Prepared by Sa diyya Hendrickson. Package Summary

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1 Introduction Prepared by Sa diyya Hendrickson Name: Date: Package Summary Understanding Variables Translations The Distributive Property Expanding Expressions Collecting Like Terms Solving Linear Equations Percentages Let s Play! (Exercises) sadiyya.hendrickson@gmail.com 1 of 17 c Sa diyya Hendrickson

2 Understanding Variables variable: a variable is a letter that is used to: (1) represent a quantity that we don t know or (2) be a placeholder for the input of one or more numbers. Because variables are just numbers in disguise, everything that we ve learned about numbers still applies! Let s look at some examples: sadiyya.hendrickson@gmail.com 2 of 17 c Sa diyya Hendrickson

3 Understanding Variables 1. area: Area is a property of two-dimensional figures and is the number of 1 1 squares needed to cover the figure. Recall that the area of a rectangle is given by: length width since this product produces the number of 1 1 squares needed to cover the rectangle. 2. volume: Volume is a property of three-dimensional figures and is the number of cubes needed to fill the same amount of space as the figure. Recall that the volume of a rectangular prism is given by: length width height. sadiyya.hendrickson@gmail.com 3 of 17 c Sa diyya Hendrickson

4 Translations Below are partial statements, with bolded terms that require us to perform one of following five operations: addition, subtraction, multiplication, division, or exponentiation. List each term under the correct category. 1. five more toys than Jasmine 2. snacks shared evenly between four people 3. seven less than Tristan s mark 4. four times the width 5. the square of the length 6. pencils distributed to eight students 7. the product of John s age and Kayley s age 8. one-third of Yan s age 9. rent increased by $ triple the cost of the shirt 11. four centimetres taller than Kamar 12. five years younger than Hamsha 13. $10 off of the regular price 14. twice as many markers 15. the distance cubed 16. the quotient of two numbers 17. the difference between Sam s and Samir s height sadiyya.hendrickson@gmail.com 4 of 17 c Sa diyya Hendrickson

5 Translations After identifying appropriate operations (if any), it s useful to translate English statements to simpler, algebraic expressions. First, name the variable using a let statement! Examples: 1. Find an expression for two consecutive integers. Solution: The word consecutive means following continuously or in a row. So, to name two consecutive integers, we simply let n = any integer. Then, n + 1 must be the integer that follows (because integers differ by 1)! 2. Find an expression for a tuition that has been increased by $500. Solution: If we let t = previous tuition costs, then t + $500 represents the new total. Q: Why do we take the time to name a variable before writing the expression, rather than just creating an expression with the actual word(s)? e.g. (any integer)+1 or (previous tuition)+$500 A: It s more efficient to write and perform operations on simpler expressions! sadiyya.hendrickson@gmail.com 5 of 17 c Sa diyya Hendrickson

6 Distributive Property The distributive property has come up time and time again, which really highlights how important it is. The property states: a(b + c) = ab + ac for integers a, b, and c a) factored to expanded form b) expanded to factored form Let s use area models to better understand this property. If a, b and c are positive integers such that a and b + c represent the length and width of a rectangle, then the area of the rectangle is given by: A = l w = a (b + c) We can also look at this in a different way. The area of this rectangle is just the sum of the areas of two smaller rectangles! sadiyya.hendrickson@gmail.com 6 of 17 c Sa diyya Hendrickson

7 Distributive Property 1. Expand the expression: 3(2x + 4) 3(2x + 4) = 3(2x) + 3(4) by the distributive property = (3 2)x + 12 by the associative property of multiplication = 6x Factor the expression: 8x x + 14 = (2 4)x + (2 7) numbers are even common factor of 2 = 2 (4x) + 2 (7) by the associative property of multiplication = 2(4x + 7) factoring out 2 by the distributive property Type 1: distributivity over subtraction (i.e. distributing over negative values) a(b c) = ab ac We will prove why the distributive property generalizes over subtraction (i.e. positive and negative integers) after we ve learned how to solve equations! Case 2: distributing negative values inside of the brackets Consider the justification below: a(b + c) = ab ac a(b + c) = ( 1)(a)(b + c) by definition of multiplying by 1 = ( 1)(a(b + c)) by the associative property of multiplication = (ab + ac) by the distributive property = the opposite of ab + ac by definition of opposites = ab + ( ac) since the opposite of ab + ac is: (the opposite of ab) plus (the opposite of ac) = ab ac by definition of subtraction as adding the opposite sadiyya.hendrickson@gmail.com 7 of 17 c Sa diyya Hendrickson

