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MATHEMATICS GRADE 7 UNIT 2 RATIONAL NUMBERS GRADE 7 β’ MODULE 2 RATIONA L NUMBERS OVERVIEW In Grade 6, students formed a conceptual understanding of integers through the use of the number line, absolute value, and opposites and extended their understanding to include the ordering and comparing of rational numbers (6.NS.C.5, 6.NS.C.6, 6.NS.C.7). This module uses the Integer Game: a card game that creates a conceptual understanding of integer operations and serves as a powerful mental model students can rely on during the module. Students build on their understanding of rational numbers to add, subtract, multiply, and divide signed numbers. Previous work in computing the sums, differences, products, and quotients of fractions serves as a significant foundation as well. In Topic A, students return to the number line to model the addition and subtraction of integers (7.NS.A.1). They use the number line and the Integer Game to demonstrate that an integer added to its opposite equals zero, representing the additive inverse (7.NS.A.1a, 7.NS.A.1b). Their findings are formalized as students develop rules for adding and subtracting integers, and they recognize that subtracting a number is the same as adding its opposite (7.NS.A.1c). Reallife situations are represented by the sums and differences of signed numbers. Students extend integer rules to include the rational numbers and use properties of operations to perform rational number calculations without the use of a calculator (7.NS.A.1d). Students develop the rules for multiplying and dividing signed numbers in Topic B. They use the properties of operations and their previous understanding of multiplication as repeated addition to represent the multiplication of a negative number as repeated subtraction (7.NS.A.2a). Students make analogies to the Integer Game to understand that the product of two negative numbers is a positive number. From earlier grades, they recognize division as the inverse process of multiplication. Thus, signed number rules for division are consistent with those for multiplication, provided a divisor is not zero (7.NS.A.2b). Students represent the division of two integers as a fraction, extending product and quotient rules to all rational numbers. They realize that any rational number in fractional form can be represented as a decimal that either terminates in 0s or repeats (7.NS.A.2d). Students recognize that the context of a situation often determines the most appropriate form of a rational number, and they use long division, place value, and equivalent fractions to fluently convert between these fraction and decimal forms. Topic B concludes with students multiplying and dividing rational numbers using the properties of operations (7.NS.A.2c). In Topic C, students problem-solve with rational numbers and draw upon their work from Grade 6 with expressions and equations (6.EE.A.2, 6.EE.A.3, 6.EE.A.4, 6.EE.B.5, 6.EE.B.6, 6.EE.B.7). They perform operations with rational numbers (7.NS.A.3), incorporating them into algebraic expressions and equations. They represent and evaluate expressions in multiple forms, demonstrating how quantities are related (7.EE.A.2). The Integer Game is revisited as students discover βif-thenβ statements, relating changes in playerβs hands (who have the same card-value totals) to changes in both sides of a number sentence. Students translate word problems into algebraic equations and become proficient at solving equations of the form ππ₯ + π = π and π(π₯ + π) = π, where π, π, and π, are specific rational numbers (7.EE.B.4a). As they become fluent in generating algebraic solutions, students identify the operations, inverse operations, and order of steps, comparing these to an arithmetic solution. Use of algebra to represent contextual problems continues in Module 3. This module is comprised of 23 lessons; 7 days are reserved for administering the Mid- and End-of-Module Assessments, returning the assessments, and remediating or providing further applications of the concepts. The Mid-Module Assessment follows Topic B, and the End-of-Module Assessment follows Topic C. 2 VOCABULARY FOR FLASHCARDS A.) Absolute Value: The absolute value of a number represents the positive distance the number is away from the origin (0). The symbol: x is used when the absolute value of a number needs to be found. B.) Additive Identity: The additive identity is 0. C.) Additive Inverse: The additive inverse