y + 34. __z__ _ 1__ 9. _ _b__ ____1____ z+1 z+1 b2 + 5b – 6 b2 + 5b – 65. _ a – b _a + b_ 10. _ 2d ___ ____ 7 ____ 77 2d2 – 11d + 14 2d2 – 11d + 14B. Subtract and express your answer in lowest term if necessary.1. 9 _r_ 6. _ _1_ ___a__ 10 2 a–b 4a – 4b2. _8_ _2_ 7. __–6 __3 _ 5 4c p2 – 4 2p + 43. _4m + 5 _m + 2_ 8. _ _1_ ___1__ 36 x2 – 9 3x + 94. _5r + 4_ _2r – 3 _ 9. _ 15 __3__ 10x 5x2 4k2 k+25. __2x _ _ 3x _ 10. _ 2t – 1 _2t + 1_ x + y 2x + 2y t+1 t–1C. Subtract and express your answer in lowest term if necessary.1. _ 1 __ 1 _ b2 – 1 b2 + 3b + 22. 8 __ 2 __3 – 7y 7y – 33. _ ___– 4____ ____1____ _ a2 – ab – 6b2 a2 + ab – 2b24. _ x _ _____3x____ x2 + xy – 2y2 x2 – 3xy + 2y25. _ – 1___ _3_ m2 + mn – 2n2 m2 – 3mn + 2n2 18

6. 4m – 3n _____4m – n_ __ 16m2 – n2 16m2 + 8mn + n27. _ _ 3a – b __ ___ – 2a + b__ _ 6a2 – 13ab – 5b2 6a2 – 19ab + 10b28. _ _ p + q ____q – p___ 3p2 + 2pq – q2 6p2 – 5pq + q29. _ m – n __ _m_+ n_ _ m2 + 2mn + n2 m2 – mn – 2n210. _ 2x + y_ ___x – 2y__ _ 2x2 + xy – y2 x2 + 3xy + 2y2D. Find the sums or differences.1. 8 _3_ 7 __m – 2 5m 5m(m – 2)2. –1 __3__ __ 4 __ 7z x + 2 7z(x + 3)3. 4 __6__ __ 1 __ r2 – r r2 + 2r r2 + r – 24. 6 __1__ __ __2_ __ k2 + 3k k2 – k k2 + 2k – 35. _ x _ ______3x___ __2x___ x2 + xy – 2y2 x2 – 3xy + 2y2 x2 – 4y2 Let’s Summarize Rational expression is the quotient of two polynomials with denominator notequal to zero. The fundamental property of rational expressions permits us to write arational expression in lowest terms, in which numerator and denominator have nocommon factor other than 1. 19

A least common denominator (LCD) of rational expressions is the leastexpression that all denominators divide into without a remainder. It is found bymultiplying together each different factor the greatest number of times it appears in anydenominator.Adding Rational ExpressionsIf _P_ and _R_ are rational expressions, then _P_ _R_ _P + R_ QQQ QQSubtracting Rational ExpressionsIf _P_ and _R_ are rational expressions, then _P_ _R_ _P – R_ QQQ QQWhat have you learned1. Give the least common denominator of __a__ , _b_ and _c_ . 36y2 12y5 6y42. Give the least common denominator of ___4__ and __10b_ . b4 – 16 b–23. Give the least common denominator of ___9 + x___ and _____y_____ . 36x2 – y4 (6x – y2)(x – y)4. Rewrite _ w4_ ____ with indicated denominator. 2d2 64d45. Rewrite __4m__ _____________ with indicated denominator. m2 – 4m m(m – 4)(m + 4)6. Add __ x___ and ____y _ .x(x2 – y2) x(x2 – y2)7. Add a and __b _ . x x+y8. Add __ x + 4_ and _ x – 5__ .(2 – x )(x + 3) (x – 2 )29. Find the difference between ____4k_____ and __ 7__ _ . (2k + 1)(3 – k) (2k + 1)(3 – k)10. Subtract 5a + 1 from ____4a_ _ . 2 – 3a 6a2 – a – 2 20

