Free Pre-Algebra Lesson 15! page 1

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1 Free Pre-Algebra esson 15! page 1 esson 15 Simplifying Algebraic Expressions You know from experience with the perimeter of a rectangle that there are sometimes many ways to write the same formula. Depends How You ook At It The formula for the perimeter of a rectangle can be written in more than one way, depending on how we group the sides. If we take apart the sides and consider them one by one, we have the formula P = If we group the sides by size, we get the formula P = And if we consider the perimeter as made of two groups of length + width, the formula is written P = 2( + ). Each formula uses the same lengths and widths, but groups them differently. The formulas are equivalent, and if you calculate the perimeter of a rectangle with any of the three versions, you will get the same result. hen you use the formula, you can use any version you find convenient in a given situation. One of the goals of algebra is to understand how numbers and variables and operations work so thoroughly that we can manipulate expressions and formulas into convenient forms without having to look at pictures.

2 Free Pre-Algebra esson 15! page 2 Rearranging Addition and Multiplication In esson 5, we looked at the arithmetic operations and noted that addition and multiplication are commutative it doesn t matter which number comes first in the operation. Adding 2+3 gives the same answer as adding 3+2, and multiplying 5 7 is the same as multiplying 7 5. Stating the commutative property is really making an observation about the operations of addition and multiplication work, based on what the operations mean physically. Using variables, the commutative property of addition and multiplication would be written mathematically like this: If a and b are any numbers, then a + b = b + a, and a b = b a. Besides the commutative property, addition and multiplication have an associative property. Suppose we add The order of operations tells us to evaluate left to right, as we read the expression. So we would add first, then add 5. But if we changed that and instead added first, then added the 2, the answer would be the same. That s the associate property in action. Note that the associative property involves three numbers being added or multiplied: If a, b, and c are any numbers, then (a + b) + c = a + (b + c), and (a b) c = a (b c). Formalizing our everyday observations about numbers into these properties is a very mathematical thing to do, but it s baffling to many people who feel that we ve just made something simple and obvious into something complicated, abstract, and difficult to understand. The trouble is that manipulating symbols can be tricky, and it s not always obvious what maneuvers are legal. Mathematicians want to clearly state what rules they are using so that they all agree. Once the rules are established, we tend to use them without thinking about them, but if we re asked to justify some argument or formula, we can look back and show which rules we used. The great thing about the commutative and associative properties is that by combining them, we can rearrange any string of numbers we are adding or multiplying into any convenient order. So if you have to add , you can group the numbers any way you like, such as (1 + 6) + (2 + 5) + (3 + 4), and you know that was ok, by some combination of the commutative and associative properties. So from now on, boldly rearrange addition and multiplication strings of numbers or variables any way you like. Carl Friedrich Gauss ( ) Gauss was one of the greatest mathematicians of all time, and contributed to many branches of mathematics. He was also a child prodigy. According to a story, young Carl s teacher required him to add all the numbers from 1 to 100. ithout writing anything down, Carl announced that the sum was hen asked how he had done this, he explained that he had rearranged the addition: ! " = = = 5050 There s a longer explanation of this story here, and more about Gauss in this ikipedia article. Example: Simplify (3a)(5a) The parentheses don t matter. This is really a string of multiplications: 3 a 5 a Rearrange it so that the numbers are together and the variables are together: = (3 5) (a a) Multiply the numbers, and rewrite the variable multiplication with an exponent: = 15a 2 Example: Simplify (5x)(5y)(2x) (5x)(5y)(2x) = 5 x 5 y 2 x = x x y = 50x 2 y