of a real number is the opposite of that number on the real number line. For example, the opposite of β3 is 3. A number and its additive inverse have the sum of 0. D.) Algebraic Equation: An algebraic equation is a collection of numbers, variables, operations, and grouping of symbols while using an equal sign. Example 1: Example 2: Example 3: Example 4: Example 5: a + b = 10 2a β 5 = 7 3abc = 100 a + a + b + b = 2a + 2b 3a = a + a + a E.) Algebraic Expression: An algebraic expression is a collection of numbers, variables, operations, and grouping of symbols without using an equal sign. Example 1: Example 2: Example 3: Example 4: Example 5: a+b 2a -5 + c 3abc (ab) β (cd) + 5 a+5 3 F.) Distance Formula: If p and q are rational numbers on a number line, then the distance between p and q is |π β π|. G.) Evaluating: Evaluating is when the variable(s) in an expression, equation or inequality is/are replaced with a number and then a numerical answer is found. H.) Inverse Operations: The use of inverse operations are essential in the topic of Algebra. Inverse operations allows the mathematician to solve for an unknown variable. The use of inverse operations is the undoing of the operation being worked with. Example 1: Example 2: Example 3: Example 4: The inverse operation of addition is subtraction. The inverse operation of subtraction is addition. The inverse operation of division is multiplication. The inverse operation of multiplication is division. I.) Irrational Numbers: A number is irrational if it canβt be written as a fraction with an integer numerator and a nonzero integer denominator. *AN EASY WAY TO REMEMBER IRRATIONAL NUMBERS* *An irrational number is a nonrepeating/nonterminating decimal (never ends!!!!!) *Three Major Categories We Study Are: 1) pi: ο° 2) Non-Perfect Square Roots: 3, 5, 6, 7,... 3) Decimals (Non-Terminating and Non-Repeating): 7.123145β¦, .03003000300003000003β¦ J.) Multiplicative Identity: The multiplicative identity is 1. 4 K.) Rational Numbers: A number is rational if it can be written as a fraction with an integer numerator and a nonzero integer denominator. Examples: 0.4, 2.9, 16 , 0.3, 0.6 5 TERMINOLGY 1.) Additive Identity: The additive identity is 0. 2.) Additive Inverse: The additive inverse of a real number is the opposite of that number on the real number line. For example, the opposite of -3 is 3. A number and its additive inverse have the sum of 0. 3.) Break-Even Point: The point at which there is neither a profit nor loss. 4.) Distance Formula: If p and q are rational numbers on a number line, then the distance between p and q is |π β π|. 5.) Loss: A decrease in the amount; as when the money earned is less than the money spent. 6.) Multiplicative Identity: The multiplicative identity is 1. 7.) Profit: A gain, as in the positive amount represented by the difference between the money earned and spent. 8.) Repeating Decimal: The decimal form of a rational number, for example π Μ . = π. π π 9.) Terminating Decimal: A decimal is called terminating if it is repeating digit is 0. 6 Subsets of Real Numbers A. Counting Numbers or Natural Numbers: 1, 2, 3, 4, 5, β¦.. B. Whole Numbers: 0, 1, 2, 3, 4, 5, β¦.. C. Integers: (negative infinity) ο ο₯,...,ο 3,ο 2,ο 1,0,1, 2,3,....., ο₯ (infinity) D. Rational Numbers: A number is rational if it can be written as a fraction with an integer numerator and a nonzero integer denominator. Examples: 0.4, 2.9, 16 , 0.3, 0.6 E. Irrational Numbers: A number is irrational if it cannot be written as a fraction with an integer numerator and a nonzero integer denominator. Examples: ο° ο» 3.141592654β¦.. 2 ο» 1.414213562β¦.. 3 ο» 1.732050808β¦.. *AN EASY WAY TO REMEMBER IRRATIONAL NUMBERS* *An irrational number is a nonrepeating/nonterminating decimal (never ends!!!!!) *Three Major Categories We Study Are: 1) pi: ο° 2) Non-Perfect Square Roots: 3, 5, 6, 7,... 3) Decimals (Non-Terminating and Non-Repeating): 7.123145β¦, .03003000300003000003β¦ 7 Integers In the world of mathematics, there are many types of numbers. In this chapter, we are going to be talking about a particular family of numbers that has been discussed before. These numbers are called integers. Let us review what they look like: (negative infinity) ο ο₯,.....,ο 5,ο 4,ο 3,ο 2,ο 1,0,1, 2,3, 4,5,.....