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Algebra 2 Summer Review Packet Summer Review Packet For Algebra 2 CP1 Name _________________________________________________________ Current Course _____________________ Math Teacher________________________ Introduction Algebra 2 builds on topics studied from both Algebra 1 and Geometry. Certain topics are sufficiently involved that they call for some review during the year in Algebra 2. However, some topics and basic skills are so fundamental to any Algebra 1 course that MASTERY of these topics is expected and required PRIOR to beginning Algebra 2. This packet covers these topics, which are: Solving simple equations in one variable, Factoring polynomials – GCF, General trinomials, perfect square trinomials and difference of squares. Graphing lines in y mx b form, and slopes of lines, Computation with fractions without the use of a calculator, (Computations using a calculator – NOT in this packet, but it will be done in class) We will be using a TI-83 or TI-84 in our classes Note about calculators: The school will issue each student who needs/wants a TI-83 or TI-84 graphing calculator, just as we issue a textbook to each student. If it is lost or broken the student must pay us the replacement value of about $90. A student may prefer to purchase his/her own. If so, we recommend the TI-84 Plus (Silver Edition optional). We then highly recommend the calculator be etched with the student’s name clearly on the front face, and on the back for security purposes. A sticker or a name written with a marker is not enough. During this summer you are to complete this review packet, following the directions for each section. Although you will have some opportunity to get some help, if you find you need a lot of help, you should consider getting it before school begins. You will be tested on this material shortly after the beginning of school. The topics on computation with and without a calculator will require true mastery – a 100% result, and will be formative up to that point. We recommend you do this work close to the onset of school. Do the work neatly on loose-leaf paper or graph paper in case your teacher decides to collect it. DO NOT LOSE THIS PACKET!! It will also be available from the school website via the math department. LHS 1 Algebra 2 Summer Review Packet COMPUTATIONS WITHOUT THE USE OF A CALCULATOR SOME NOTES….. You are expected to know how to do operations involving fractions, mixed numbers and signed numbers without the use of a calculator. Here are a few basic principles: A mixed number like 42 3 really means 4 2 , which becomes 4 3 3 12 3 14 3 3 2 3 2 3 Adding and subtracting fractions requires that all the fractions have the same denominator. Some examples: 7 1 7 3 1 8 21 8 29 1) DO NOT change this to a mixed number. 