3 Free Pre-Algebra esson 15! page 3 Combining ike Terms To find the perimeter, add the sides of the rectangle, Rearrange the addition, and group the like variables together, Now, we have two sides with length, and two sides with length, so another way to write the expression is The process of adding + to get 2 is called combining like terms. If we invest some time in learning a little technical vocabulary it will be easier to understand how this works. The vocabulary will continue to be useful in your entire study of algebra. I m going to make up an algebraic expression: 6x 2 + 3x + y + 2x + 5. It s made up of numbers and variables and operations, combined in a meaningful way, but I can t really call it a formula, because it doesn t represent any physical model, nor is it an equation, because it doesn t have an equals sign and two sides. However, formulas and equations are made up of algebraic expressions. 6x 2 + 3x + y + 2x + 5 This algebraic expression has five terms, separated by the + signs. Terms are what is being added. Each term is a number, a variable, or a product of numbers and variables. terms According to the order of operations, the exponents and multiplications should be done first. But since we don t know what numbers the variables stand for, we can t go any further with the multiplications. So instead, we look for terms with the same variable part, called 6x 2 + 3x + y + 2x + 5 like terms. In this expression, the like terms are 3x and 2x, since their variable part, x, is the same. The term 6x 2 does not have the same variable part, because x 2 is considered to like be different from x. terms Since we have 3 x s and 2 more x s, we combine them into 5 x s. 3x + 2x = 5x. Mathematically it works out because multiplication stands for repeated addition: 3 x is the same as x + x + x and 2 x is the same as x + x, and so 3x + 2x = (x + x + x) + (x + x) = 5x. hen we combine like terms, our expression has been simplified. Terms that are not like cannot be combined. 6x 2 + 3x + y + 2x + 5 = 6x 2 + 5x + y + 5 The number part of a term is called the coefficient. hen we combine like terms, we add the coefficients, and keep the variable part the same. The variable part represents the things we are adding, and the coefficient tells how many of them there are to add up. You can think of 3x + 2x as adding three xylophones and two more xylophones to get five xylophones. The coefficients 3 and 2 are added to make 5, and the variable part, x, stays the same. Students often want to acknowledge somehow that there were two x s to start with, and respond by putting an exponent with the variable, but this is incorrect. An exponent stands for a repeated multiplication. The variables are not being multiplied by each other, so there is no change in the variable or exponent when you combine like terms. No! 3x + 2x = 5x 2 No! Example: Simplify 4N + 5M + 2N + 7. ook for like terms, and if you find any, combine them: 4N + 5M + 2N + 7 4N + 5M + 2N + 7 = 6N + 5M + 7

4 Free Pre-Algebra esson 15! page 4 Example: Simplify 2a + a + 5b 2 + 2b The like terms are shown in like colors: 2a + a + 5b 2 + 2b = 3a + 7b The Distributive Property The distributive property is another property of numbers, combining the operations of multiplication and addition. It corresponds to the way we naturally use numbers in everyday life. et s say a donut costs 10. You re at the donut shop with your friends and you buy them each a donut. 50 There are 3 pink and 2 chocolate, which makes 5 donuts, so it cost (3 + 2) = 10(5) = 50 Next time you re there, you only have a couple of friends with you, so you buy 3 donuts. 30 But then, your other friends come in, so you buy 2 more (3) + 10(2) = = 50 Either way, you still paid 50 for 5 donuts. 10(3 + 2) = 10(3) + 10(2) The arrows give a visual image of the distributive property. Instead of doing the addition inside the parentheses first, we will distribute the multiplication to each of the numbers being added: 10(3 + 2) = 10(3) + 10(2) = = 50 ritten mathematically: If a, b, and c are any numbers, then a(b + c) = a(b) + a(c). This is true, but somewhat pointless when we re just using numbers. There s no real reason to use the distributive property when we can follow the order of operations. But if the addition in parentheses includes any variables, and doesn t have like terms, the distributive property can help with simplification. Example: Simplify 2(3a + 4). 2(3a + 4) = 2(3a) + 2(4) = 6a + 8