ο₯ (infinity) Integers on a number line look like this: smaller negative numbers larger zero positive numbers *When negative numbers are written, the symbol (-) is placed before the number. Example 1: Example 2: ο ο 3 is read as: βnegative threeβ 5.9 is read as: βnegative five point nineβ or βnegative five and nine tenthsβ F. Opposite Number: The value of -5 and 5 on the graph below are an equal distance from each other. These are called opposite numbers. The reason for this is because the two values are equal distance from the origin (0). 8 G. Absolute Value: The absolute value of a number represents the positive distance the number is away from the origin (0). The symbol: x is used when the absolute value of a number needs to be found. Example 1: 2 = 2 Example 2: ο 2 = 2 Example 3: 0 = 0 Example 4: Example 5: ο ο 2 = ο 2 Example 6: ο 2 = ο2 ο 6ο2 6-2 4 Example 7: ο 7 ο 3 ο 7β3 -10 Example 8: Example 9: 0 ο ο 2 Example 10: ο 8 ο« ο9 8+9 17 ο¨ 2ο© ο ο¨2ο© 0β2 ο 2 4 9 2 2 H. Additive Inverse(Opposite): A number and its opposite have a sum of zero. Example 1: g + (-g) = 0 Example 2: 5 + (-5) = 0 Example 3: 1 ο¦ ο1 οΆ ο«ο§ ο· ο½ 0 2 ο¨ 2 οΈ I. Addition Property of Zero: A number added to zero leaves the number unchanged. Example 1: g + 0 = g Example 2: 5 + 0 = 5 Example 3: 1 1 ο«0 ο½ 2 2 J. Commutative Property: Addition and Multiplication PLEASE NOTE: CHANGE IN ORDER Example 1: x + y = y + x Example 2: 3 + 8 = 8 + 3 Example 3: x β’ y = y β’ x Example 4: ο3 1 1 ο3 ο· ο½ ο· 4 3 3 4 K. Associative Property: Addition and Multiplication PLEASE NOTE: CHANGE IN GROUPING Example 1: a + (b + c) = (a + b) + c Example 2: 4 + (6 + 7) = (4 + 6) + 7 Example 3: a β’ (b β’ c) = (a β’ b) β’ c Example 4: 2 x (3 x 4) = (2 x 3) x 4 10 L. Polynomial: A polynomial is an expression that has a sum or difference of one or more terms. M. Algebraic Expression: An algebraic expression is a collection of numbers, variables, operations, and grouping of symbols without using an equal sign. Example 1: a + b Example 2: 2a -5 + c Example 3: 3abc Example 4: (ab) β (cd) + 5 Example 5: a + 5 N. Terms: Terms are the parts that are added or subtracted in a mathematical expression. axn a: represents any real number (any number we have used this year) x: represents the variable being used but could be any letter (c, d, m, r, ...) n: is an exponent/power that is a whole number (0, 1, 2, 3, 4, β¦) Examples of Polynomials Name Number of Terms Examples Monomial 1 x, -3a, 3, 5y, 0.5c, Binomial 2 3a β 5, b2 + 9b Trinomial 3 x2 β 6x + 7, 4c2 + c - 2 *Please Note: There can be polynomials with four or more terms. 11 O. Distributive Property of Multiplication Over Addition and Subtraction 1) Distributive Property of Multiplication Over Addition(βsum ofβ): a(b + c) = ab + ac OR ab + ac = a(b + c) 2) Distributive Property or Multiplication Over Subtraction(βdifference ofβ): a(b β c) = ab β ac OR ab β ac = a(b - c) HOW THE DISTRIBUTIVE PROPERTY CAN BE USED FINDING PRODUCTS Example 1: 4(3 + 2) = (4 x 3) + (4 x 2) = 12 + 8 = 20 Example 2: 2(8 β 3) = (2 x 8) β (2 x 3) = 16 β 6 = 10 Example 3: 9 x 27 = (9 x 20) + (9 x 7) = 180 + 63 = 243 Example 4: 5 1 x6= 2 1 1οΆ ο¦ ο§ 5 ο« ο· 6 = (5 x 6) + ( x 6) = 30 + 3 = 33 2 2οΈ ο¨ Example 5: 4 x 29 = (4 x 30) β (4 x 1) = 120 β 4 = 116 Example 6: 7 x 102 = (7 x 100) + (7 x 2) = 700 + 14 = 714 Example 7: 3.5 x 20 = (3 + .5)20 = (3 x 20) + (.5 x 20) = 60 + 10 = 70 CHANGING THE FORM OF AN ALGEGRAIC EXPRESSION Example 1: 8(x + y) = 8x + 8y Example 2: 3.4(a β b) = 3.4a β 3.4b Example 3: 1 1 1 (9 x ο 6 x) ο½ ( ο· 9 x ο ο· 6 x) ο½ 4.5 x ο 3 x ο½ 1.5 x 2 2 2 12 USING THE NUMBER LINE TO MODEL ADDITION OF INTEGERS Example 1: -4 + 9 Example 2: -5 + -2 Example 3: 8 + -13 FINAL THOUGHT: A.) ADD POSITIVE INTEGERS BY COUNTING UP/RIGHT. B.) ADD NEGATIVE INTEGERS BY COUNTING DOWN/LEFT. C.) ON A NUMBER LINE, ARROWS ARE USED TO REPRESENT INTEGERS, THEY SHOW LENGTH AND DIRECTION. D.) THE LENGTH OF AN ARROW ON THE NUMBER LINE IS THE ABSOLUTE VALUE OF THE INTEGER. E.) THE SUM OF SEVERAL ARROWS IS THEF FINAL POSITION OF THE LAST ARROW. 13 ADDITION OF RATIONAL NUMBERS A. Adding Two Rational Numbers: 2 Situations Situation 1: SAME SIGNS {βSAME SIGNS, FIND SUM OF ABSOLUTE VALUES, KEEP SIGNβ} Rule: Add the absolute value of each number and keep the sign Example 1: 4 + 2 = ? 4 ο½4 2 ο½2 Therefore: 4 + 2 = 6 ANSWER IS POSITIVE BECAUSE SIGN OF ORIGINAL NUMBERS ARE POSITIVE Example 2: ο 2 ο½2 ο 3 ο½3 ο 2 + ο3 = ? Therefore: ο 2 + ο 3 = ο 5 THE ANSWER IS NEGATIVE BECAUSE THE SIGN OF ORIGINAL NUMBERS ARE NEGATIVE Example 3: -3x + -5x = ? ο 3x ο½ 3x ο 5x ο½ 5x Therefore: -3x + -5x = -8x THE ANSWER IS NEGATIVE BECAUSE THE SIGN OF ORIGINAL NUMBERS ARE NEGATIVE 14 Situation 2: DIFFERENT SIGNS {βFIND THE DIFFERENCE OF ABSOLUTE VALUES, KEEP SIGN OF HIGHER ORIGINAL ABSOLUTE VALUEβ} Rule: 1. Take absolute value of each number 2. Subtract the smaller absolute value from the bigger absolute value 3. The sign of the original number with the larger absolute value will be the sign of the answer Example 1: ο ο 4x + 2x = ? 