Regina Summer Math Review. For students who will be taking. Algebra 2. Completed review packet due the first day of classes

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Regina Summer Math Review For students who will be taking Algebra 2 Completed review packet due the first day of classes

Algebra 2 Summer Review Packet Welcome to Algebra 2! In the following pages, you will find review materials that will prepare you for Algebra 2. Please take the exercises seriously as this will allow us to hit the ground running in the fall. If the examples preceding the practice problems are not enough of a reminder of a concept, please remember that Kahn Academy, YouTube, and math.com are very useful resources! The review materials are separated into weeks. These weeks are only a suggestion. You will have the most benefit from this material if you work on it throughout the summer and do a final review of your work a week or two before school starts. This packet must be complete the first day of class. Table of Contents: Week 1: Real numbers and Operations Week 2: Algebraic Expressions and Models Week 3: Solving Linear Equations Week 4: Properties of Exponents Week 5: Solving Linear Inequalities Week 6: Solving Absolute Value Equations and Inequalities Week 7: Lines Week 8: Graphing Linear Equations Week 9: Graphing Linear Inequalities Materials Needed for Algebra *Graphing calculator (Texas Instruments models TI-83, TI-84 families) 3-Ring Binder (Required) Loose-Leaf Paper (Required) Graphing calculators are needed in this course and subsequent courses, even courses in college. Invest in one now, keep good care of it, and use it for many years to come.

Algebra 2 Summer Review Week 1: Real numbers and Operations Types of Real Numbers: Natural (Counting): 1, 2, 3, 4, Whole: 0, 1, 2, 3, Integers: -2, -1, 0, 1, 2, Rational: Can be written as a fraction or a repeating or terminating decimal Irrational: Not rational Properties of Addition and Multiplication (Addition shown first, Multiplication second) 1. Commutative Property: 5+6=6+5 5 6=6 5 2. Associative Property: (5+6)+9=5+(6+9) 5(6 9)=(5 6)9 3. Identity Property: 5+0=5 5 1=5 4. Inverse Property: 5+( 5)=0 5 " : =1 5. Distributive Property: 5(6+9)=56+69 Classify each variable according to the set of numbers that best describes its values. 1. the area of the circle A found by using the formula πr 2 1. 2. the number n of equal slices in a pizza; the portion p of the pizza 2. in one slice 3. the air temperature t in Saint Paul, MN, measured to the nearest 3. degree Fahrenheit 4. the last four digits s of a Social Security number 4. Plot the following values on the number line provided. Please label each by its exercise number. 5. 5 " 6. 4 7. 2.25 8. 6 " 9. 8 # ) Compare the two numbers. Use >, <, or =. A calculator should not be used. 10. 2 2 11. 29 5 12. 4 17 13. 50 6.8 Name the property of real numbers illustrated by each equation. 14. 2 3+ 5 =2 3+2 5 15. 16+ 13 = 13+16 16. 7 " 4 =1 17. 5 0.2 7 = 5 0.2 7 14. 15. 16. 17.

Week 2: Algebraic Expressions and Models Order of Operations Parentheses Exponents Multiplication/Division Addition/Subtraction Vocabulary Terms: Parts added together to make an expression. Coefficients: The number located in front of the variable. Constant: Numbers in an expression without a variable. Write an algebraic expression that models each word phrase. 1. seven less than the number ; 2. the sum of 11 and the product of 2 and a number < 1. 2. Write an algebraic expression that models each situation. 3. Arin has $520 and is earning $75 each week babysitting. 4. You have 50 boxes of raisins and are eating 12 boxes each month. 1. 2. Evaluate each expression for the given values of the variables. 5. 4=+3 >+2= 5> == 2 >=4 6. 9 3 5 9 # 5=4 9= 1 7. 2 3? 5@ +3?+4@?=3 @= 5 5. 6. 7. Simplify by combining like terms. 8. 5A 3A # +16A # 9. )(:BC) D + E D 6 10. ;+ FG # +;# +; 11. 45 5(5+1) Identify the following components from the expression 5A 4 8A+47 12. The number of terms 13. Leading coefficient 14. Constant Term 8. 9. 10. 11. 12. 13. 14.

