# Algebra Tiles Activity 1: Adding Integers

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2 The Zero Pair When two tiles of unlike signs are combined, they form a zero pair ( = 0) + - =0 When you combine positive and negative tiles, find the zero pairs and remove them. The tiles that remain show the answer. C. Try the following additions with the tiles. Draw the tiles, remove the zero pairs (cross them out) and find the integer answer. Model Answer In every example above, what type of numbers were you adding? 7. What operation did you perform with the number parts? 8. What happened with the sign? 9. What rule can you develop for adding positive and negative integers? Conclusion: When adding integers with the same sign, you the numbers and the sign. When adding integers with different signs, you the numbers and use the sign of the absolute value.

3 Algebra Tiles Activity 2: Subtracting Integers We are going to use positive (yellow) and negative (red) tiles to discover the rules for adding and subtracting integers. Each tile has a value of 1 or 1. When you subtract with tiles, you remove the tiles that are being subtracted. What remains is your answer. 5 3 = = 2 A. Try the following subtractions with the tiles. Draw the tiles and find the integer answer. Model Answer ( 2) Using tiles, can you do the following? 4 7 At first it appears not, but you can subtract more than you have by introducing zero pairs until you have enough. Step 1: Start with 4 tiles. Step 2: Introduce 3 zero pairs = Step 3: Remove the 7 positive tiles. This leaves 3 negative tiles, which is your answer

4 B. Try the following subtractions with the tiles. Draw the tiles and find the integer answer. Model Answer 1. 2 ( 4) 2. 4 ( 6) ( 2) 6. What did you notice about subtracting integers? 7. What rule can you develop for subtracting integers? The rule for subtracting integers is often stated as to subtract integers you add the opposite. C. Try the following subtractions with the tiles by adding the opposite. Draw the tiles and find the integer answer. Model Answer ( 4) ( 4)

5 Write a paragraph explaining to a friend how to subtract integers.

6 EXPLORING THE DISTRIBUTIVE PROPERTY x 2 = x 2 = x = x = +1 = 1 = blue red green red yellow red The Distributive Property combines multiplication with addition or subtraction. Example: 2(x + 3) 2(x + 3) = 2x + 6 Model each of the following problems to simplify the variable expression. 1. 2(x + 4) 2. 3(x + 2) 3. 2(2x + 1) 4. 3(x 1) 5. 4(x 2) 6. 2(3x + 1) 7. 2(-x + 1) 8. 3(x 4) 9. 5(-2x + 3) 10. In your own words explain how the distributive property works or write an algebraic expression to represent the concept.

7 EXPLORING SIMPLIFYING VARIABLE EXPRESSIONS x 2 = x 2 = x = x = +1 = 1 = blue red green red yellow red You simplify variable expressions by combining like terms. Use zero pairs, where needed, and perform as many operations as possible within the expression. Example. 3a + ( a) = + = = 2a 0 2y y 3 = + + = 3y Use models to simplify these expressions. 1. 4x + 7x = 2. 5b 2b = 3. 3a + (-2a) = 4. 5y ( 2y) = 5. 3c c 2 = x + 3x x = Remember 2(x + 3) means + = 2x + 6 Use models to simplify these expressions. 7. 2(y 2 4) = 8. 3(a + 3) = 9. 2(z 4) + 9 = 10. 2(2b 2 1) + 7 = c 2 + 4(c 2 + 3c) 5c = 12. 8y + 2(3 y) 4 =

8 NY Standards: 7.A.4; A.A.22 EXPLORING SOLVING EQUATIONS x = x = +1 = 1 = green red yellow red IMPORTANT: Equations are like a scale in balance: to keep it balanced you must always do the operations on both sides of the equal sign at the same time. Ex. x + 5 = 8 = Remove 5 yellow squares from each side. Your answer x = 3. Now you try a few using your algebra tiles. 1. x + 4 = x + 3 = x + 6 = 9 4. x + 5 = x + 2 = 12 + x x = 5 + 2x Sometimes you need to group the tiles to find the value of the variable. Think about these examples. Use your tiles and group them to find the values. 7. 2x = x = x = x = 20 Combine the two methods to solve these equations x + 4 = x + 5 = 2x x = 3x x + 1 = 4x + 10 =