4x = 4x 2x = 2x 4x β 2x = 2x Therefore: ο 4 x ο« 2 x ο½ ο 2 x FINAL ANSWER IS NEGATIVE BECAUSE ο 4x HAS A LARGER ABSOLUTE VALUE Example 2: 8 + ο 3 = ? 8 =8 ο 3 =3 8β3=5 Therefore: 8 + ο 3 = 5 FINAL ANSWER IS POSITIVE BECAUSE 8 HAS A LARGER ABSOLUTE VALUE Example 3: 1 ο ο« 5ο½? 2 1 1 = 2 2 ο5 = 5 5- 1 10 1 9 1 = - = ο½4 2 2 2 2 2 Therefore: 1 ο 1 ο« 5 ο½ ο4 FINAL ANSWER IS NEGATIVE BECAUSE 2 2 -5 HAS A LARGER ABSOLUTE VALUE 15 SUBTRACTION OF RATIONAL NUMBERS Rule: 1. Follow addition rules 2. Change signs only with double negatives DOUBLE NEGATIVES BECOME POSITIVE Example 1: 4 β 8 = ? 4 ο½4 ο8 ο½ 8 8β4=4 Therefore: 4 β 8 = ο 4 FINAL ANSWER IS NEGATIVE BECAUSE -8 HAS A LARGER ABSOLUTE VALUE Example 2: 4 - ο 4 = ? EXAMPLE OF DOUBLE NEGATIVES REWRITE: 4 + 4 = ? 4 ο½4 4 ο½4 4+4=8 Therefore: 4 - ο 4 = 8 FINAL ANSWER IS POSITIVE BECAUSE THE SIGNS ARE THE SAME Example 3: -6p β 14p = ? ο6 p ο½ 6 p ο14 p ο½ 14 p 6p + 14p = 20p Therefore: -6p β 14p = -20p FINAL ANSWER IS NEGATIVE BECAUSE THE SIGNS ARE THE SAME 16 THE DISTANCE BETWEEN TWO RATIONAL NUMBERS Example 1: What is the distance between 3 and 8? |π β π| = |βπ| = π Example 2: What is the distance between -3 and 8? |βπ β π| = |βππ| = ππ Example 3: What is the distance between 10 and -6? |ππ β (βπ)| = |ππ + π| = ππ FINAL THOUGHTS: 1.) TO FIND THE DISTANCE BETWEEN TWO RATIONAL NUMBERS ON A NUMBER LINE, YOU CAN COUNT THE NUMBER OF UNITS BETWEEN THE NUMBERS. 2.) USING A FORMULA, THE DISTANCE BETWEEN RATIONAL NUMBERS, p and q, IS |π β π|. 3.) DISTANCE IS ALWAYS POSITIVE. 4.) CHANGE MAY BE POSITIVE OR NEGATIVE. FOR EXAMPLE, THERE IS -4O CHANGE WHEN THE TEMPERATURE GOES FROM 7O TO 3O. 17 DIVISION AND MULTIPLICATION OF RATIONAL NUMBERS Rule 1: Even number of negatives will result in a positive answer Rule 2: Odd number of negatives will result in a negative answer PLEASE NOTE: RULES FOR DIVISION AND MULTIPLICATION OF RATIONAL NUMBERS ARE THE SAME! Example 1: 5 x 5 = 25 (no negative signs) Example 2: ο 3 x 9 = ο 27 (one negative sign) Example 3: ο 4 x 4 = ο 16 (one negative sign) Example 4: ο 8 x ο 2 = 16 (two negative signs) ο Example 5: 4 x 3 x 2 = ο 24 (one negative sign) Example 6: -2 x ο 3 x 2 = 12 (two negative sign) Example 7: ο 1 x ο 2 x 3 x 2 x ο 4 = ο 48 (three negative signs) MULTIPLYING FRACTIONS AND INTEGERS: NO CALCULATOR, NO PROBLEM!!!!! Example 8: 2 ο· 15 IS THE SAME AS: 3 β2 MULTIPLIED BY 15 DIVIDED BY 3β OR β15 DIVIDED BY 3 MULTIPLIED BY 2β ANSWER: 10 Example 9: 3 (10 x ) IS THE SAME AS: 5 β3 MULTIPLIED BY 10x DIVIDED BY 5β OR β10x DIVIDED BY 5 MULTIPLIED BY 3β ANSWER: 6x 18 Different Variations of Multiplication Forms a) A dot can be used: 3 ο· 6 = 18 (2.1) ο· 4 = 8.4 b) Two parenthesis next to each other with numbers inside: (3)(6) = 18 c) Two or more variables next to each other: ab or xyz Example 10: Example 11: Example 12: 8 = 4 (no negative signs) 2 7 1 ο 9 ο½ 3 (no negative signs) 3 ο Example 13: Example 14: = ο 7 (one negative sign) ο 8 ο½ 4 (two negative signs) 2 0 ο½ 0 (no negative signs) 2 10 10 2 5 2 Example 15: 9 ο½ οΈ ο½ ο½ 1 (no negative signs) 2 9 3 3 3 3 ο 4 ο 4 3 ο4 13 ο4 1 ο 2 1 ο2 26 Example 16: (one negative sign) ο½ οΈ ο½ ο΄ ο½ ο΄ ο½ ο΄ ο½ 3 26 13 26 3 2 3 1 3 3 13 Please Remember: It is impossible to divide by zero, the answer is called βundefined.β Examples 15: 7 οΈ 0 = undefined Example 16: 0 7 = undefined Example 17: 0 οΈ 0 = undefined 19 WRITING OF NEGATIVE SIGN WITH DIVISION ο¦ p οΆ ο¨ο pο© p οο§ ο· ο½ ο½ q ο¨ οq ο© ο¨qοΈ 1 ο¦ 1 οΆ ο¨ ο1ο© οο§ ο· ο½ ο½ 3 ο¨ ο3ο© ο¨3οΈ a ο¦ a οΆ ο¨ οa ο© οο§ ο· ο½ ο½ 4 ο¨ ο4 ο© ο¨4οΈ 7 ο¦ 7 οΆ ο¨ ο7 ο© οο§ ο· ο½ ο½ 9 ο¨ ο9 ο© ο¨9οΈ 20 CHANGING DECIMALS TO FRACTIONS OR MIXED NUMBERS METHOD OF CHANGING DECIMALS TO FRACTIONS 1. Determine the place value the terminating decimal ends. 2. Write the decimal part over the place holder and simplify. PLEASE NOTE: IF A NUMBER APPEARS TO THE LEFT OF THE DECIMAL, WRITE IT NEXT TO THE SIMPLIFIED FRACTION AND IT WILL RESULT IN A MIXED NUMBER. Example 1: 0.215 πππ ππ = ππππ πππ Example 2: 2.35 ππ πππ 2 = π ππ π ππ CONVERTING FRACTIONS TO DECIMALS-FRACTIONS WITH DENOMINATORS HAVING FACTORS OF ONLY 2 AND/OR 5 Example: Write the number π