8 3 8 3 3 8 24 24 24 2) 5 1 2 5 (5 1 ) (2 5 ) 4 3 4 3 21 4 13 5 21 4 ( 5 ) 13 ( 4 ) 105 52 5 5 4 20 20 53 20 a c a c Multiplying fractions: The rule is , and you MUST reduce as possible. b d bd Some examples: 7 12 7 12 1 4 4 3) It is always better to reduce BEFORE you multiply 9 21 9 21 3 3 9 4) (7 1 )(3 5 ) (7 1 )(3 5 ) ( 15 )( 5 ) (15)( 17) (3)(2 17) 51 2 2 2 2 2 17 (2)(5) 2 Dividing fractions: The rule is to change the division problem into a multiplication problem by multiplying the numerator by the reciprocal of the denominator, and then multiply as above. a a c a d ad That is: b , and then reduce as possible. d c b d b c bc Some examples: 11 11 7 77 5) 4 2 7 4 2 8 2 2 2 3 2 8 8 5 (8)(5) (1)(5) 5 5 6) 3 3 or 31 5 3 1 5 16 5 3 16 (3)(16) (3)(2) 6 6 LHS 2 Algebra 2 Summer Review Packet COMPUTATIONS WITHOUT THE USE OF A CALCULATOR WORKSHEET… For the worksheet on this topic, you will be given an algebraic expression using variables, and the values for these variables. You will then substitute the values for the variables and find the final simplified answer as a reduced fraction or an integer. NO MIXED NUMBERS as answers. Instructions: Given the following values or the six variables, find the value of each expression below in simplest, reduced form. 1 4 5 a , b , c 2 3 , x 1, 4 y 2, z 3 5 6 1) a b y 2) bz c a y 3) 4) b cb z z b 5) 6) a b c 7) 3 y 12 z 3a 8) y c 5b 8c 18az 9) 10) 6a 3 x x a b 11) 12) a 2 b 2 z y (3 x y ) 2 y x 13) 14) 4a 2 7 a b x z y 15) 16) b a y x z 23 2 55 140 And the final answers are: 1) 2) 3) 4) 15 3 16 117 25 2 3 18 5) 6) 7) 17 8) 9) 10) 5 34 165 4 5 4 169 3 8 169 56 11) 12) 13) 14) 15) 16) Solved 5 225 4 7 60 15 Examples LHS 3 Algebra 2 Summer Review Packet Solving Equations 1 Example 1: Solve x2 9 3 1 x 2 9 3 2 2 Eliminate the +2 by subtracting 2 from both sides. 1 Solution: x 7 3 1 3 x 3 7 Eliminate the 1 3 by multiplying by its reciprocal, 3, on both sides. 3 x 21 Example 2: Solve 3y 8 7 11 3y 8 7 11 3y 24 7 11 Use the distributive property and combine to simplify the left side 3y 17 11 of the equation. Solution: - 17 - 17 Eliminate the +17 by subtracting 17 from both sides. 3y -6 Eliminate the 3 by dividing both sides by 3. 3 3 y -2 Example 3: Solve 7a 2 6 2a 8 a 7 a 2 6 2a 8 a 7 a 14 6 2a 8 a Use the distributive property and combine to simplify both 7 a 20 3a 8 sides of the equation. 3a 3a Eliminate the variable term on the right by subtracting 3a. Solution: 4a 20 8 20 20 Eliminate the –20 by adding 20 to both sides. 4a 28 Eliminate the 4 by dividing both sides by 4. 4 4 a 7 Example 4: Solve 3(1 n) 5n 2(n 1) 3(1 n) 5n 2(n 1) Continue as in the previous example. 3 3n 5n 2n 2 Solution: 3 2n 2n 2 2n 2n Correct steps have resulted in a false statement. 32 Conclude that the equation has no solution. no solution or Ø or {} Solved Examples LHS 4 Algebra 2 Summer Review Packet Factoring (Common Monomial Factor) Example 5: Factor 15x 3 y 2 z 20x 2 yz 2 Solution: Find the greatest common factor on the terms in the polynomial. GCF is 5x 2 yz . Divide each term in the polynomial by the GCF to find the other factor. 