Week 3: Solving Linear Equations Remember to solve equations, you can add, subtract, multiply or divide by any number or variable as long as you do the same operation to the other (entire) side. 2 5 (A 3)=A 2 Original Problem 2 5 A 6 5 =A 2 Distribute the # to each term on the left side of the = J 5K 2 5 A 6 5 L=5(A 2) 2A 6=5A 10 5A 5A Multiply both sides of the equation by 5 to get rid of the fractions on the left Subtract 5A on each side 3A 6= 10 +6 +6 Add 6 on each side 3A = 4 3 3 Divide both sides by 3 A= 4 3 Solve for A. Check your solution by plugging the value into the original equation. Solve each equation. 1. 9 H 3 =12H 1. 4. 3 A+1 =2(A+11) 4. 2. 7I+5=6I+11 3. 5>+8 12>=16 15> 2. 3. 5. " ) I 2 =I+4 6. 4 # ) A= 7 5. 6. Write an equation and solve each problem. 7. Two brothers are saving money to buy tickets to a concert. Their combined savings is $55. One brother has $15 more than the other. How much has each saved? 8. What three consecutive numbers have a sum of 126? 9. Two trains left a station at the same time. One traveled north at a certain speed and the other traveled south at twice that speed. After 4 hours, the trains were 600 miles apart. How fast was each train traveling?

Week 4: Properties of Exponents Properties of Exponents Assume that 5,6,Z,[ are real numbers. 5 P =1 5 Bc = 1 1 5 c =5c 5Bc 5 d 5 c =5 dmc (5 d ) c =5 d c (56) c =5 c 6 c 5 d 5 c =5dBc e 5 c 6 f = 5c 6 c Simplify and rewrite each expression using only positive exponents. a. 55 ) 35 BE 456 O 9 ) b. =5 3 5 ) 5 BE 5 J 69 ) =45 "BJ 6 OB" 9 )B) = 155 )M BE = 155 B" = 15 5 =45 BE 6 J 9 P = 46J 5 E Simplify each expression. Use only positive exponents. 1. (25 ) )(55 E ) 2. ( 3A # )( 4A B# ) 3. (3A # I ) ) # 4. (3A BE I ) ) # 5. E:Q #: R 6. "#ST U V ES WX 7. OSV Y )SU G 8. #S R ) ) 9. ( 4Z # [ ) )(2Z[) 10. 2A ) I 4 B# 11. )\WG ] V F Y WV )\] 12. h 4 _ ) P 13. \G ] R F` \ V ] R F W` 14. SG U E "OS U 15. a E ; # a;

Week 5: Solving Linear Inequalities *Remember, when you multiply or divide each side of an inequality by a negative, you must reverse the inequality symbol. *Closed dot represents 5[k. This means the value is included in the solution. *Open dot represents > 5[k <. This means our value is not included in the solution. Compound Inequalities: Two simple inequalities joined by the words and or or. AND: 2 A<1 OR: A< 1 l< A 2 2 1 1 2 3< 6A+1 13 3< 6A+1 13 1 1 1 2 < 6A 12 6 6 6 1 3 >A 2 Original Problem Subtract 1 on each side Divide each side by 6 **Remember to reverse each inequality sign Final Answer Graph Write the inequality that represents the sentence. 1. Five less than a number is at least 28. 2. The product of a number and four is at most 10. 3. Six more than a quotient of a number and three is greater than 14. Solve each inequality. Graph the solution. 4. 55 10>5 6. 2 [+2 +6 16 5. 25 2I 33 7. 2(75+1)>25 10 Solve each compound inequality. Graph the solution. 8. 8<4A<12 10. 2A+3<12 or 4A 7>21 9. 2 3A 8 10 11. 2A>3 A or 2A<A 3