ππ = π ππ β’π π ππ as a decimal without a calculator. The denominator needs a factor of 5 to be a power of 10. Multiply both the numerator and denominator by 5 to arrive at decimal form is 0.35 ππ πππ . The FINAL THOUGHT: ANY TERMINATING DECIMAL CAN BE CONVERTED TO A FRACTION USING PLACE VALUE. A FRACTION WHOSE DENOMINATOR INCLUDES ONLY FACTORS FO 2 AND 5 CAN BE CONVERTED TO A DECIMAL BY WRITING THE DENOMINATOR AS A POWER OF TEN. 21 CHANGING PERCENTS TO FRACTIONS METHOD OF CHANGING PERCENTS TO FRACTIONS: 1. Write the percent over 100 (percent means per 100) 2. Simplify if asked Example 1: π% = π πππ Example 2: πππ% = ππ πππ πππ π π π π π πππ Example 3: π % = π = = = ππ ππ ππ π β’ =π π πππ π ππ = π ππ CHANGING PERCENTS TO DECIMALS METHOD OF CHANGING PERCENTS TO DECIMALS 1. Move the decimal point two places to the left (means dividing by 100) Example 1: 90% = .9 Example 2: 0.2% = 0.002 π Μ Example 3: π % = π. ππ π Example 4: 5% = 0.05 22 CHANGING DECIMALS TO PERCENT METHOD OF CHANGING DECIMALS TO PERCENTS: 1. Move the decimal point two places to the right (means multiplying by 100) Example 1: 0.8 = 8% Example 2: 1.95 = 195% Example 3: 1 = 100% Example 4: 0.089 = 8.9% CHANGING FRACTIONS TO PERCENTS METHOD OF CHANGING FRACTIONS TO PERCENTS: 1. Convert the fraction to a decimal (numerator divided by the denominator) 2. Move the decimal point two places to the right (means multiplying by 100) Example 1: π Example 2: π Example 3: Example 4: = π. ππ = ππ% π Μ = ππ. π Μ % = π. π π π ππ = π. ππ = ππ% πππ π = ππ = ππππ% 23 CONVERTING A RATIONAL NUMBER TO A DECIMAL USING LONG DIVISION Procedure: Long division has basically 4 steps. Step 1: Step 2: Step 3: Step 4: Divide Multiply Subtract Drop down next digit REPEAT STEPS 1 β 4 UNTIL THE DECIMAL FORM OF A RATIONAL NUMBER TERMINATES IN 0βS OR REPEATS HELPFUL TIP STARTING WITH A FRACTION IN LOWEST TERMS, IF THE DENOMINATOR HAS A PRIME FACTOR OTHER THAN 2 OR 5, THEN THE DECIMAL FORM OF THE NUMBER WILL BE REPEATING Example 1: 1.363 11 15.000 -11 40 -33 70 -66 40 -33 7 15 11 ANSWER = 1.36 *The decimal repeats and could have been determined before division because the denominator 11 is a prime factor other than 2 or 5. 24 Algebra Terms A. Algebraic Equation: An algebraic equation is a collection of numbers, variables, operations, and grouping of symbols while using an equal sign. Example 1: Example 2: Example 3: Example 4: Example 5: a + b = 10 2a β 5 = 7 3abc = 100 a + a + b + b = 2a + 2b 3a = a + a + a B. Algebraic Expression: An algebraic expression is a collection of numbers, variables, operations, and grouping of symbols without using an equal sign. Example 1: Example 2: Example 3: Example 4: Example 5: a+b 2a -5 + c 3abc (ab) β (cd) + 5 a+5 C. Coefficient: When a constant and variables are factors of a product, the constant is the number that is in front. This is called the coefficient. D. Evaluating: Evaluating is when the variable(s) in an expression, equation or inequality is/are replaced with a number and then a numerical answer is found. E. Factor: If two or more numbers are to be multiplied, each of the numbers, as well as the product of any of them is called a factor. A factor may involve variables also. 25 F. Inverse Operations: The use of inverse operations are essential in the topic of Algebra. Inverse operations allows the mathematician to solve for an unknown variable. The use of inverse operations is the undoing of the operation being worked with. Example 1: The inverse operation of addition is subtraction. Example 2: The inverse operation of subtraction is addition. Example 3: The inverse operation of division is multiplication. Example 4: The inverse operation of multiplication is division. 26 G. Polynomial: A polynomial is an expression that can be formed from a variable and numerical coefficients, using only the operations of addition, subtraction, and multiplication. The variable is never in the denominator of a fraction. All exponents of the variable must be positive integers. axn a: represents any real number (any number we have used this year) x: represents the variable being used but could be any letter (c, d, m, r, ...) n: is an exponent/power that is a whole number (1, 2, 3, 4, β¦) Terms: Terms are the parts that are added or subtracted in a polynomial. Also, terms are the parts that make up an expression, equation or inequality. Examples: 2, 5a, c, -7c, ab, β¦.. Terms that are (like or same) can be combined together. Examples of Polynomials Name Number of Terms Examples Monomial 1 x, -3a, 3, 5y, 0.5c, Binomial 2 3a2 β 5, b2 + 9b Trinomial 3 x2 β 6x + 7, 4c2 + c - 2 *Please Note: There can be polynomials with four or more terms. H. Tape Diagrams: A tape diagram can be used to model and identify the sequence of operations to find a solution algebraically. I. Substitution: Substitution is when the variable(s) in an expression, equation, or inequality is/are replaced with a number. J. Variable: A variable is a letter that is used to represent one or more numbers. K. Values: The numbers that represent variables are called values. 