8 Expanding Expressions Expand the following expressions: 1. 7(4x 2 5) 7(4x 2 5) = 7(4x 2 ) 7(5) by the distributive property = (7 4)x 2 35 by the associative property of multiplication = 28x x( 3x 3 + 9x) 2x( 3x 3 + 9x) = ( 2x)( 3x 3 ) + ( 2x)(9x) by the distributive property = ( 2)( 3)(x 3 x 1 ) + ( 2)(9)(x x) by the associative property = 6 + ( )x 2 by properties of exponents = by the definition of subtraction 3. 4x( 6x 2 + x 8) 4x( 6x 2 + x 8) = ( 4x)( 6x 2 ) + ( 4x)(x) ( 4x)(8) by the distributive property = ( 4)( 6)(x 1 x 2 ) + ( 4)(x x) ( 4)(8)x by the associative property = x 3 + ( 4)x 2 ( 32)x by properties of exponents = by definition of subtraction Can you identify the solutions using the distributive property and mental math? 1. 6x(2x 3) (a) 24x 2 8x x 2 ( 2x 2 + x 1) (b) 12x ( 3x 2 x + 5) (c) 14x 4 + 7x 3 + 7x x( 2x 3) (d) 24x 2 8x x 2 (2x 2 + x + 1) (e) 12x 2 18x 6. 8(3x 2 x 5) (f) 14x 4 7x 3 + 7x 2 sadiyya.hendrickson@gmail.com 8 of 17 c Sa diyya Hendrickson

9 Collecting Like Terms Suppose we are asked to expand an expression that yields the following result: 4x x + x 2 5x x 3 x The terms in an expression are separated by addition symbols (all subtraction symbols can be rewritten so that we are adding the opposite). The original expression becomes: 4x 2 + ( 4) + 2x + x 2 + ( 5x 3 ) ( x 3 ) + ( x) Above, there are terms: 4x 2, 4, 2x,, 5x 3, 7,, and x It s all About the Powers Simplifying an expression requires that we group terms based on their powers. Given some variable, say x, we have like terms if they belong to the same powers family. 1. The Constants (x 0 ): e.g. 1, 2, 5, etc. 2. First family (x 1 ): e.g. x, 2x, 5x, etc. 3. Squared family (x 2 ): e.g. x 2, 2x 2, 5x 2, etc. 4. Cubed family (x 3 ): e.g. x 3, 2x 3, 5x 3, etc. and so on... How Many do We Have? To simplify, we add/subtract terms that belong to the same family! For example, in the Cubed family we have: five negative x 3 s and one positive x 3. So, one of the negatives will cancel with the positive, leaving behind four negative x 3 s (i.e. 4x 3 ). Algebraically: 5x 3 + x 3 = 5 x 3 + x 3 x 3 is a common factor = ( 5 + 1)x 3 by the distributive property = 4x 3 So, simplifying: ( 5x 3 + x 3 ) + (4x 2 + x 2 ) + (2x x) + ( 4 + 7) grouping families = ( 5 + 1)x 3 + (4 + 1)x 2 + (2 1)x + 3 by the distributive property = 4x 3 + 5x 2 + x + 3 sadiyya.hendrickson@gmail.com 9 of 17 c Sa diyya Hendrickson