9 EXPLORING S0LVING EQUATIONS 2 x = x = +1 = 1 = green red yellow red IMPORTANT: Equations are like a scale in balance: to keep it balanced you must always do the operations on both sides of the equal sign at the same time. REMEMBER: When subtracting integers you add the opposite. Ex. x 3 = 6 becomes x + 3 = 6 = You must add positive 3 to both sides then remove the zero pairs. Your answer x = 9. Now you try a few using your algebra tiles. Write the steps you used below the problem. 1. x 4 = x 3 = 7 = 3. x 6 = 4 4. x 5 = x 3 = x 2 = x 5 = x 8 = 14

10 EXPLORING ADDING AND SUBTRACTING POLYNOMIALS NY Standards: 8.A.5, 7, 8; A.A.13 x 2 = x 2 = x = x = +1 = 1 = blue red green red yellow red Add and subtract by combining like terms and by using adding zero where needed. Ex. (2x 2 + x + 3) + (x 2 3x + 1) + Form a zero pair with +x and x. = 3x 2 2x Try these: 1. (2x 2 + 3x + 3) + (x 2 + 2x 2) 2. (3x 2 + 4x 3) + (4x 2 2x 4) 3. (x 2 + 3x 4) (2x 2 + 2x 3) 4. (5x 2 2x + 1) (x 2 +2x 4) 5. (2x 2 + x + 3) (x 2 + 2x 2) 6. (-3x 2 + 4x + 3) + (3x 2 3x 5)

11 EXPLORING MULTIPLYING POLYNOMIALS x 2 = x 2 = x = x = +1 = 1 = blue red green red yellow red ex. 1. x(x + 1) 2. 2x( x + 3) x + 1 x 2 x x x x x 2 x = x 2 + x = 2x 2 + 6x x 2 x x x Try these examples: 1. 2(3x + 1) 2. x(x + 2) 3. 2x(x + 2) 4. x(2x 1) 5. 3x(2x + 1) 6. 2x( x + 4)

12 EXPLORING MULTIPLYING BINOMIALS x 2 = x 2 = x = x = +1 = 1 = blue red green red yellow red ex. 1. (x + 1)(x + 1) 2. (x + 1)( x + 3) x + 1 x x 2 x x x x x 2 x = x 2 + 2x + 1 = x 2 + 2x x x 1 Try these examples: 1. (x + 2)(x + 1) 2. (x + 2)(x + 2) 3. ( 2x 1)(x + 2) 4. (x 3)(x - 2) 5. (-x + 3)(2x + 1) 6. (x 4)(x + 4)

13 NY Standards: A.A.14 EXPLORING DIVIDING POLYNOMIALS In order to divide polynomials, put your divisor on the left side and arrange your algebra tiles into a rectangular array so that the left edge matches your divisor. You may need to add zero pairs to complete your model. Ex. Divide x 2 + 5x + 6 by x + 2 What is the factor across the top? x 2 + 5x + 6 divided by x + 2 = x + 3, or (x + 2)(x + 3) = x 2 + 5x + 6 Divide x 2 3x + 2 by x 1 Divide x 2 2x 3 by x + 1 x 2 3x + 2 divided by x 1 = x 2, or (x 1)(x 2) = x 2 3x + 2 Notice I had to add a zero -pair (x & x) in order to match the +1 in the divisor. x 2 2x 3 divided by x + 1 = x 3, or (x + 1)(x 3) = x 2 2x 3

14 Divide each problem. 1. x 2 5x + 6 divided by x 2 2. x 2 4 divided by x x 2 6x + 8 divided by x 2 4. a 2 4a + 4 divided by x 2 5. x 2 + 8x + 15 divided by x y 2 5y 3 divided by 2x m 2 + m 1 divided by 2x x 2 7x + 12 divided by x 3