Week 6: Solving Absolute Value Equations and Inequalities *The absolute value of a number A is written A, is the distance the value is from zero. The absolute value of a number is always positive. 5 = 5 =5 5 Solve the absolute value equation. 2A 5 =9 Solve each equation. Check for extraneous solutions. 1. 3A =18 Break original problem into two since the value of 2A 5 could have been a positive 9 or 9 2A 5=9 2A 5= 9 Solve each smaller equation as normal A=7 A= 2 **Always check answers in the original equation to make sure they make true statements. If an answer comes from correct algebra, but does not work in the original equation, it is called an extraneous solution. *To solve an absolute value inequality, it is important to remember that a < or represents an AND statement and a > or represents an OR statement. Example Example 2A+7 <11 3A 2 8 11<2A+7<11 3A 2 8 Or 3A 2 8 18<2A <4 A 2 Or A 10 3 9<A<2 Distance of 5 0 5 Distance of 5 3. 3 H+7 =12 2. ;+5 =8 4. 4 2I +5=9 Solve each inequality. Graph the solution. 5. 5 I+3 <15 7. " # 2A 1 3 1 6. 46 3>9 8. 2 4A+1 5 1

Week 7: Lines Finding Slope Z Z= I A =<oap <q[ =I # I " A # A " Finding Slope from a Graph Finding Slope from Two Points Find the slope from the two points ( 2,7) and (3, 1) 1 7 3 ( 2) = 8 5 Z= 8 5 I=1 Positive because it went up A=2 Positive because it went right Therefore, Z= " #. Find the slope from the following lines or points. 1. 2. 3. 4. 5. 8,10, ( 7,14) 7. 18, 20, ( 18,5) Z= 6. 19,6, (15,16) Z= 8. 4,7, (8,7) Z= Z=

Week 8: Graphing Linear Equations Slope-Intercept Form I=ZA+6 Z is slope of the line 6 is the y-intercept (0,6) Standard Form ra+si=t x-intercept is u (where I=0) v y-intercept is u (where A=0) w Graph both intercepts and connect with a line Point-Slope Form I I " =Z(A A " ) Z is the slope of the line (A ",I " ) is a point on the line Graph the point, then use the slope to graph more points Graph I= # ) A 4 y-intercept is 4 or (0, 4) slope is # à move up 2, right 3 ) Graph 2A+I= 2 x-intercept is B# =1 à (1,0) B# y-intercept is B# = 2 à (0, 2) " Graph I 1= " # (A 1) Point: (1,1) Slope: " à move up 1, right 2 # Identifying the A- and I-intercepts in Any Form A-intercept is the point on the A-axis where the graph crosses. This is also the line where I= 0. Substitute I=0 to find the A-value of this point. I-intercept is the point on the I-axis where the graph crosses. This is also the line where A= 0. Substitute A=0 to find the A-value of this point. Find the A- and I-intercepts. I 3=3(A+1) A-intercept: I=0: 0 3=3(A+1) Solve 3=3A+3 6=3A A-intercept: ( 2,0) A = 2 I-intercept: A=0: I 3=3(0+1) I 3=3(1) I 3=3 I=6 I-intercept: (0,6)

Graph each equation. Fill-in the appropriate information. 1. A+I=3 A-intercept: I-intercept: Slope: 2. I= 2A 3 A-intercept: I-intercept: Slope: 3. I=5A 2 A-intercept: I-intercept: Slope: 4. I 4= " # (A+3) Point: Slope: A-intercept: I-intercept: 5. I 5=2(A 3) Point: Slope: A-intercept: I-intercept: 6. A+4I=4 A-intercept: I-intercept: Slope:

Week 9: Graphing Linear Inequalities 1. Graph the line the same way you would any other linear equation. 2. Remember < or > represents a dashed line and or represents a solid line. 3. Choose a test point on the graph to see if it satisfies the inequality. If it does, shade to cover the test point as it is a solution. If it is not, shade away from it. Graph I< " E A+3 1. Graph I= " E A+3 2. Graph has a dashed line. 3. Test Point: (0,0): 0< " E (0)+3 0<3 à TRUE Shade to cover (0,0). Graph the linear inequalities. 1. 3A+I 6 2. A+I< 2 3. A+4I 8 4. I ) E A+1 5. I< A+4 6. I # J A 2