10 Equations An equation is a mathematical statement involving an equal sign: a = b The equal sign separates the left side (L.S.) of the equation from the right side (R.S.). Below are some important properties of equations. Let a, b and c be any numbers. Then: 1. Reflexive Property: If a = b, then b = a e.g. If x + 3 = 5, then 5 = x If a = b, then a ± c = b ± c In other words, as long as we add or subtract the same thing on both sides, the equality still holds! 3. If a = b, then a c = b c In other words, as long as we multiply by the same thing on both sides, the equality still holds! 4. If a = b and c 0, then a c = b c In other words, as long as we divide by the same thing on both sides, the equality still holds! Q: Suppose we were asked to find a number such that when you double it and then increase the result by 1, the answer is 5. How would you find this number? A: One way is to work backwards from 5! We must undo everything that was done to create the number 5 (i.e. do the opposite)! First, subtract the 1 that was added to increase the result to 5. This returns 4. Then, take the number 4 and divide it by 2, to undo the doubling (i.e. multiplication by 2). So, the answer is 2. sadiyya.hendrickson@gmail.com 10 of 17 c Sa diyya Hendrickson

11 Equations The previous question was simple enough that we could work backwards fairly easily to find the desired number. However, if the problem becomes more complicated, this is not an efficient method. Let s consider using an algebraic expression: Let x = {some number} such that: 2x + 1 = 5. Below, we will do the opposite to both sides to balance the equation and transform 2x + 1 = 5 into x = {some number} sadiyya.hendrickson@gmail.com 11 of 17 c Sa diyya Hendrickson

12 Equations Summarizing our Strategies: S1 Identify which side of the equation x is on. If there is more than one x, decide which side you want to move all of them to. S2 Do the opposite to rearrange the equation so that x is on one side (by itself) and numbers are on the other side. Addition and are opposite operations! Multiplication and are opposite operations! S3 Sneak up on x! The ultimate goal is to isolate x, and the best approach is to get everything out of x s way, starting from the numbers furthest away! Finally, we check that our solution is correct. In other words, we need to make sure that when x = 1, L.S. = R.S. 2 L.S. = 4 x ( = 4 1 ) 2 = R.S. = 5x + 7 ( = 5 1 ) = = 9 2 = 9 2 Therefore, L.S. = R.S. sadiyya.hendrickson@gmail.com 12 of 17 c Sa diyya Hendrickson

13 Equations Solve the same equation: 4 x = 5x + 7 by moving the x s to the R.S. You should get the same answer! Proving the Distributive Property over Subtraction Now we have the skills to prove the distributive property over subtraction: a(b c) = ab ac Proof: First, we let k = b c. k = b c k +c = b c +c adding c to both sides k + c = b since c + c = 0 a (k + c) = a (b) multiplying both sides by a ak + ac = ab by the distributive property on L.S. ak + ac ac = ab ac subtracting ac from both sides ak = ab ac since ac ac = 0 a(b c) = ab ac since k = b c sadiyya.hendrickson@gmail.com 13 of 17 c Sa diyya Hendrickson

14 Percents percent: percent comes from the latin phrase per centum and means per hundred. In other words: p % = p 100 Example 1: 33 % means 33 per hundred, which is just the fraction Example 2: When we purchase an item from a store and are charged 13 % in taxes, we are paying 113 % of the retail price. This is because we are paying 100 % of the retail price plus an additional 13 % in taxes. Rounding When asked to round a decimal number to a nearest place value, the convention is to: Look at the place value immediately to the right, and... If the digit in that position is greater than or equal to 5 (i.e. 5 9), round up If the digit in that position is less than 5 (i.e. 0 4), we round down Example: Round % to the nearest hundredth. First we identify the place value to the right of the hundredths place (i.e. ). Let s underline it: Because the number is 6 (which is greater than 5), we round 7 up to 8, resulting in: %. Let s explore this in detail on the next page! sadiyya.hendrickson@gmail.com 14 of 17 c Sa diyya Hendrickson

