VARIABLES AND COMBINING LIKE TERMS and 2.1.2


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1 VARIABLES AND COMBINING LIKE TERMS 2.. and 2..2 Algebraic epressions can be represented b the perimeters and areas of algebra tiles 2 (rectangles and squares) and combinations of algebra tiles. The dimensions of each tile are shown along its sides and the area is shown on the tile itself in the figures at left. When using the tiles, perimeter is the distance around the eterior of a figure. The area of 2 a figure is the sum of the areas of the individual pieces (length times width of each piece). Combining terms that have the same area to write a simpler epression is called combining like terms. When working without tiles or pictures of the tiles, simpl add or subtract the coefficients of the like terms. The Math Notes boes on pages 49 and 57 eplain the standard order of operations and their use with algebra tiles. Using Algebra Tiles Eample Eample 2 Eample Perimeter = Perimeter = 6 8 Perimeter = 8 6 Area = 2 2 Area = Area = Combining Like Terms Eample = Eample = Eample 6: 3 2! 4 3! 2 3! 7 = 2 2!! 4 6 Algebra Connections Parent Guide
2 Problems Determine the area and the perimeter of each figure Simplif each epression b combining like terms ! ! ! 3! 4 2! ! ! (4 2 4! ) ( 2! 7) 7. ( ) (2 2 4! 2 ) 8. (7 2! 6! 9)! (9 2 3! 4) 9. (3 2! 8! 4)! (5 2 ) Answers. P = 4 4 A = P = 4 4 A = P = 8 6 A = P = 4 6 A = P = 4 4 A = P = 6 A = 2 3. P = 4 8 A = P = A = P = 6 4 A = ! 5 3.! !2 2! 9! 5 9.!2 2! 9! 5 Chapter 2: Variables and Proportions 7
3 WRITING AND SIMPLIFYING 2..3 through 2..7 ALGEBRAIC EXPRESSIONS An epression mat is an organizing tool that is used to represent algebraic epressions. Pairs of epression mats can be modified to make an equation mat (see net section). The upper half of an epression mat is the positive region and the lower half is the negative region. Positive algebra tiles are shaded and negative tiles are blank. A matching pair of tiles with one tile shaded and the other one blank represents zero (0). Tiles ma be removed from or moved on an epression mat in one of three was: () removing the same number of opposite tiles in the same region; (2) flipping a tile from one region to another. Such moves create opposites of the original tile, so a shaded tile becomes unshaded and an unshaded tile becomes shaded; and (3) removing an equal number of identical tiles from both the and  regions. See the Math Notes bo on page 60. Eamples 3! 4 can be represented various was. = = The epression mats at right all represent zero. = = Eample 3 2! (2! 3) Epressions can be simplified b moving tiles to the top (change the sign) and looking for zeros. = = 3 2! (2! 3) 3 2! Algebra Connections Parent Guide
4 Eample 2! (2! 3)! 2 = =! (2! 3)! 2! 2 3! 2! 2 Problems Simplif each epression ! 4! 7 8.!! 4! 7 9.!(! 3) 0. 4! ( 2). 5! (!3 2) 2.! 5! (2! ) 3.! 2! 2 4.!3 5 5! 5. 3! ( 5) 6.!( ) ! 7! (3! 7) 8.!( 2 3)! 3 Answers ! ! 6.! 5 7.!2! 8.!5! 7 9.! ! 2. 8! ! 7 3.! !! !4!! 3 Chapter 2: Variables and Proportions 9
5 SOLVING EQUATIONS 2..8 and 2..9 Combining two epression mats into an equation mat creates a concrete model for solving equations. Practicing solving equations using the model will help students transition to solving equations abstractl with better accurac and understanding. In general, and as shown in the first eample below, the negative in front of the parenthesis causes everthing inside to flip from the top to the bottom or the bottom to the top of an epression mat, that is, all terms in the epression change signs. After simplifing the parentheses, simplif each epression mat. Net, isolate the variables on one side of the equation mat and the nonvariables on the other side b removing matching tiles from both sides. Then determine the value of the variable. Students should be able to eplain their steps. See the Math Notes bo on page 69. Procedure and Eample Solve 2! (!2) = 5! (! 3). First build the equation on the equation mat. Second, simplif each side using legal moves on each epression mat, that is, on each side of the equation mat. Isolate terms on one side and nonterms on the other b removing matching tiles from both sides of the equation mat. Finall, since both sides of the equation are equal, determine the value of. 0 Algebra Connections Parent Guide 2 2 = 5! = 8 3 = 6 = 2