Key to correctionHow much do you know?1. d. 180y42. d. a or b since (8 – a) is the same as – (a – 8)3. b. x(x – 9)(x – 5) since x2 – 9x = x( x – 9)4. 17 17(4r) 68r 9r 36r2 36r25. _12__ _12( r + 7)__ _12r + 84__ since r2 – 7r = r(r – 7) r2 – 7r r(r – 7)(r + 7) r(r – 7)(r + 7)6. __a___ __ _b _ ___ a + b___ ___1___ 2(a2 – b2) 2(a2 – b2) 2(a – b)(a + b) 2(a – b)7. _p_ __q _ _p(q – p) + q (p –q) _qp – p2 + qp – q2_ p–q qp – p2 – q2 + qp q–p (p – q)(q – p)– p2 + 2qp – q2 1 or just take any of (p – q) or (q – p) since (p – q) = – (q – p)– p2 + 2qp – q2_p + –(1)(q) p – q 1(p – q) p-q8. r – 3s_ ____r – s__ ___r – 3s___ ____r – s_ _ r2 – 4s2 r2 – 4rs + 4s2 (r – 2s)(r + 2s) (r – 2s)(r – 2s) (r – 3s)(r – 2s)__ _ (r – s)(r + 2s) __(r – 2s)(r – 2s)(r + 2s) (r – 2s)(r – 2s)(r + 2s) r2 – 5rs + 6s2__ r2 +rs – 2s2 __(r – 2s)(r – 2s)(r + 2s) (r – 2s)(r – 2s)(r + 2s) 2r2 – 4rs + 4s2__ or 2(r2 – 2rs + 2s2) __(r – 2s)(r – 2s)(r + 2s) (r – 2s)(r – 2s)(r + 2s)9. ____2_____ ______m_ _ ____2 – m___ ___1___ (m + 3)(2 – m) (m + 3)(2 – m) (m + 3)(2 – m) (m + 3)10. ____3x___ _____x_ _ ____3x___ ____x____ _ x2 – 3xy + 2y2 x2 + xy – 2y2 (x – 2y)(x – y) (x + 2y)(x – y) 21

_ _ 3x(x + 2y)___ __ _ x(x – 2y) _ _(x + 2y)(x – 2y)(x – y) (x + 2y)(x – 2y)(x – y)____3x2 + 6xy______ __ _ x2 – 2xy _ _(x + 2y)(x – 2y)(x – y) (x + 2y)(x – 2y)(x – y)3x2 + 6xy – x2 + 2xy_ ____2x2 + 8 xy_____ or _ 2( x2 + 4xy)____(x + 2y)(x – 2y)(x – y) (x + 2y)(x – 2y)(x – y) (x + 2y)(x – 2y)(x – y)Try this outReview 1 2. 6, 5  30 3. 9, 7  63 4. 12, 60  60 6. 45, 18  90 7. 35, 28 140 8. 14, 26  1821. 4, 8  85. 9, 27 27Review 2Add: Express each sum in lowest term.1. 15 or 2 1 6. 15 + 22 37 or 1 _1_ 62 36 36 362. 1 3 7. 36 + 33 69 1 25 8 44 44 443. _3 + 14_ 17 1 1_ 8. _11 + 12 23 1 _7_ 16 16 16 16 16 164. 28 + 10 38 1_3_ 9. 42 + 13 55 1 1_ 35 35 35 54 54 545. 20 + 27 47 1 23 10. 16 + 70 86 43 24 24 24 98 98 49 _8_ _1_ _2_ 15 15 5 _1_ _1_ _7_ 5 3 15 _4_ _3_ _2_ 15 5 15 22

Subtract. Express each difference in lowest term.1. 5 6. 12 – 11 1_ 6 48 482. 45 – 20 25 5_ 7. 28 – 14 14 2 90 90 18 49 49 73. 21 – 20 1 8. 13 1 2 24 24 20 44. 25 – 24 _1_ 9. 12 – 12 = 0 60 60 365. _20 – 27 –7 10. 27 – 39 –12 – 2 24 24 30 30 5Take a break and solve.Place the digits 2, 4, 6 and 8 in the boxes at the right to make1. The smallest possible sum. 2 4 5 68 62. The largest possible sum. 8 6 5 1 24 23. The smallest possible difference. 4 2 1 86 64. The largest possible difference. 8 6 2 1 24 2Lesson 1A. Find the least common denominator for the following rational expressions.1. _2_ , _8_ LCD = 60k 15k 4k2. _5_ , _2_ LCD = 15y5 5y3 15y53. _ 1__ , __7_ LCD = 50m7 25m3 10m7 23