5 Free Pre-Algebra esson 15! page 5 Example: Simplify 15(x + 9). 15(x + 9) = 15(x) + 15(9) = 15x Example: Simplify 2( + ). 2( + ) = The next examples use a combination of simplification methods. Example: Simplify 5(2x + 6) + 3(5x + 7). The steps written in gray are often done mentally. You do not have to write them out when showing your work. 5(2x + 6) + 3(5x + 7) = 5(2x) + 5(6) + 3(5x) + 3(7) = 10x x + 21 = (10x + 15x) + ( ) = 25x + 51 Example: Simplify 4x + 3(5 + 6x + 2x). 4x + 3(5 + 6x + 2x) = 4x + 3(5 + 8x) = 4x + 3(5) + 3(8x) = 4x x = (4x + 24x) + 15 = 28x + 15 hat About Subtraction and Division? If this material is a review for you, you may wonder what happened to the problems with subtractions and divisions. Addition and multiplication are commutative and associative operations, and those properties make the simplification rules work. Subtraction and division are neither commutative nor associative, and yet we d like to be able to simplify algebraic expressions involving those operations. The way to do this is by using addition with negative numbers, and multiplication with reciprocals. These topics are coming up soon, but for now, we ll just use the natural numbers in algebraic expressions.!

6 Free Pre-Algebra esson 15! page 6 esson 15: Simplifying Algebraic Expressions orksheet Name Below are the answers. hat were the questions? Rearrange to Multiply Combine ike Terms Distributive Property Mixed Practice 10x 3 10x 2x x x 2 7x + 8y 10x + 4 7x x 11x + 15y 5x x xy 10x x + 2 9x x 2 y 5x + 7y 10x + 5 3x + 10 Here are the problems to simplify that correspond to the answers above. They are in the correct columns, but not in the correct order. Simplify. Then match the answers to the questions. (5x)(2x) 5x + 7y 5(x + 2) x + 2(x + 5) (5x)(2) 2x + 8x 5(2x + 1) 5 + 5(2x + 5) (5x)(2x 2 ) 5x + 6x + 7y + 8y 2(5x + 1) 2x + 5(x + 2) (5x)(x)(2y) 4 + 7x + 3x 2(x + 5) 5(2x + 1) + 2(x + 5) (5x)(2y) 8y + 6x + x 2(5x + 2) 2(2x + 1) + 5(x + 5)

7 Free Pre-Algebra esson 15! page 7 Concept Questions 1. Another way to illustrate the natural action of the distributive property is with the area of rectangles. rite the area of each rectangle as a multiplication of the sides. How should you write the length of the green rectangle to illustrate the distributive property? + = 2. Fill in the blanks with vocabulary from the lesson. a. The division problem (16 8) 2 has a different answer than the problem 16 (8 2). This shows that division is not. b. hen a number is multiplied by a variable, as in the expresson 5x, the number is called the. c. In the 6N + 8M + 2N + 1, the are 6N, 8M, 2N, and 1. d. e call 6N and 2N because they have the same variable part. expression terms like terms coefficient commutative associative distributive e. e can justify rearranging the multiplication (5x)(4x) to (5 4) (x x) by combining the and properties of multiplication. f. If you think of a fraction as a division, the fact that 3/4 is not the same as 4/3 shows that division is not.

8 Free Pre-Algebra esson 15! page 8 esson 15: Simplifying Algebraic Expressions Homework 15A Name 1. Find all the factor pairs of the given numbers Find the prime factorizations of the numbers Simplify the fractions using the prime factorizations. Convert improper fractions to mixed numbers. Simplify the fractions using the GCF. Convert improper fractions to mixed numbers Multiply the fractions. 4. Simplify the fractions. a a. 45 rotations 1 second 3 minutes 60 seconds 1 1 minute b a 16a 45 b. 45a2 b 90ab 2 5. Solve the equations. a. a 5 = Use the numbers you know in the formula to solve the problems. a. A rectangle has area 575 square feet. The length is 23 feet. hat is the width? b. 4b = 12

9 Free Pre-Algebra esson 15! page 9 5. continued Solve the equations. c. c! 22 = 114 d. d + 77 = continued Use the numbers you know in the formula to solve the problems. b. A jet traveled 3,030 miles at a rate of 505 mph. How long did the trip take? 7. Rearrange the multiplication to simplify. a. (3a)(8a) 8. Simplify by combining like terms where possible. a. 3x + 2y + 5x + 12y b. (5x)(x 2 ) c. (12y)(3xy) b c n + 4n + n Use the distributive property to simplify the expressions. a. 5(n + 1) 10. Simplify. a. 2a + 3(a + 1) b. 6(3x + 2) b (3x + 7) c. 7x(y + 9) c. 2(3x + 1) + 5(3x + 1) 11. Estimate the number of sailboats. 12. Find the area and perimeter of the figure (centimeters).