15x 3 y 2 z 20 x 2 yz 2 3xy and 4z 5x 2 yz 5 x 2 yz 15x 3 y 2 z 20x 2 yz 2 5x 2 yz3xy 4 z Factoring (Difference of Squares) Example 6: Factor 16x 2 25 Solution: A term is a square if the exponents on all variables in it are even and the coefficient is the square of an integer. Both 16x2 and 25 are squares. This will only factor if the two squares are subtracted – thus a difference of squares. In particular, 16x2 = (4x)2 and 25 = (5)2. The binomial factors as the sum and the difference of the squares roots: 16x2 – 25 = (4x + 5)(4x – 5) Factoring (Perfect Square Trinomials) Example 7: Factor 4x2 – 20xy + 25y2 Solution: A perfect square trinomial is a polynomial with three terms that results from squaring a binomial. In other words, perfect square trinomial = (some binomial) 2. To factor, we must first check that the trinomial is a perfect square by answering the following three questions: 1. Is the first term a square? Answer: Yes, 4x2 = (2x)2 2. Is the last term a square? Answer: Yes, 25y2 = (5y)2 3. Ignoring the sign at this time, is the middle term twice the product of 2x and 5y? Answer: Yes, 20xy = 2 (2x5y) Having established that the trinomial is a perfect square, it is easy to find the binomial of which it is a square; its terms are in the parenthesis above. 4x2 – 20xy + 25y2 = (2x - 5y)2 Since the middle term in the trinomial is negative, the factored result is a subtraction. Factoring (Product – Sum) Example 8: Factor y2 + 14y + 40 Solution: If, as above, the trinomial is in standard form, the product number is the numerical term, 40, and the sum number is the coefficient of the linear term, 14. The unique pair of number whose sum is 14 and product is 40 is 4 and 10. Therefore, the trinomial factors as y2 + 14y + 40 = (y + 4)(y + 10) (Note: all factoring problems can be checked by multiplying out) Solved Examples LHS 5 Algebra 2 Summer Review Packet Graphing Example 9: Find the slope and the y-intercept of the line whose equation is 3y = 2x + 9 and use them to graph the equation. 3y 2 x 9 3y 2 x 9 Solution: Divide by 3 to solve for y. 3 3 3 2 y x3 3 Now the equation is in the familiar form y = mx + b. 2 The slope is the coefficient of the x-term, , and the y-intercept is the constant 3 2 term, 3. To graph the line, plot the y-intercept, (0 , 3). Then, since the slope is , move 2 3 units up(rise) and 3 units right (run) to locate a second point. (See below.) y +3 +2 x Example 10: a) Find the slope of the line passing through the points (-2 , 5) and (4 , 8). b) Given that the y-intercept is 6, write an equation for the line. y 2 y1 85 3 1 Solution: a) slope x 2 x1 42 6 2 b) In a linear equation of the form y = mx + b, the slope is the coefficient of the x-term or linear term, namely m. The numerical term or constant term is the y-intercept, namely b. So, starting with y = mx + b, substitute ½ for m and 6 for b. Therefore, the equation is 1 y x6 2 LHS 6 Algebra 2 Summer Review Packet Exercises Solve each equation. 1. 2x – 1 = 11 2. -4n + 9 = -3 1 2 3. x7 6 4. x8 2 3 2x 1 1 x 5. 5 6. 7 3 2 7. 7x – 4x = 56 8. -3n – 5n = 240 9. y – 7 + 4y = 13 10. 0 = n – 13 – 4n 11. 35 = 5 (n + 2) 12. 2 (4 – x) = 22 13. y + 5 – 4y = -10 14. 15 = 8x – 5 + 2x 1 15. (x 4) 16 16. 3 (a – 5) + 19 = -1 2 17. -3 = 4 (k + 7) - 15 18. 4c + 3 (c – 2) = -34 8 2x 3 19. 0 20. 1 (v 2) 5 5 4 21. -9 – 3 (2q – 1) = -18 22. -2 = 4 (s + 8) – 3s 23. c – (1 – 2c) + (c – 3) = -4 24. 