4. __1__ , _3_ LCD = 5a2b5 5a2b3 15a55. _ 7___ , ___9___ LCD = 6(w – 2), 9(w – 2) = 54(w – 2) 6w – 12 9w – 186. _ 2 , __3__ LCD = 9(2x3), 9(x – 4) = 18x3 (x – 4) 18x3 9x – 367. _ 7 , __1___ LCD = 5(y – 6), 6(y – 6) = 30(y – 6) 5y – 30 6y – 368. _ 6 , __3__ LCD = 2(4n), 12(n – 2) = 24(n – 2) 8n 12n – 249. _ 5 , __3___ LCD = 12( p + 5), p(p + 5) = 12p(p + 5) 12p + 60 p2 + 5p10. _ 1 , __10__ LCD = d( d + 7), 5(d + 7) = 5d(d + 7) d2 + 7d 5d + 35B. Find the least common denominator for the following rational expressions.1. _ 3__ , ____2____ LCD = 8(y + 2), (y + 2)(y + !) = 8(y + 2)(y + 1) 8y + 16 y2 + 3y + 22. __3_ , __5__ LCD = p – 3 or 3 – p p–3 3–p3. _ _2__ , ____5_____ LCD = k(k – 5), (k – 5)(k + 3) = k(k – 5)(k + 3) k2 – 5k k2 – 2k – 154. ____4_____ , ____7_____ LCD = (x + 7)(x – 5), (x – 5)(x + 8) = (x + 7)(x – 5)(x + 8) x2 + 2x – 35 x2 + 3x – 405. _ __1____ , ____9____ LCD = (z + 6)(z – 2), (z – 5)(z + 6) = (z – 2)(z – 5)(z + 6) z2 + 4z – 12 z2 + z – 306. _ _ 2_ , ____7_____ LCD = 9(a – 2), (a – 5)(a – 2) = 9(a – 2)(a – 5) 9a – 18 a2 – 7a + 107. _ 2 , ____7___ LCD = r(r + 6), (r – 3)(r + 6) = r(r + 6)(r – 3) r2 + 6r r2 + 3r –188. _ ___7__ , _____3__ LCD = (2y – 1)(y + 4), (2y – 1)(y – 3)2y2 + 7y – 4 2y2 – 7y + 3 = (y + 4)(2y – 1)(y – 3) 24

9. _ _1___ , _____3____ LCD = (5q – 2)(q + 3), (5q – 2)(q – 4) 5q2 + 13q – 6 5q2 – 22q + 8 = (q + 3)(5q – 2)(q – 4)10. _ 8 ___ , ____10____ LCD = (2d – 7)(d – 2), (2d – 7)(d + 3) 2d2 – 11d + 14 2d2 – d – 21 = (d – 2)(2d – 7)(d + 3)C. Rewrite each rational expression with the given denominator.1. _–11 _ – 88_ 6. _ –5p_ _– 20p__ k 8k 6p + 18 24p + 722. _12_ _24y2 7. _ 6 6m + 6____ 35y 70y3 m2 – 4m m(m – 4)(m + 1)3. 15m2 _60m2k3 8. __15_ __15a + 40___ 8k 32k4 a2 – 9a a(a – 9)(a + 8)4. _19z_ _27z__ 9. _ _36r __36r2 + 36r____ 2x – 6 6x – 18 r2 – r – 6 (r – 3)(r + 2)(r + 1)5. _ –2a__ _– 4a__ 10. _ 4m ___ ____4m2 + 8m______ 9a – 18 18a – 36 m2 – 8m + 15 (m – 5)(m – 3)(m + 2)Lesson 2A. Add and express your answer in lowest term if necessary.1. 3 _2_ 5 6. _ _3s_ __5s__ 8s___ ppp 9s – 8 9s – 8 9s – 82. _8_ _5_ 13 7. _ r _ ___6_ ___ r + 6__ 1__r2 r2 r2 r2 + 3r –18 r2 + 3r –18 (r + 6)(r – 3) r – 33. _ y__ _2y_ 3y 8. _ 2y__ ____1____ 2y + 1___ k–5 k–5 k–5 2y2 + 7y + 3 2y2 + 7y + 3 (2y + 1)(y + 3) __1__ y+34. __z__ _ 1__ z + 1 1 9. _ _6__ ____b____ 6+b 1__z+1 z+1 z+1 b2 + 5b – 6 b2 + 5b – 6 (b + 6)(b – 1) b – 15. a – b _a + b_ 2a 10. _ 2d ___ ____– 7 ___ __ 2d____ 7 77 2d2 – 11d + 14 2d2 – 11d + 14 (2d – 7)(d – 2) 1___ d–2 25