10 Free Pre-Algebra esson 15! page 10 esson 15: Simplifying Algebraic Expressions Homework 15A Answers 1. Find all the factor pairs of the given numbers. Simplify the fractions using the GCF. Convert improper fractions to mixed numbers = = = 14 = 7 6 = Find the prime factorizations of the numbers. 95 = = Simplify the fractions using the prime factorizations. Convert improper fractions to mixed numbers = = = = 6 5 = Multiply the fractions. a. b. = = = a 16a a a = Simplify the fractions. a. 45 rotations 1 second 3 minutes 60 seconds 1 1 minute 45 rotations 1 second 3 minutes 1 = 8100 rotations 60 seconds 1 minute b. 45a2 b 90ab 2 = a a b a b b = a 2b 5. Solve the equations. a. a 5 = 10 5 a = a = 50 b. 4b = 12 4b 4 = 12 4 b = 3 6. Use the numbers you know in the formula to solve the problems. a. A rectangle has area 575 square feet. The length is 23 feet. hat is the width? A = = = 23 = 25 feet

11 Free Pre-Algebra esson 15! page continued Solve the equations. c. c! 22 = 114 c! = c = 136 d. d + 77 = 111 d + 77! 77 = 111! 77 d = Rearrange the multiplication to simplify. a. (3a)(8a) = (3 8) (a a) = 24a2 6. continued Use the numbers you know in the formula to solve the problems. b. A jet traveled 3,030 miles at a rate of 505 mph. How long did the trip take? d = rt 505t 505 = = 505t t = 6 hours 8. Simplify by combining like terms where possible. a. 3x + 2y + 5x + 12y = 8x + 14y b. (5x)(x 2 ) = 5 (x x x) = 5x 3 b = c. (12y)(3xy) = (12 3) x (y y) = 36xy c n + 4n + n + 2 = 10n Use the distributive property to simplify the expressions. a. 5(n + 1) = 5n Simplify. a. 2a + 3(a + 1) = 2a + 3a + 3 = 5a + 3 b. 6(3x + 2) = 18x + 12 b (3x + 7) = x + 28 = 12x + 33 c. 7x(y + 9) = 7xy + 63x 11. Estimate the number of sailboats. c. 2(3x + 1) + 5(3x + 1) = 6x x + 5 = 21x Find the area and perimeter of the figure (centimeters). There are about 400 sailboats. Area 18 square cm; Perimeter 26 cm

12 Free Pre-Algebra esson 15! page 12 esson 15: Simplifying Algebraic Expressions Homework 15B Name 1. Find all the factor pairs of the given numbers Find the prime factorizations of the numbers Simplify the fractions using the prime factorizations. Convert improper fractions to mixed numbers. Simplify the fractions using the GCF. Convert improper fractions to mixed numbers Multiply the fractions. 4. Simplify the fractions. a a b. 75x x b. 7ab 21a 5. Solve the equations. a. 3a = Use the numbers you know in the formula to solve the problems. a. A rectangle has area 120 square feet. The length is 15 feet. hat is the width? b. b! 7 = 21

13 Free Pre-Algebra esson 15! page continued Solve the equations. c. c + 7 = continued Use the numbers you know in the formula to solve the problems. b. A biker traveled 60 miles at a rate of 20 mph. How long did the trip take? d. d 7 = Rearrange the multiplication to simplify. a. (8a 2 )(a) 8. Simplify by combining like terms where possible. a. 8a + 7a b. (8a)(2a) c. (8a)(8ab) b. 8a + a c. 8a 2 + 2a 2 9. Use the distributive property to simplify the expressions. a. 3(7x + 3) b. 7(7x + 3) c. 2(x + 1) 10. Simplify. a. 2(3x + 11) + 9 b. 4(3x + 5) + 8x c. 3(4x + 5) + 6(7x + 8) 11. Estimate the number of pigeons. 12. Find the area and perimeter of the figure (centimeters).