27 Distributive Property of Multiplication Over Addition and Subtraction 1) Distributive Property of Multiplication Over Addition(βsum ofβ): a(b + c) = ab + ac OR ab + ac = a(b + c) 2) Distributive Property or Multiplication Over Subtraction(βdifference ofβ): a(b β c) = ab β ac OR ab β ac = a(b - c) HOW THE DISTRIBUTIVE PROPERTY CAN BE USED FINDING PRODUCTS Example 1: 4(3 + 2) = (4 x 3) + (4 x 2) = 12 + 8 = 20 Example 2: 2(8 β 3) = (2 x 8) β (2 x 3) = 16 β 6 = 10 Example 3: 9 x 27 = (9 x 20) + (9 x 7) = 180 + 63 = 243 Example 4: 5 1 x6= 2 1 1οΆ ο¦ ο§ 5 ο« ο· 6 = (5 x 6) + ( 2 x 6) = 30 + 3 = 33 2οΈ ο¨ Example 5: 4 x 29 = (4 x 30) β (4 x 1) = 120 β 4 = 116 Example 6: 7 x 102 = (7 x 100) + (7 x 2) = 700 + 14 = 714 Example 7: 3.5 x 20 = (3 + .5)20 = (3 x 20) + (.5 x 20) = 60 + 10 = 70 CHANGING THE FORM OF AN ALGEGRAIC EXPRESSION Example 1: 8(x + y) = 8x + 8y Example 2: 3.4(a β b) = 3.4a β 3.4b Example 3: 1 1 1 (9 x ο 6 x) ο½ ( ο· 9 x ο ο· 6 x) ο½ 4.5 x ο 3 x ο½ 1.5 x 2 2 2 28 EVALUATING VARIABLE EXPRESSION INVOLVING THE DISTRIBUTIVE PROPERTY 1A. Simplify the variable expression. -2(x β 7y) + 3(2x β 2y) -2x + 14y + 6x - 6y 4x + 10y 1B. Using Part A, evaluate when x = 2 and y = -3. 4x + 8y 4(2) + 8(-3) 8 β 24 -16 SIMPLIFYING VARIABLE EXPRESSIONS 1. 5x + 15 β 10 2. 3(2 β 3c) + 5c 6 β 9c + 5c 6 β 4c 5x + 5 3. 9 β (5r + 7) 4. 7x β 4y β 2(5x + 6y) 9 β 5r β 7 2 β 5r 5. ο¨ 7x β 4y - 10x β 12y -3x β 16y ο© ο¨ 2 2 14 x 2 ο« 21 ο 40 x 2 ο« 25 7 5 ο© 6. ο¨ ο© ο¨ 4 3 2 x 100 x 2 ο« 20 ο x 9 x 2 ο« 6 5 3 4 x 2 ο« 6 ο 16 x 2 ο 10 80 x 5 ο« 16 x 3 ο 6 x 3 ο 4 ο 10 x 2 ο 4 80 x 5 ο« 10 x 3 ο 4 29 ο© SIMPLIFYING VARIABLE EXPRESSIONS CONTINUED 1 3 1 1 7. ο¦ο§ 2 x ο οΆο· ο ο¦ο§1 x ο« οΆο· ο¨ 3 4οΈ ο¨ 2 3οΈ 1 1 3 1 2 x ο1 x ο ο 3 2 4 3 1 2 8 2 x ο½ 2 x ο½1 x 3 6 6 1 2 3 6 3 6 5 x 6 - 1 x ο½ ο1 x ο½ ο1 x 3 9 ο½ 4 12 1 3 + ο½ Answer = 30 4 12 13 1 ο½1 12 12 5 1 x ο1 6 2 One Step Equations Method: 1. Simplify the left side of the equal sign (if possible). 2. Simplify the right side of the equal sign (if possible). Please Note: When simplifying, this could involve combining like terms, distributive property, etc. a) Distributive Property of Multiplication over Addition: a(b + c) = ab + ac b) Distributive Property of Multiplication over Subtraction: a(b β c) = ab β ac 3. Gather all variables to one side of the equal sign by using inverse operations. 4. Gather all constants to the opposite side of the variables by using inverse operations. 5. Isolate (get by itself) the variable by using inverse operations. 6. Check *substitute solution into original equation *perform steps for order of operations Examples 1. a + 5 = 10 - 5 -5 a = 5 Check a + 5 = 10 (5) + 5 = 10 10 = 10 Check 3. x β 7 = -10 x β 7 = -10 +7 +7 (-3) β 7 = -10 x = -3 -10 = -10 2. b + 1.2 = 3.7 - 1.2 -1.2 b = 2.5 Check b + 1.2 = 3.7 (2.5) + 1.2 = 3.7 3.7 = 3.7 Check 4. b β 7.98 = 9.00 b β 7.98 = 9.00 + 7.98 +7.98 (16.98) β 7.98 = 9.00 b = 16.98 9.00 = 9.00 31 One Step Equations Continued ο3a 12 ο½ 5. ο3 ο3 a = -4 Check -3a = 12 -3(-4) = 12 12 = 12 Check 6. d οΈ 5.5 = 33 d x 5.5 x 5.5 = 181.5 32 d οΈ 5.5 = 33 (181.5) οΈ 5.5 = 33 33 = 33 Two-Step Equations Method: 1. Simplify the left side of the equal sign (if possible). 2. Simplify the right side of the equal sign (if possible). Please Note: When simplifying, this could involve combining like terms, distributive property, etc. a) Distributive Property of Multiplication over Addition: a(b + c) = ab + ac b) Distributive Property of Multiplication over Subtraction: a(b β c) = ab β ac 3. Gather all variables to one side of the equal sign by using inverse operations. 4. Gather all constants to the opposite side of the variables by using inverse operations. 5. Isolate (get by itself) the variable by using inverse operations. 6. Check *substitute solution into original equation *perform steps for order of operations Examples 1. 2x + 6 =12 - 6 -6 2x 6 ο½ 2 2 Check 2x + 6 = 12 2(3) + 6 = 12 6 + 6 = 12 12 = 12 2. 