B. Add and express your answer in lowest term if necessary.1. 1 _2 1 + 2(2) 5_2y y 2y 2y2. _3_ _5_ 3(4) + 5(3) _27_3c 4c 12c 12c3. _4_ _1_ 4y + 33y y2 3y24. _ 2_ _ 3_ 2x + 3(2) 2x + 610x 5x2 10x2 10x25. _2xb _a – b 2xb + 2x(a – b) 2xb + 2xa – 2xb _2xa_ _a_ 4x2 2x 4x2 4x2 4x2 2x6. _ _1_ _1_ m + m – 8 _ 2m – 8_m – 8 m m(m – 8) m(m – 8)7. _ 2 5_ 2(p2) + 5(p – 3) 2p2 + 5p – 15p – 3 p2 p2(p – 3) p2(p – 3)8. _ 2x __3x__ 2(2x) + 3x 7x___ x+y 2x + 2y 2(x + y) 2(x + y)9. _ 5 __4__ 5(b + 6) + 4(b – 6) 5b + 30 + 4b – 24 __9b + 6___b – 6 b + 6 (b – 6)(b + 6) (b – 6)(b + 6) (b – 6)(b + 6)10. t – 1 t + 1 (t – 1)(t – 1) + (t + 1)(t + 1) t2 – 2t + 1 + t2 + 2t + 1 2t2 + 2__t+1 t–1 (t + 1)(t – 1) (t + 1)(t – 1) (t + 1)(t – 1)C. Add and express your answer in lowest term if necessary.1. _ _ 3 _ __– 2 __ 3(b + 1) _ _–2(b – 3)__ 3b + 3 – 2b +6__b2 – 5b + 6 b2 – b – 2 (b – 3)(b – 2) (b – 2)(b + 1) (b – 3)(b – 2)(b + 1) __ b + 9 __ (b – 3)(b – 2)(b + 1)2. 2m + 3_ __m – 4 _ 2m + 3 m – 4_ _ m2 – 4m + 4 m2 + m – 6 (m – 2)(m – 2) (m + 3)(m – 2) (2m + 3)(m + 3) + (m – 4)(m – 2) (m + 3)(m – 2)(m – 2) 26

2m2 + 9m + 9 + m2 – 6m + 8 3m2 + 3m + 17___ (m + 3)(m – 2)(m – 2) (m + 3)(m – 2)(m – 2)3. _r – 3s_ + ____r – s__ _ r – 3s____ + ____r – s____ r2 – 4s2 r2 – 4rs + 4s2 (r – 2s)(r + 2s) (r – 2s)(r – 2s) (r – 3s)(r – 2s) +(r – s)(r + 2s) (r + 2s)(r – 2s)(r – 2s) r2 – 5rs + 6s2 + r2 + rs – 2s2 2r2 – 4rs + 4s2 __ (r + 2s)(r – 2s)(r – 2s) (r + 2s)(r – 2s)(r – 2s)4. _ 3p + 2q__ + ____3p – 2q___ _ 3p + 2q__ + ____3p – 2q___ 3p2 + pq – 2q2 3p2 + 5pq + 2q2 (3p – 2q)(q + 1) (3p + 2q)(q + 1) (3p + 2q)(3p + 2q) + (3p – 2q)(3p – 2q) (3p – 2q)(3p + 2q)(q + 1) 9p2 + 12pq + 4q2 + 9p2 – 12pq + 4q2 (3p – 2q)(3p + 2q)(q + 1) 18p2 + 8q2 _____ (3p – 2q)(3p + 2q)(q + 1)5. _ 3m ___ + _4m_– 1_ 3m ___ + _ 4m_– 1 _m2 + mn – 2n2 m2 – n2 (m + 2n)(m – n) (m + n)(m – n) 3m(m + n) + (4m_– 1)(m + 2n)_ (m + 2n)(m – n)(m + n) 3m2 + 3mn + 4m2_– m + 8mn – 2n_ (m + 2n)(m – n)(m + n) 7m2 + 11mn – m – 2n_ (m + 2n)(m – n)(m + n)6. 2x + y_ + ___x – 2y_ __ 2x + y_ + ___x – 2y_ __2x2 + xy – y2 x2 + 3xy + 2y2 (2x – y)(x + y) (x + 2y)(x +y)(2x + y)(x + 2y) + (x – 2y)(2x – y) 2x2 + 5xy + 2y2 + 2x2 – 5xy + 2y2 (2x – y)(x + 2y)(x +y) (2x – y)(x + 2y)(x +y) 4x2 + 4y2 _____ or ___4(x2 + y2)_____(2x – y)(x + 2y)(x +y) (2x – y)(x + 2y)(x +y) 27

