Algebra Tiles Activity 1: Adding Integers

Algebra TilesActivity 1: Adding IntegersNY Standards: 7/8.PS.6,7; 7/8.CN.1; 7/8.R.1; 7.N.13We are going to use positive (yellow) and negative (red) tiles to discover the rules foradding and subtracting integers. Each tile has a value of 1 or –1.3 = + + + –2 = – –A. Try the following additions with the tiles. Draw the tiles and find the integer answer.1. 3 + 12. 3 + 53. 2 + 4ModelAnswer4. In every example above, what type of numbers were you adding?5. What operation did you perform with the number parts?6. What happened with the sign?7. What rule can you develop for adding positive integers?B. Try the following additions with the tiles. Draw the tiles and find the integer answer.ModelAnswer1. –2 + –12. –4 + –33. –2 + –54. In every example above, what type of numbers were you adding?5. What operation did you perform with the number parts?6. What happened with the sign?7. What rule can you develop for adding negative integers?

The Zero PairWhen two tiles of unlike signs are combined, they form a zero pair (+1 + –1 = 0)+-=0When you combine positive and negative tiles, find the zero pairs and remove them. Thetiles that remain show the answer.C. Try the following additions with the tiles. Draw the tiles, remove the zero pairs (cross themout) and find the integer answer.ModelAnswer1. –3 + 12. –4 + 53. –2 + 64. 7 + –45. 2 + –56. 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.

Algebra TilesActivity 2: Subtracting IntegersWe are going to use positive (yellow) and negative (red) tiles to discover the rules foradding 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 remainsis your answer.5 – 3 = + + + + += 2A. Try the following subtractions with the tiles. Draw the tiles and find the integer answer.ModelAnswer1. 3 – 12. –5 – (–2)3. 6 – 44. Using tiles, can you do the following? 4 – 7At first it appears not, but you can subtract more than you have by introducing zero pairs untilyou 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.+ + + + +–+–+–

B. Try the following subtractions with the tiles. Draw the tiles and find the integer answer.ModelAnswer1. –2 – (–4)2. –4 – (–6)3. 2 – 54. –2 – 45. 5 – (–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 theopposite.”C. Try the following subtractions with the tiles by adding the opposite. Draw the tiles and findthe integer answer.ModelAnswer1. –3 – 42. 4 – (–4)3. –2 – 64. –3 – (–4)5. 2 – 5

EXPLORING THE DISTRIBUTIVE PROPERTYx 2 = –x 2 = x = –x = +1 = –1 =blue red green red yellow redThe Distributive Property combines multiplication with addition or subtraction.Example: 2(x + 3)2(x + 3) = 2x + 6Model 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 algebraicexpression to represent the concept.

EXPLORING SIMPLIFYING VARIABLE EXPRESSIONSx 2 = –x 2 = x = –x = +1 = –1 =blue red green red yellow redYou 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) = + = = 2a02y + 7 + y – 3 = + + = 3y + 4–3Use models to simplify these expressions.1. 4x + 7x = 2. 5b – 2b =3. 3a + (-2a) = 4. 5y – (–2y) =5. 3c 2 – 8 + 2c 2 = 6. 8 – 5x + 3x 2 – 2 + 3x =Remember 2(x + 3) means + = 2x + 6Use 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 =11. 10c 2 + 4(c 2 + 3c) – 5c = 12. 8y + 2(3 – y) – 4 =

NY Standards: 7.A.4; A.A.22EXPLORING SOLVING EQUATIONSx = –x = +1 = –1 =green red yellow redIMPORTANT: Equations are like a scale in balance: to keep it balanced youmust 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 = 11 2. x + 3 = 103. x + 6 = 9 4. x + 5 = 95. 2x + 2 = 12 + x 6. 4 + 3x = 5 + 2xSometimes 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 = 8 8. 3x = 129. 4x = 8 10. 5x = 20Combine the two methods to solve these equations.11. 3x + 4 = 10 12. 4x + 5 = 2x + 1113. 8 + 5x = 3x + 12 14. 7x + 1 = 4x + 10=

EXPLORING S0LVING EQUATIONS 2x = –x = +1 = –1 =green red yellow redIMPORTANT: Equations are like a scale in balance: to keep it balanced youmust 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 = 6becomesx + –3 = 6=You must add positive 3 to both sides then remove the zero pairs. Your answerx = 9.Now you try a few using your algebra tiles. Write the steps you used below theproblem.1. x – 4 = 10 2. x – 3 = 7=3. x – 6 = 4 4. x – 5 = 65. 2x – 3 = 1 6. 3x – 2 = 107. 4x – 5 = 7 8. 2x – 8 = 14

EXPLORING ADDING AND SUBTRACTING POLYNOMIALSNY Standards: 8.A.5, 7, 8; A.A.13x 2 = –x 2 = x = –x = +1 = –1 =blue red green red yellow redAdd 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 + 40Try 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)

EXPLORING MULTIPLYING POLYNOMIALSx 2 = –x 2 = x = –x = +1 = –1 =blue red green red yellow redex. 1. x(x + 1) 2. 2x(–x + 3)x + 1–x 2 x x xx x 2 x = x 2 + x = –2x 2 + 6x–x 2 x x xTry 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)

EXPLORING MULTIPLYING BINOMIALSx 2 = –x 2 = x = –x = +1 = –1 =blue red green red yellow redex. 1. (x + 1)(x + 1) 2. (x + 1)(–x + 3)x + 1x –x 2 x x xx x 2 x = x 2 + 2x + 1 = –x 2 + 2x + 3+ 1 –x 1 1 11 x 1Try 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)