3x = 27 – 15x 25. 51a – 56 = 44a 26. -7a = -12a - 65 27. 3p – 8 = 13 – 4p 28. 5n + 1 = 5n -1 2 29. 8 (5 – n) = 2n 30. x7 x 3 9 2y 23 11c 31. y 32. 5c 7 7 33. 3 (30 + s) = 4 (s + 19) 34. 2 (g – 2) – 4 = 2 (g – 3) 35. 4 (3y – 1) + 13 = 5y + 2 36. 4 (a + 2) = 14 – 2 (3 – 2a) LHS 7 Algebra 2 Summer Review Packet Exercises Factor. 1. 15a – 25b + 20 2. 6x2 + 10x 3. 6p2q – 9pq 4. 7y3 – 21y2 – 14y 5. 6ab2 – 8a2b 6. -15x2y2 – 6xy2 7. 5ax2 + 10a2x – 15a3 8. 48a3b2 + 72a2b3 9. c2 - 36 10. 9a2 - 100 11. 169m2 - 225 12. 144 – y2 13. m6 – 25 14. 25m4 – n2 15. 81p8 – 121q6 16. 49a2 – 9b10 17. x2 + 5x + 4 18. c2 – 10c + 16 19. q2 + 16q + 15 20. a2 + 10ab + 24b2 21. u2 – 50uv + 49v2 22. y2 + 5y - 6 23. x2 + 2x - 8 24. x2 – 25x - 54 25. x2 – 10xy – 75y2 26. a2 + 5ab – 84b2 27. 4w2 + 12w + 9 28. 25m2 – 60mn + 36n2 29. 49a2 + 28ab + 4b2 30. 4s2 – 36st + 81t2 LHS 8 Algebra 2 Summer Review Packet Exercises Find the slope and the y-intercept of the line determined by each equation. Use them to graph the equation on the graph paper provided. You may put two problems on a graph. 2 1. y x8 2. y = 8 – 2x 3. y = -x 3 4. y=4 5. 3x + y = 9 6. 4x – 3y = 9 7. x + 5y = 5 8. 6x + 4y = 8 Find the slope of the line passing through each pair of points. 9. (0 , 7) and (1 , 9) 10. (5 , 2) and (7 , 0) 11. (8 , 1) and (-8 , 1) 12. (3 , -1) and (6 , 7) 13. (-2 , 0) and (2 , -3) 14. (-2 , -6) and (-2 , 4) Find the equation of the line passing through each given point with the given slope and y-intercept. 2 15. slope = 3, y-intercept = 7 16. slope = , y-intercept = -5 3 17. slope = 0, y-intercept = -6 LHS 9 Algebra 2 Summer Review Packet LHS 10 Algebra 2 Summer Review Packet Key to Exercises Solve Equations 1. x=6 2. n=3 3. x = -2 4. x = 12 5. x=8 6. x = -13 56 7. x 8. n = -30 9. y=4 3 13 10. n 11. n=5 12. x = -7 3 13. y=5 14. x=2 15. x = -36 5 16. a 17. k = -4 18. c = -4 3 19. x=4 20. v=6 21. q=2 3 22. s = - 34 23. c=0 24. x 2 25. a=8 26. a = -13 27. p=3 28. No solution 29. n=4 30. x = -21 1 31. y=1 32. c 33. s = 14 2 34. No solution 35. y = -1 36. Any Real Number Factor 1. 5 (3a – 5b + 4) 2. 2x (3x + 5) 3. 3pq (2p – 3) 4. 7y (y2 – 3y – 2) 5. 2ab (3b – 4a) 6. -3xy2 (5x + 2) 7. 5a (x2 + 2ax – 3a2) 8. 24a2b2 (2a + 3b) 9. (c + 6)(c – 6) 10. (3a + 10)(3a – 10) 11. (13m + 15)(13m – 15) 12. (12 + y)(12 – y) 3 3 13. (m + 5)(m – 5) 14. (5m2 + n)(5m2 – n) 15. (9p4 + 11q3)(9p4 – 11q3) 16. (7a + 3b5)(7a – 3b5) 17. (x + 1)(x + 4) 18. (c – 2)(c – 8) 19. (q + 1)(q + 15) 20. (a + 6b)(a + 4b) 21. (u – 49v)(u – v) 22. (y + 6)(y – 1) 23. (x + 4)(x – 2) 24. (x – 27)(x + 2) 25. (x – 15y)(x + 5y) 26. (a – 7b)(a + 12b) 27. (2w + 3)2 28. (5m – 6n)2 29. (7a + 2b)2 30. (2s – 9t)2 LHS 11 Algebra 2 Summer Review Packet Key to Exercises Graph 2 1. slope = , y-intercept = 8 5. slope = -3, y-intercept = 9 3 4 2. slope = -2, y-intercept = 8 6. slope = , y-intercept = -3 3 1 3. slope = -1, y-intercept = 0 7. slope = , y-intercept = 1 5 3 4. slope = 0 , y-intercept = 4 8. slope = , y-intercept = 2 2 #3 #2 y #1 #8 #5 y #6 #4 #7 x 9. 2 10. -1 11. 0 8 3 12. 13. 14. no slope 3 4 2 15. y = 3x + 7 16. y x5 17. y = -6 3 LHS 12