7. _ __x + 3y__ ___x – y__ _ _ __x + 3y__ ___x – y__ x2 + 2xy + y2 x2 + 4xy + 3y2 (x + y)(x + y) (x + 3y)(x + y) _ (x + 3y)(x + 3y) + (x – y)(x + y) (x + y)(x + y)(x + 3y) _x2+ 6xy+ 9y2 + x2 – y2 (x + y)(x + y)(x + 3y) ___2x2+ 6xy+ 8y2 __ or 2(x2 + 3xy + 4y2)_ (x + y)(x + y)(x + 3y) (x + y)(x + y)(x + 3y)8. _ _r + 1_ ___r – 1__ _r + 1_ ___r – 1__r2 – 3r – 10 r2 + r – 30 (r – 5)(r + 2) (r + 6)(r – 5) _(r + 1)(r + 6) + (r – 1)(r + 2)_ (r – 5)(r + 2)(r + 6) _(r + 1)(r + 6) + (r – 1)(r + 2)_ (r – 5)(r + 2)(r + 6) _r2 + 7r + 6+ r2+r – 2 _ (r – 5)(r + 2)(r + 6) 2r2 + 8r + 4 _ or 2(r2 + 4r + 2) _ (r – 5)(r + 2)(r + 6) (r – 5)(r + 2)(r + 6)9. _ m __m_– 1__ m _ _m_– 1_ _ (m + 1)(m – 1) (m + 1)(m + 1)m2 – 1 m2 + 2m + 1 m(m + 1) + (m_– 1)(m – 1) (m – 1)(m + 1)(m + 1) m2 + m + m2 – 2m + 1_ (m – 1)(m + 1)(m + 1) _ 2m2 – m + 1 _ (m – 1)(m + 1)(m + 1)10. _ 2w + 1_ ___3w + 5___ _ 2w + 1_ ___3w + 5___3w2 +10w – 8 2w2 + 5w – 12 (3w – 2)(w + 4) (2w – 3)(w + 4) (2w + 1)(2w – 3) + (3w + 5)(3w – 2) (3w – 2)(w + 4)(2w – 3) 28

(2w + 1)(2w – 3) + (3w + 5)(3w – 2) (3w – 2)(w + 4)(2w – 3) 4w2 – 4w – 3 + 9w2 + 9w – 10 (3w – 2)(w + 4)(2w – 3) 13w2 + 5w – 13 _ (3w – 2)(w + 4)(2w – 3)Lesson 3A. Subtract and express your answer in lowest term if necessary.1. 3 _2_= 1 6. _ _3s_ __5s__ _– 2s_ p pp 9s – 8 9s – 8 9s - 82. _8_ _5 3_ 7. _ r _ _ __3_ __ ___r – 3___ __1___ r2 r2 r2 r2 + 3r –18 r2 + 3r –18 (r + 6)(r – 3) r + 63. _ y__ _2y_ _– y_ 8. _ 2y__ ___1_____ ___2y – 1___ _1_ k–5 k–5 k–5 2y2 – 7y + 3 2y2 – 7y + 3 (2y – 1)(y – 3) y – 34. __z__ _ 1__ z – 1 9. _ _b__ ____1___ __ b – 1___ _ 1__ z+1 z+1 z+1 b2 + 5b – 6 b2 + 5b – 6 (b – 1)(b + 6) b + 65. _ a – b _a + b_ –b2 77710. _ 2d ___ ___ 7 __ 2d – 7___ 1__ d–22d2 – 11d + 14 2d2 – 11d + 14 (2d – 7)(d – 2)B. Subtract and express your answer in lowest term if necessary.1. 9 _r 9 – 5r 10 2 102. _8_ _2_ 32c – 105 4c 20c3. _4m + 5 _m + 2_ 2(4m + 5) – m – 2 8m + 10 – m – 2 7m + 836 6 664. _5r + 4_ _2r – 3 x(5r + 4) – 2(2r – 3) 5rx + 20 – 4r + 6 5rx – 4r + 26 10x 5x2 10x2 10x2 10x25. __2x _ _ _3x __ 2(2x) – 3x __ x_ _ x + y 2x + 2y 2(x + y) 2(x + y) 29

6. _ _1_ ___a__ 4 – a__ a–b 4a – 4b 4(a – b)7. __–6 __3 _ _– 12 – 3(p – 2)_ __– 6 + 3p___ p2 – 4 2p + 4 2(p + 2)(p – 2) 2(p + 2)(p – 2) __3(–2 + p)___ ___ 3___ 2(p + 2)(p – 2) 2(p + 2)8. _ _1_ ___1__ _____1____ ___1___ _3_– (x – 3)_ ____6 – x_ __ x2 – 9 3x + 9 (x + 3)(x – 3) 3(x + 3) 3(x + 3)(x – 3) 3(x + 3)(x – 3)9. _ 15 __3__ 15(k + 2) – 3(4k2) 15k + 30 – 12k2 4k2 k+2 4k2(k + 2) 4k2(k + 2)10. 2t – 1 2t + 1 (2t – 1)(t – 1) – (2t + 1)(t + 1) (2t2 – 3t + 1) – (2t2 + 3t + 1) (t + 1)(t – 1)t+1 t–1 (t + 1)(t – 1) 2t2 – 3t + 1 –2t2 – 3t – 1 – 6t____ (t + 1)(t – 1) (t + 1)(t – 1)C. Subtract and express your answer in lowest term if necessary.1. _ 1 __ 1 _ _____1____ _____1_____b2 – 1 b2 + 3b + 2 (b – 1)(b + 1) (b + 2)(b + 1) _(b + 2) – ( b – 1) _ b + 2 – b + 1___ (b – 1)(b + 1)(b + 2) (b – 1)(b + 1)(b + 2) 3 _____ (b – 1)(b + 1)(b + 2)2. 8 __ 2 __ 8 __ 2 __ 8 – ( – 2) 10__ – (7y – 3) 7y – 3 – (7y – 3) – (7y – 3)3 – 7y 7y – 33. _ ___– 4____ ____1____ _ _ ___– 4__ __ _____1____ _ a2 – ab – 6b2 a2 + ab – 2b2 (a – 3b)(a + 2b) (a – b)(a + 2b) _– 4(a – b) – (a – 3b)__ (a – 3b)(a + 2b)(a – b) _– 4a + 4b – a + 3b_ ____– 5a + 7b_ ____ (a – 3b)(a + 2b)(a – b) (a – 3b)(a + 2b)(a – b)4. _ x _ ______3x___ _ x _ ______3x___x2 + xy – 2y2 x2 – 3xy + 2y2 (x + 2y)(x – y) (x – 2y)(x – y) 30