15 NY Standards: 8.A.11; A.A.20; A2.A.7 EXPLORING FACTORING POLYNOMIALS In order to factor polynomials, put your algebra tiles into a rectangular array. You may need to add zero pairs to complete your model. Ex. Factor x 2 + 4x + 3 What should be on the left hand side? above? x 2 + 4x + 3 = (x + 1)(x + 3) Factor 2x 2 + 3x 2x 2 + 3x = x(2x + 3) Factor x 2 + 3x + 4 Notice I had to add a zero -pair (x & x). x 2 + 3x + 4 = (x + 1)( x + 4)

16 Factor each problem. 1. 3x 2 + 2x 2. 5x 2 15x x 2 + 6x a 2 + 4a x 2 6x y 2 2y m 2 + m x 2 + 7x + 2

17 EXPLORING SQUARE TRINOMIALS In order to factor square trinomials, put your algebra tiles into a square array. Ex. Factor x 2 + 4x + 4 What are the factors on the left hand side? above? (They should be the same.) x 2 + 4x + 4 = (x + 2)(x + 2) Factor 4x 2 4x + 1 Factor p 2 2p + 1 4x 2 4x + 1 = (2x 1)(2x 1) p 2 2p + 1 = (p 1)(p 1)

18 Determine if any of the following are square trinomials. Explain why or why not. Can you tell without using the algebra tiles? 1. x 2 + 6x x 2 + 6x p 2 8p a 2 + 4a q 2 10q y y m m x 2 6x + 9

19 NY Standards: A.A.28; A2.A.24, 25 COMPLETING THE SQUARE Completing the square enables you to solve equations that are not easily factorable. It will lead to the quadratic formula for solving quadratic equations. Method 1: In the equation ax 2 + bx +c = 0, if a is a square and b is evenly divisible by 2 a, then arrange ax 2 + bx into the beginning of a square. Add as many zero pairs as needed to make a square. Combine the extra +1's and 1's together. Ex. Solve x 2 + 4x + 3 = 0 x 2 + 4x + 3 = (x + 2) 2 1 = 0, or (x + 2) 2 = 1, x + 2 = ±1, x = 1, 3 Solve 4x 2 4x 3 = 0 4x 2 4x 3 = (2x 1) 2 4 = 0, or (2x 1) 2 = 4, 2x 1 = ±2, x = 3 2, " 1 2

20 Method 2: In the equation ax 2 + bx +c = 0, if a is not a square, then divide the equation by a. Divide the b a in half and arrange the x2 + b x into the beginning of a square. Add as many zero a pairs as needed to make a square. Combine the extra +1's and 1's together. Solve 3p 2 6p 2 = 0 p 2 2p 2 3 = p 2 6p 2 = 0 p 2 2p 2 3 = 0, (p 1)2 5 3 = 0, p = 1± 5 3 Solve: 2x 2 5x = 3 x 2 " 5 2 x = = = Notice that in this example we did not add a zero pair of and equation, but instead added + 25 to both sides of the equation. 16 2x 2 5x = 3 x 2 " 5 2 x = 3 2, # x " 5 & % ( \$ 4' Solve the following equations by completing the square. 1. x 2 2x 3 = 0 2 to the left side of the = 49 16, x " 5 4 = ± 7 4, x = 3, " x 2 + 4x = 2

21 3. p 2 6p + 3 = 0 4. a 2 + 4a = q 2 10q 5 = y y = 4 7. m 2 7m + 3 = x 2 6x = 1 The Quadratic Formula Solve: a " x 2 + b " x + c = 0 x 2 + b a x + c a = 0 x2 + b a x = " c a b 2a b 2a = " c a b 2a = " c a + b 2 4a 2 b 2a b 2 4a 2 a " x 2 + b " x + c = 0 " x + b % \$ ' # 2a& 2 = b2 ( 4ac 4a 2, x + b 2a = ± b2 " 4ac 2a, or x = "b ± b2 " 4ac 2a.

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