NY Standards: A.A.14EXPLORING DIVIDING POLYNOMIALSIn order to divide polynomials, put your divisor on the left side and arrange your algebra tilesinto a rectangular array so that the left edge matches your divisor. You may need to add zeropairs to complete your model.Ex. Divide x 2 + 5x + 6 by x + 2RearrangeWhat is thefactoracross thetop?x 2 + 5x + 6 divided by x + 2 = x + 3, or (x + 2)(x + 3) = x 2 + 5x + 6Divide x 2 – 3x + 2 by x – 1RearrangeDivide x 2 – 2x – 3 by x + 1x 2 – 3x + 2 divided by x – 1 = x – 2, or(x – 1)(x – 2) = x 2 – 3x + 2RearrangeNotice I had toadd 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

Divide each problem.1. x 2 – 5x + 6 divided by x – 22. x 2 – 4 divided by x + 23. x 2 – 6x + 8 divided by x – 24. a 2 – 4a + 4 divided by x – 25. x 2 + 8x + 15 divided by x + 36. 2y 2 – 5y – 3 divided by 2x + 17. 6m 2 + m – 1 divided by 2x + 18. x 2 – 7x + 12 divided by x – 3

NY Standards: 8.A.11; A.A.20; A2.A.7EXPLORING FACTORING POLYNOMIALSIn order to factor polynomials, put your algebra tiles into a rectangular array. You may need toadd zero pairs to complete your model.Ex. Factor x 2 + 4x + 3RearrangeWhat should be on theleft hand side?above?x 2 + 4x + 3 = (x + 1)(x + 3)Factor 2x 2 + 3xRearrange2x 2 + 3x = x(2x + 3)Factor –x 2 + 3x + 4RearrangeNotice I had toadd a zero-pair (x & –x).–x 2 + 3x + 4 = (x + 1)(–x + 4)

Factor each problem.1. –3x 2 + 2x2. 5x 2 –15x + 103. x 2 + 6x + 84. a 2 + 4a + 45. x 2 – 6x + 96. y 2 – 2y – 157. m 2 + m – 128. 6x 2 + 7x + 2

EXPLORING SQUARE TRINOMIALSIn order to factor square trinomials, put your algebra tiles into a square array.Ex. Factor x 2 + 4x + 4RearrangeWhat are the factorson the left hand side?above? (They shouldbe the same.)x 2 + 4x + 4 = (x + 2)(x + 2)Factor 4x 2 – 4x + 1RearrangeFactor p 2 – 2p + 14x 2 – 4x + 1 = (2x – 1)(2x – 1)Rearrangep 2 – 2p + 1 = (p – 1)(p – 1)

Determine if any of the following are square trinomials. Explain why or why not. Can you tellwithout using the algebra tiles?1. x 2 + 6x + 92. x 2 + 6x + 363. p 2 – 8p + 164. a 2 + 4a + 645. q 2 – 10q – 256. 9y 2 + 12y + 47. 16m 2 + 40m + 258. 4x 2 – 6x + 9

NY Standards: A.A.28; A2.A.24, 25COMPLETING THE SQUARECompleting the square enables you to solve equations that are not easily factorable. It will leadto 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 makea square. Combine the extra +1's and –1's together.Ex. Solve x 2 + 4x + 3 = 0!Rearrangex 2 + 4x + 3 = (x + 2) 2 – 1 = 0, or (x + 2) 2 = 1, x + 2 = ±1, x = –1, –3Solve 4x 2 – 4x – 3 = 0Rearrange4x 2 – 4x – 3 = (2x – 1) 2 – 4 = 0, or (2x – 1) 2 = 4, 2x – 1 = ±2, x =32 , " 1 2!

Method 2: In the equation ax 2 + bx +c = 0, if a is not a square, then divide the equation by a.bDivide the in half and arrange the x 2 + b x into the beginning of a square. Add as many zeroa apairs as needed to make a square. Combine the extra +1's and –1's together.!Solve 3p 2 – 6p – 2 = 0 ⇒ p 2 – 2p –!23 = 0 –!Rearrange23–233p 2 – 6p – 2 = 0 ⇒ p 2 – 2p –!Solve: 2x 2 – 5x = 3 ⇒!54!54=x 2 " 5 2 x = 3 2!3223 = 0, (p – 1)2 –!!= 4916 ,!54542516!53 = 0, p =!=!!321± 5 3! !25Notice that in this example we did not add a zero pair of +16 and – 25to the left side of the1625! !equation, but instead added + to both sides of the equation.162x 2 – 5x = 3 ⇒x 2 " 5! 2 x = 3 2 ,#x " 5 &% ($ 4'2+2516!x " 5 4 = ± 7 4 , x = 3, " 1 2 .Solve the following equations by completing the square.1. x 2 ! 2x – 3 = 0 !!!2. x 2 + 4x = – 2

3. p 2 – 6p + 3 = 04. a 2 + 4a = 145. q 2 – 10q – 5 = 06. 9y 2 + 12y = – 47. m 2 – 7m + 3 = 08. 4x 2 – 6x = 1The Quadratic FormulaSolve:a " x 2 + b " x + c = 0 ⇒x 2 + b a x + c a = 0 ⇒x 2 + b a x = " c a!!!b2ab2a!!=" c a!b2a=" c a+b 24a 2!b2ab 24a 2!!!a " x 2 + b " x + c = 0 ⇒!"x + b %$ '# 2a&2= b2 ( 4ac,4a 2 !x + b 2a = ± b2 " 4ac2a!, or x ="b ± b 2 " 4ac2a.!!