_x(x – 2y) – 3x(x + 2y)_ _x2 – 2xy – 3x2 – 6xy_ (x + 2y)(x – y)(x – 2y) (x + 2y)(x – y)(x – 2y) ____ – 2x2 – 8xy____ (x + 2y)(x – y)(x – 2y)5. _ – 1___ _ 3 _ – 1 ___ _3_m2 + mn – 2n2 m2 – 3mn + 2n2 (m + 2n)(m – n) (m – 2n)(m – n)– (m – 2n) – 3(m + 2n) _– m + 2n – 3m – 6n__(m + 2n)(m – n)(m – 2n) (m + 2n)(m – n)(m – 2n) _– 4m – 4n _ (m + 2n)(m – n)(m – 2n)6. 4m – 3n _____4m – n_ __ ____ 4m – 3n__ _____4m – n_ __ 16m2 – n2 16m2 + 8mn + n2 (4m + n)(4m – n) (4m + n)(4m + n) (4m – 3n)(4m + n) – (4m – n)(4m – n) (4m + n)(4m + n)(4m – n) 16m2 – 8mn – 3n2 – (16m2 – 8mn + n2) (4m + n)(4m + n)(4m – n) 16m2 – 8mn – 3n2 –16m2 + 8mn – n2 (4m + n)(4m + n)(4m – n) ________– n2_________ (4m + n)(4m + n)(4m – n)7. _ _ 3a – b __ ___ – 2a + b__ _ _ _ 3a – b __ ___ – 2a + b_ _ 6a2 – 13ab – 5b2 6a2 – 19ab + 10b2 (3a + b)(2a – 5b) (3a – 2b)(2a – 5b)_(3a – b)(3a – 2b) – (– 2a + b)(3a + b)_ (3a + b)(2a – 5b)(3a – 2b)_9a2 – 9ab + 2b2 – (– 6a2 + ab + b2)_ (3a + b)(2a – 5b)(3a – 2b)_9a2 – 9ab + 2b2 + 6a2 – ab – b2)_ (3a + b)(2a – 5b)(3a – 2b)_ 15a2 – 10ab + b2 _(3a + b)(2a – 5b)(3a – 2b) 31

8. _ _ p + q ____q – p___ _ _ p + q ____q – p_ __3p2 + 2pq – q2 6p2 – 5pq + q2 (3p – q)(p + q) (2p – q)(3p – q) _ (p + q)(2p – q) – (q – p)(p + q)_ (3p – q)(p + q)(2p – q) _ 2p2 + pq – q2 – (q2 – p2)_ _ 2p2 + pq – q2 – q2 + p2_ (3p – q)(p + q)(2p – q) (3p – q)(p + q)(2p – q) _ ___3p2 + pq – 2q2 _____ (3p – q)(p + q)(2p – q)9. _ m – n __ _m_+ n_ _ _ m–n __ _m_+ n _ _m2 + 2mn + n2 m2 – mn – 2n2 (m + n)(m + n) (m – 2n)(m + n) (m – n)(m – 2n) – (m_+ n)(m + n)_ (m + n)(m + n)(m – 2n) m2 – 3mn + 2n2 – (m2_+ 2mn + n2)_ (m + n)(m + n)(m – 2n) m2 – 3mn + 2n2 – m2_– 2mn – n2_ (m + n)(m + n)(m – 2n) – 5mn + n2_____ (m + n)(m + n)(m – 2n)10. _ 2x + y_ ___x – 2y__ _ _ 2x + y_ ____x – 2y__ _2x2 + xy – y2 x2 + 3xy + 2y2 (2x – y)(x + y) (x + 2y)(x + y) _(2x + y)(x + 2y) – (x – 2y)(2x – y) (2x – y)(x + y)(x + 2y) _2x2 + 5xy + 2y2 – (2x2 – 5xy + 2y2) (2x – y)(x + y)(x + 2y) _2x2 + 5xy + 2y2 – 2x2 + 5xy – 2y2) (2x – y)(x + y)(x + 2y) _______10xy_ _ (2x – y)(x + y)(x + 2y)D. Find the sums or differences.1. 8 _3_ 7 __ 8(5m) + 3(m – 2) + 7 5m(m – 2)m – 2 5m 5m(m – 2) 32