3y β 12 = 9 + 12 +12 3 y 21 ο½ 3 3 y = 7 x=3 33 Check 3y β 12 = 9 3(7) β 12 = 9 21 β 12 = 9 9 = 9 Two-Step Equations Continued Check 3. -3a + 10 = -20 -3a + 10 = -20 - 10 -10 -3(10) + 10 = -20 ο3a ο30 ο½ -30 + 10 = -20 ο3 ο3 a = 10 -20 = -20 4. 5x β 22 = 78 + 22 +22 5 x 100 ο½ 5 5 x = 20 Check 5x β 22 = 78 5(20) β 22 = 78 100 β 22 = 78 78 = 78 Solving Equations By Combining Like Terms Method: 1. Simplify the left side of the equal sign (if possible). 2. Simplify the right side of the equal sign (if possible). Please Note: When simplifying, this could involve combining like terms, distributive property, etc. a) Distributive Property of Multiplication over Addition: a(b + c) = ab + ac b) Distributive Property of Multiplication over Subtraction: a(b β c) = ab β ac 3. Gather all variables to one side of the equal sign by using inverse operations. 4. Gather all constants to the opposite side of the variables by using inverse operations. 5. Isolate (get by itself) the variable by using inverse operations. 6. Check *substitute solution into original equation *perform steps for order of operations 34 Solving Equations By Combining Like Terms Continued Examples 1. b + 10 -3 = 44 οΈ 2 b+7 = 22 -7 -7 b = 15 2. 4x + 8x β 2x + 7 = -13 10x + 7 = -13 - 7 = -7 10 x ο20 ο½ 10 10 x = -2 3. 33 = 4n β 7 + 6n 33 = 10n β 7 +7 +7 40 10n ο½ 10 10 4= n Check b + 10 β 3 = 44 οΈ 2 (15) + 10 -3 = 44 οΈ 2 22 = 22 Check 4x + 8x β 2x + 7 = -13 4(-2) + 8(-2) β 2(-2) + 7 = -13 -8 + -16 - -4 + 7 = -13 -13 = -13 Check 33 = 4n β 7 + 6n 33 = 4(4) β 7 + 6(4) 33 = 16 - 7 + 24 33 = 33 35 Translating Algebraic Expressions and Equations + - Add Difference Sum Minus Plus Remainder Total less than increased by decreased by x : = multiply product times double triple divide quotient remainder equals is equal to Is 1. Six more than a number b 1. 6 + b 2. Eight decreased by a number m 2. 8 - m 3. The product of four and a number x 3. 4x 4. The quotient when a number z is divided by fourteen 4. z οΈ 14 5. The value in cents of k dimes 5. 10k 6. Twice a number p increased by ten 6. 2p +10 7. Twelve more than a number x is seventeen 7. 12 + x = 17 8. Thirty-two less than a number m is eighteen 8. m β 32 = 18 9. Mark has q quarters worth a total of $12.50 9. 25q = 1250 9. .25q = 12.50 (decimal form) 10. Mary has d dimes and p pennies worth a total of $12.50 10. .10d + .01p = 12.50 (decimal form) 36 10. 10d + 1p = 1250 Distributive Property Method: 1. Simplify the left side of the equal sign (if possible). 2. Simplify the right side of the equal sign (if possible). Please Note: When simplifying, this could involve combining like terms, distributive property, etc. a) Distributive Property of Multiplication over Addition: a(b + c) = ab + ac b) Distributive Property of Multiplication over Subtraction: a(b β c) = ab β ac 3. Gather all variables to one side of the equal sign by using inverse operations. 4. Gather all constants to the opposite side of the variables by using inverse operations. 5. Isolate (get by itself) the variable by using inverse operations. 6. Check *substitute solution into original equation *perform steps for order of operations Examples Check 1. -4(y + 8) = 16 -4y β 32 = 16 + 32 +32 ο4 y 48 ο½ ο4 ο4 y = -12 -4(y + 8) = 16 -4(-12 + 8) = 16 -4(-4) = 16 16 = 16 105 ο½ 3(4 ο 2b) 2. 5 21 = 12 β 6b -12 -12 9 ο6b ο½ ο6 ο6 Check 105 ο½ 3(4 ο 2b) 5 21 = 3(4 β 2(-1.5)) 21 = 3(4 + 3) 21 = 3(7) 21 = 21 -1.5 = b 37 Distributive Property Continued 3. -2(x β 3) = -3(-8) -2x + 6 = 24 ο2 x 18 ο½ ο2 ο2 x = -9 Check -2(x β 3) = -3(-8) -2(-9 β 3) = -3(-8) -2(-12) 24 = 4. 9 β 4(x -3) = 17 9 β 4x +12 = 17 24 = 24 21 β 4x -21 = 17 = -21 ο4 x ο4 ο½ ο4 ο4 x= 1 5. Three times the sum of x and 8 is 30. Find the number x. Let x = the number The number = x =2 3(x + 8) = 30 3x + 24 = 30 -24 -24 3x 6 ο½ 3 3 x=2 6. Five time the difference of x and 10 is 40. Let x = the number The number = x = 18 5(x - 10) = 40 5x - 50 = 40 +50 +50 5 x 90 ο½ 5 5 x = 18 38 Check 9 β 4(x β 3) = 17 9 β 4(1 β 3) = 17 9 β 4(-2) = 17 9+ 8 = 17 17 = 17 Solving Equations With Variables On Both Sides Method: 1. Simplify the left side of the equal sign (if possible). 2. Simplify the right side of the equal sign (if possible). Please Note: When simplifying, this could involve combining like terms, distributive property, etc. a) Distributive Property of Multiplication over Addition: a(b + c) = ab + ac b) Distributive Property of Multiplication over Subtraction: a(b β c) = ab β ac 3. Gather all variables to one side of the equal sign by using inverse operations. 