6 Once students understand how to solve equations using an equation mat, the ma use the visual eperience of moving tiles to solve equations with variables and numbers. The procedures for moving variables and numbers in the solving process follow the same rules. Note: When the process of solving an equation ends with different numbers on each side of the equal sign (for eample, 2 = 4), there is no solution to the problem. When the result is the same epression or number on each side of the equation (for eample, 2 = 2 ) it means that all numbers are solutions. See Section in this guide. Eample Solve 3 3! = 4 9 Solution 3 3! = 4 9 6! = = 0 = 5 problem simplif add, subtract 4 on each side divide Eample 2 Solve!2! (!3 3) =!4 (!! 2) Solution Problems!2! (!3 3) =!4 (!! 2)!2 3! 3 =!4!! 2! 2 =!! 6 2 =!4 =!2 problem remove parenthesis (flip) simplif add, add 2 to each side divide Solve each equation.. 2! 3 =! ! =! ! 3 = 2! ! (! 2) = !( 3) = 2! 6 6.!4 3! = ! 3 = ! 3 2 = ! 8! 2 = ! (! 3) = 4! (3! ). 2! 7 =!! 2.!2! 3 =! 2! 4 3.!3 7 =! 4. 2! 4 =!3! (!) 5. 2!! =! 3! (!5 ) 6.!4! 3 =!! = !(! 2) =! 5! !! 3 = 4! ! (! 3) =! 5 Answers !6 7.!7 8. no answer ! all numbers no answer 7. 8.! !4 3 Chapter 2: Variables and Proportions 2 5
7 PROPORTIONS 2.2. through Students solve proportional reasoning (ratio) problems in a variet of was. The ma find the number or cost per one unit and then multipl b the number of units. The ma also organize work in a table. Later in the course students will use ratio equations or proportions. See Section for this topic. Eample Mrs. Salem s student assistant can correct 8 homework papers in 0 minutes. At this rate how man papers can she correct during a 55 minute class? 8 0 =.8 papers per minute.8 55 = 99 papers Eample 2 Tob uses 7 tubes of toothpaste ever 0 months. How man tubes will he use in 24 months? How long will 35 tubes last him? Tubes Months or 7 tubes 35 tubes = so = 0 months 50 months 5 about 7 tubes 50 months 2 Algebra Connections Parent Guide
8 Problems Solve and eplain our reasoning. Alice knows si cups of rice will make enough Spanish rice to feed 5 people. How much rice is needed to feed 75 people? 2. Elaine can plant 6 flowers in 0 minutes. How man can she plant in 25 minutes? 3. Ivanna needs to bu 360 cherries for a large salad. She can bu nine cherries for $0.57. How much will 360 cherries cost? 4. A plane travels 3400 miles in eight hours. How far would it travel in si hours at the same rate? 5. Leslie can write a 500word essa in a hour. If she writes an essa in 20 minutes, approimatel how man words should the essa contain? 6. About eight out of ever 00 people in the state have red hair. If a tpical classroom in the state has 25 students, how man would ou epect to have red hair? If the tpical middle school has 650 students, how man would ou epect to have red hair? 7. When Carlos rides his bike to school, it takes 5 minutes to go 8 blocks. If he rides at the same speed, how long should it take him to travel 30 blocks? 8. Simba the cat is on a diet. Ten pounds of special lowfat food costs $ How much would 30 pounds cost? How much would 36 pounds cost? 9. Elizabeth came to bat 0 times in 20 games. How man times should she epect to bat in 62 games? 0. L can deliver 32 newspapers in 25 minutes on his bike. Net week he needs to deliver 80 newspapers in the same neighborhood. How long should it take him if he works at the same rate as he did for 32 newspapers? Answers. 30 cups flowers 3. $ miles 5.!67 words 6. 2 & 52 people min. 8. $67.50; $ atbats min. Chapter 2: Variables and Proportions 3
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