15 Percents We have the following four cases: I. Percents to Fractions II. Percents to Decimals III. Decimals to Percents IV. Fractions to Percents I. Percents to Fractions By the definition of percent, we simply drop the % symbol write the value over 100! e.g. 26% = = in reduced form II. Percents to Decimals Recall that: p% = p. In the previous lesson, we learned that whenever we divide a 100 number by 10 k, we simply move the decimal point k places to the left! Here, we have: 100 = Strategies: e.g. Express 435 % as a decimal Drop the % symbol: e.g. 435 Identify where the decimal is (if there isn t one, place one at the end): e.g Then move the decimal two places to the left! e.g III. Decimals to Percents To go from percents to decimals, we have to divide by 100, so if we want to go from decimals to percents, we simply have to do the opposite (i.e. multiply by 100)! In the previous lesson, we learned that whenever we multiply a number by 10 k, we simply move the decimal point k places to the right! Here, we have: 100 = Strategies: e.g. Express as a percent Identify where the decimal is (if there isn t one, place one at the end): e.g (the decimal is between the 0 and the 3) Move the decimal two places to the right! e.g Place the % symbol at the end: e.g % sadiyya.hendrickson@gmail.com 15 of 17 c Sa diyya Hendrickson

16 Percents IV. Fractions to Percents To go from fractions to percents, we must satisfy the following equation: a b = p%, which is equivalent to a b = p 100 (where a and b are integers) If the fraction already has a denominator of 100, there is nothing to do! Otherwise, we simply need to see that above, we have an equation where we have to solve for p! Notice that to isolate p, we simply need to do the opposite of dividing by 100 (i.e. by 100)! Strategies: e.g. Express 5 8 as a percent Multiply the fraction by 100 and find the equivalent decimal: p = = by reducing the pair (8, 100) = by definition of fraction multiplication = 62.5 by long division or mental math Place the % symbol at the end: e.g % 1. A store is offering 30% off of a $75 item that you d like to purchase. How much will you save? S1: Write the percent as a decimal i.e. 30% = 0.3 S2: Translate the English statement into an algebraic equation i.e. Let s = amount saved. Then s = since of means to multiply here! S3: Calculate i.e. s = = 22.5, meaning that you save $ Calculate the 20% tip of a $15 dine-in meal using mental math. First notice that you get 20% by doubling 10%. Now we can use the fact that: 10% = 10 = 1, so 10% of 15 translates to (15) = = 1.5 (since dividing by 10 means moving the decimal point one place to the left! Doubling $1.50 produces a tip of $3! sadiyya.hendrickson@gmail.com 16 of 17 c Sa diyya Hendrickson

17 Let s Play! 1. Give translations for the English statements on page 4. Be sure to begin with the appropriate let statements. 2. Use the distributive property to expand the following expressions: (a) 5(x + 4) (b) 2(6x + 5) (c) x(8x + 2) (d) 6x(2x + 3) (e) x(3x 2 + 4) (f) 2(x 2 + 3x + 4) (g) 7x(2x 2 + x + 1) (h) x(3x 2 + 4x + 9) (i) 2x(4x 2 + 5x + 7) 3. Simplify the following expressions (i.e. first expand if necessary (using the distributive property) and then collect like terms): (a) 2x 5x (b) x (3x + 5) (c) 4x + 3 (x + 6) (d) 2x 2 x x 2 (3x + 4) (e) x 3 + 2x 5 x(x 2 x + 4) (f) 5x 4 x 3 + 2x 2 + x 2 (x 2 3x 2) 4. Solve the following equations: (a) 4x + 2 = 14 (b) 3 2x = 11 (c) 3x 8 = 17 (d) 5 x = 3x 9 (e) x 4 = 5 7x (f) 2x + 9 = x 5. Find the fraction, decimal and/or percent representations: (a) 1 20 (b) 3.57 (c) 76.3% (d) 25 4 (e) 42.8 (f) 45% 6. Suppose you wanted to purchase a new winter jacket that has a retail price of $300. (a) Calculate how much additional money you need to cover the taxes (13%). (b) If you received $170 on your birthday to put towards this purchase, what percentage of the purchase price has been covered? (be sure to create a let statement for the unknown percentage, and then translate the english statement: what percent of $300 equals $170 into an equation that you can solve.) sadiyya.hendrickson@gmail.com 17 of 17 c Sa diyya Hendrickson

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