40m + 3m – 6 + 7 43m + 1_ 5m(m – 2) 5m(m – 2)2. –1 __3__ __ 4 __ –(x + 2)(x + 3) + 3(7z)(x + 3) + 4(x + 2)_ 7z x + 2 7z(x + 3) 7z(x + 2)(x + 3) –(x2 + 5x + 6) + 21xz + 63z + 4x + 8_ 7z(x + 2)(x + 3) –x2 – 5x – 6 + 21xz + 63z + 4x + 8_ 7z(x + 2)(x + 3) –x2 – x + 2 + 21xz + 63z_ 7z(x + 2)(x + 3)3. 4 __6__ __ 1 __ 4 ___6__ __ 1 __ r2 – r r2 + 2r r2 + r – 2 r(r – 1) r(r + 2) (r + 2)(r – 1) 4(r + 2) + 6(r – 1) + r_ r(r – 1)(r + 2) 4r + 8 + 6r – 6 + r __11r + 2___ r(r – 1)(r + 2) r(r – 1)(r + 2)4. 6 __1__ _ __2_ __ 6 ____1___ __ __2_ __k2 + 3k k2 – k k2 + 2k – 3 k(k + 3) k(k – 1) (k + 3)(k – 1) 6 ____1___ __ __2_ __ k(k + 3) k(k – 1) (k + 3)(k – 1) 6(k – 1) –_(k + 3)_ + 2k__ k(k + 3)(k – 1) 6k – 6 – k – 3 + 2k_ ___7k – 9___ k(k + 3)(k – 1) k(k + 3)(k – 1)5. _ x _ ____3x_____ __2x___ x2 + xy – 2y2 x2 – 3xy + 2y2 x2 – 4y2 _ x_ ______3x___ _____2x_____ (x – y)(x + 2y) (x – y)(x – 2y) (x + 2y)(x – 2y) _x(x – 2y) – 3x(x + 2y) + 2x(x – y)_ (x – y)(x + 2y)(x – 2y) 33

_x2 – 2xy – 3x2 – 6xy + 2x2 – 2xy_ (x – y)(x + 2y)(x – 2y) _ – 10xy _ (x – y)(x + 2y)(x – 2y)What have you learned1. 36y5 2. b4 – 16 or (b2 + 4)(b + 2)(b – 2)3. (6x – y2)(6x + y2)(x – y) 4. _w4 _32d2w45. __4m__ ___4m2 + 16m__ 2d2 64d4 m2 – 4m m(m – 4)(m + 4) 6. __ x + y __ ___1___ x(x + y)(x – y) x(x – y)7. a _b _ a(x+ y) + bx ax + ay + bx x x + y x (x + y) x (x + y)8. _ x + 4 _ _ __x – 5___ _ x + 4 _ _ __x – 5___-(x – 2)(x + 3) (x – 2)(x – 2 ) -(x – 2)(x + 3) (x – 2)(x – 2) (x + 4)(x – 2) + –(x – 5)(x+ 3) -(x – 2)(x – 2)(x + 3) x2 + 2x – 8 + – (x2 – 2x – 15) -(x – 2)(x – 2)(x + 3) _____4x + 7__ _ -(x – 2)(x – 2)(x + 3)9. ____4k_____ __ 7__ _ __ __ 4k – 7 __ (2k + 1)(3 – k) (2k + 1)(3 – k) (2k + 1)(3 – k)10. – 4a_– (5a + 1)(2a + 1) – 4a_– (10a2 + 7a + 1) – (3a – 2)(2a + 1) – (3a – 2)(2a + 1) – 4a_–10a2 – 7a – 1_ – (3a – 2)(2a + 1) –10a2 – 11a – 1__ – (3a – 2)(2a + 1) 10a2 + 11a + 1_ (3a – 2)(2a + 1) 34