4. Gather all constants to the opposite side of the variables by using inverse operations. 5. Isolate (get by itself) the variable by using inverse operations. 6. Check *substitute solution into original equation *perform steps for order of operations Examples Check 1. 7m = 2m + 10 7m = 2m + 10 -2m -2m 7(2) = 2(2) + 10 5m 10 ο½ 14 = 4 + 10 5 5 m= 2 14 = 14 2. y β 5 = 2y +10 -y -y Check y β 5 = 2y + 10 (-15) β 5 = 2(-15) + 10 -5 = y + 10 -20 = -30 + 10 -10 -15 = -20 = - 10 y 39 -20 Solving Equations With Variables On Both Sides Continued 3. 3x = x + 2 -x -x 2x 2 ο½ 2 2 x= 1 Check 3x = x + 2 3(1) = (1) + 2 3 5. 10y β (5y + 8) = 42 10y β 5y β 8 = 42 5y β 8 = 42 +8 = +8 5 y 50 ο½ 5 5 y = 10 6. 3(a β 5) = 2(2a + 1) 3a β 15 = 4a + 2 -3a -3a -15 = a + 2 -2 -2 -17 = a = 3 4. 5b β 2 = b + 10 -b -b 4b β 2 = +2 Check 5b β 2 = b + 10 5(3) β 2 = (3) + 10 10 15 - 2 = +2 4b 12 ο½ 4 4 b =3 Check 10y β (5y + 8) = 42 10(10) β (5 ο· 10 + 8) = 42 100 β 58 = 42 42 = 42 Check 3(a β 5) = 2(2a + 1) 3(-17 β 5) = 2(2 ο· -17 + 1) 3(-22) = 2(-33) -66 = -66 40 13 = 13 13 Solving Equations With Fractions Method: 1. Determine the Lowest Common Denominator (L.C.D) *Lowest Common Denominator (L.C.D) = Lowest Common Multiple (L.C.M) 2. Multiply each part of the equation by the L.C.D/L.C.M. *This eliminates the denominators CAUTION: If the L.C.D/L.C.M is being multiplied by a binomial, trinomial, etc. write parenthesis around the polynomial first 3. Follow normal algebraic procedures Examples 1. 2x ο« 7 2x ο 9 ο ο½3 6 10 ο¦ 2x ο« 7 οΆ ο¦ 2x ο 9 οΆ 30ο§ ο· ο 30ο§ ο· ο½ 30ο¨3ο© ο¨ 6 οΈ ο¨ 10 οΈ L.C.D/L.C.M (6, 10, 1) = 30 10x + 35 β 6x + 27 = 90 4x + 62 = 90 -62 -62 4 x 28 ο½ 4 4 x=7 Check 2x ο« 7 2x ο 9 ο ο½3 6 10 2ο·7 ο«7 2ο·7ο9 ο ο½3 6 10 21 5 ο ο½3 6 10 105 15 ο ο½3 30 30 3=3 41 Solving Equations With Fractions Continued 2. 2 1 11 1 ο« ο½ ο 3x 3 6 x 4 L.C.D/L.C.M (3x, 3, 6x, 4) = 12x ο¦ 2οΆ ο¦1οΆ ο¦ 11 οΆ ο¦1οΆ 12 xο§ ο· ο« 12 xο§ ο· ο½ 12 xο§ ο· ο 12 xο§ ο· ο¨ 3x οΈ ο¨ 3οΈ ο¨ 6x οΈ ο¨4οΈ 8 + 4x = 22 β 3x +3x +3x 8 + 7x = 22 -8 -8 7 x 14 ο½ 7 7 x=2 Check 2 1 11 1 ο« ο½ ο 3x 3 6 x 4 2 1 11 1 ο« ο½ ο 3ο· 2 3 6ο· 2 4 2 1 11 1 ο« ο½ ο 6 3 12 4 4 4 11 3 ο« ο½ ο 12 12 12 12 8 8 ο½ 12 12 2 2 ο½ 3 3 42 Algebraic Equations and Solutions and Tape Diagrams 1.) Randy and Elisa went to a football game. Each person purchased a ticket and a $4 water. If the total cost of the tickets and water was $48, find the price of each ticket. Algebraic Equations and Solution Let t = each ticket price Bottle water: $4 2(t + 4) = 48 2t + 8 = 48 -8 -8 ππ π = ππ π t = 20 Each ticket is $20 Tape Diagram t+4 t+4 48 β 8 = 40 2t = 40 t = 20 43 Algebraic Equations and Solutions and Tape Diagrams 2.) Tara needed to rent a big screen television for a family reunion. The initial cost was $50 and 10 for each hour she rented the big screen television. If Tara spent a total of $200, how many hours did she use the television? Algebraic Equation and Solution Let h = hours used big screen television Initial cost = $50 10h + 50 = 200 -50 -50 πππ ππ = πππ ππ h = 15 hours Total hours = 15 + 1 = 16 Tape Diagram 10 10 10 10 10 10 10 10 10 200 β 50 = 150 150 ÷ 10 = 15 15 + 1 = 16 hours total 44 10 10 10 10 10 10 50 Discount Prices and Sales Tax 1.) An item that costs $120 is discounted 20%. Then an 8% tax is applied to the sale price. Find the sale price including the tax. Sale price (without tax) = 120(1 - .20) = 120 (.8) = 96 Sale price (with tax) = 96(1 + .08) = 96(1.08) = 103.68 2.) An item that costs $120 has an 8% tax applied to the cost. Then a 20% discount is applied to the price with tax. Find the sale price including tax. Price of item (including tax) = 120(1 + .08) = 120(1.08) = 129.60 Sale price (with discount applied) = 129.60(1 - .2) = 129.60(.8) = 103.68 PLEASE NOTE THAT WHETHER THE DISCOUNT IS APPLIED FIRST OR THE TAX IS APPLIED FIRST, THE SALE PRICE WILL BE THE SAME. 45 Investment Terminology L. Deposit: Deposits are added to an account balance, money is deposited into the account. M. Gains: Gains are added to an account balance, as they are positive returns on the investment. N. Withdrawals: Withdrawals are subtracted from an account balance; money is taken out of the account. O. Losses: Losses are subtracted from an account balance; as they are negative returns on the investment. P. Fees: Fees are subtracted from an account balance; as the bank/financial company is charging you for a service. 46