Module 2 Systems of Linear Equations and Inequalities What this module is about This module is about finding the solution of systems of linear equations andinequalities algebraically. In Module 1, you were able to find the solution graphically,identify the kind of graphs and the number of solutions that you will generate from thesystem. Try to find the similarity in solving the solution algebraically and graphically. What you are expected to learn This module is designed for you to find the solution algebraically, given a system oflinear equation in two variables by: • Elimination • Substitution How much do you knowA. Tell whether the given ordered pair is the solution of the given system: 1. (6, -8) x + 2y + 10 = 0 2x – 3y + 30 = 0 2. (3, -2) 2x – y = 8 y + 3x = 7 3. (5, 2) 4x + 3y = 26

3x + 7y = 29 4. (4, -2) x+y=2 2x + 5y = -2 5. (5, -2) x–5=0 y+2=0B. Solve the following systems by elimination: 1. 2x + y = 14 x–y=4 2. x + y = 2 2x – y = 4 3. x + y = 7 x – y = -3 4. 3m – 4n = 1 3m – 2n = -1 5. 7a – b = -1 7a + 3b = 3C. Solve the following systems by substitution: 1. x + 3y = -4 y = -x 2. x = y + 4 x+y=6 3. y – x = 4 x = -y – 2 4. 4s – 3t = 8 2s + t = -1 5. 2x + y = 1 3x + y = 4 2

What you will do Lesson 1 Solution of Systems of Linear Equations Solved by Elimination You can solve systems of linear equations in two variables by several methods.One is by the elimination method using addition. Here you will solve for the value of thevariable in the given system.Example 1: Solve the given system by elimination.5x + 2y = 11 Equation 13x – 2y = 13 Equation 28x = 24 by addition the variable y is eliminated8x = 24 multiply both sides of the equation by 1.88 8 x =3 3 is the value of xNow solve for y: The variable y in the system can be solved by substituting the value of xin Equations 1 or 2.Use Equation 1:5x + 2y = 11 Substitute the value of x5(3) + 2y = 11 The value of y is – 2 15 + 2y = 11 2y = 11 – 15 2y = - 4 2y = - 4 22 y=-2Check the answer by substituting the values of x and y in the two equations.Equation 1: Equation 2:5x + 2y = 11 3x – 2y = 135(3) + 2(-2) = 11 3(3) – 2(-2) = 13 15 – 4 = 11 9 + 4 = 13 11 = 11 13 = 13 3

Since the values of x and y satisfy the two equations, then, the solution of thesystem is (3, - 2). In this example, addition of the two equations immediately eliminates one of thevariables. But this is not always the case. The next example will illustrate that you must multiply one or both of the equationsby a non-zero constant to get an equivalent system before elimination method is applied.Example 2: Solve the system by elimination:4x + 3y = -1 Equation 12x – 5y = 19 Equation 2Solution:To eliminate x, multiply Equation 2 by –2. Call the result Equation 3.-2(2x – 5y) = 19(-2) by distributive property of multiplication over addition-4x + 10y = - 38 Equation 3Now add Equation 1 and Equation 3: 4x + 3y = -1 Equation 1- 4x + 10y = -38 Equation 3 13y = -39 y = -39 13 y = -3Substitute the value of y in Equation 1 or 2 and solve for x. Use Equation 1.4x + 3y = -14x + 3(-3) = -14x – 9 = -1 4x = -1 + 9 4x = 8 x=8 4 x=2Therefore the solution is (2, -3). Sometimes each equation of the system of equations must be multiplied by aconstant so that the coefficients of one of the variables are opposite in signs but equal inabsolute value. The next example will illustrate. 4

Example 3: Solve the system by elimination:3x – 5y = 19 Equation 15x + 2y = 11 Equation 2 It is clear that when you add the two equations, you will not eliminate one of thevariables. Therefore, you must use multiplication to form an equivalent system. Thechoice multiplier will depend on the variable you decide to eliminate. Suppose you decide to eliminate y. You must multiply Equation 1 by 2 andEquation 2 by 5.2(3x – 5y) = 2(19) 6x – 10y = 385(5x + 2y) = 5(11) 25x + 10y = 55 31x = 93 x = 93 31 x =3Note that –10 and 10 are equal in absolute values but opposite in sign. You can now substitute 3 to x in any of the two equations to solve for y. Try usingEquation 2.5x + 2y = 11 Equation 25(3) + 2y = 1115 + 2y = 11 2y = 11 – 15 2y = - 4 y=-4 2 y = -2Therefore the solution for the system is (3, -2). Again, the solution should be checked by substituting 3 for x and 2 for y in the twoequations.Equation 1 Equation 2 3x – 5y = 19 Substitute 3 for x 5x + 2y = 113(3) – 5(-2) = 19 and –2 for y. 5(3) + 2(-2) = 11 Simplify both 9 + 10 = 19 equations. 15 – 4 = 11 19 = 19 11 = 11Since the solution satisfy both equations, the solution (3, -2) is correct. 5