Summer Prep Packet for students entering Algebra The following skills and concepts included in this packet are vital for your success in Algebra. The Mt. Hebron Math Department encourages all students to keep up with their math skills over the summer. We believe that worksheets contained in this summer packet will be helpful to you as you prepare for a great math year in Algebra. *Please note: Completion of the following packet is strongly encouraged although it is not a requirement.
Solving Multi-Step Equations Solve Multi-Step Equations To solve equations with more than one operation, often called multi-step equations, undo operations by working backward. Reverse the usual order of operations as you work. Example Solve 5x + 3 = 3. 5x + 3 = 3 5x + 3-3 = 3-3 5x = 0 5x 5 = 0 5 x = 4 Original equation. Subtract 3 from each side. Simplify. Divide each side by 5. Simplify. Solve each equation. Check your solution. 1. 5x + = 7. 6x + 9 = 7 3. 5x + 16 = 51 4. 14n - 8 = 34 5. 0.6x - 1.5 = 1.8 6. 7 8 p - 4 = 10 7. 16 = d - 1 14 8. 8 + 3n 1 = 13 9. g -5 + 3 = -13 10. 4b + 8 = 10 11. 0.x - 8 = - 1. 3.y - 1.8 = 3-13. -4 = 7x - (-1) -8 14. 8 = -1 + k -4 Write an equation and solve each problem. 16. Find three consecutive integers whose sum is 96. 17. Find two consecutive odd integers whose sum is 176. 18. Find three consecutive integers whose sum is -93. 15. 0 = 10y - 40 Chapter Glencoe Algebra 1
Solving Equations with the Variable on Each Side Variables on Each Side To solve an equation with the same variable on each side, first use the Addition or the Subtraction Property of Equality to write an equivalent equation that has the variable on just one side of the equation. Then solve the equation. Example 1 Solve 5y - 8 = 3y + 1. Example 5y - 8 = 3y + 1 5y - 8-3y = 3y + 1-3y y - 8 = 1 y - 8 + 8 = 1 + 8 y = 0 y = 0 y = 10 The solution is 10. Solve -11-3y = 8y + 1. -11-3y = 8y + 1-11 - 3y + 3y = 8y + 1 + 3y -11 = 11y + 1-11 - 1 = 11y + 1-1 -1 = 11y -1 11 = 11y 11-1 1 11 = y The solution is -1 1 11. Solve each equation. Check your solution. 1. 6 - b = 5b + 30. 5y - y = 3y + 3. 5x + = x - 10 4. 4n - 8 = 3n + 5. 1.x + 4.3 =.1 - x 6. 4.4m + 6. = 8.8m - 1.8 7. 1 b + 4 = 1 8 b + 88 8. 3 4 k - 5 = 1 k - 1 9. 8-5p = 4p - 1 4 10. 4b - 8 = 10 - b 11. 0.x - 8 = - - x 1. 3y - 1.8 = 3y - 1.8 13. -4-3x = 7x - 6 14. 8 + 4k = -10 + k 15. 0 - a = 10a - 16. 3 n + 8 = 1 n + 17. 5 y - 8 = 9-3 y 18. -4r + 5 = 5-4r 5 19. -4-3x = 6x - 6 0. 18-4k = -10-4k 1. 1 + y = 10y - 1 Chapter 3 Glencoe Algebra 1
Multiplying Monomials Monomials A monomial is a number, a variable, or the product of a number and one or more variables with nonnegative integer exponents. An expression of the form x n is called a power and represents the product you obtain when x is used as a factor n times. To multiply two powers that have the same base, add the exponents. Product of Powers For any number a and all integers m and n, a m a n = a m + n. Example 1 Simplify (3x 6 )(5x ). Example (3x 6 )(5x ) = (3)(5)(x 6 x ) Group the coeffi cients and the variables = (3 5)(x 6 + ) Product of Powers = 15x 8 Simplify. The product is 15x 8. Simplify each expression. Simplify (-4a 3 b)(3a b 5 ). (-4a 3 b)(3a b 5 ) = (-4)(3)(a 3 a )(b b 5 ) = -1(a 3 + )(b 1 + 5 ) = -1a 5 b 6 The product is -1a 5 b 6. 1. y(y 5 ). n n 7 3. (-7x )(x 4 ) 4. x(x )(x 4 ) 5. m m 5 6. (-x 3 )(-x 4 ) 7. (a )(8a) 8. (rn)(rn 3 )(n ) 9. (x y)(4xy 3 ) 10. 1 3 ( a 3 b)( 6b 3 ) 11. (-4x 3 )(-5x 7 ) 1. (-3j k 4 )(jk 6 ) 13. ( 5a bc 3 ) ( 1 5 abc 4 ) 14. (-5xy)(4x )(y 4 ) 15. (10x 3 yz )(-xy 5 z) Chapter 7 89 Glencoe Algebra 1
Multiplying Monomials Simplify Expressions An expression of the form (x m ) n is called a power of a power and represents the product you obtain when x m is used as a factor n times. To find the power of a power, multiply exponents. Power of a Power For any number a and all integers m and n, (a m ) n = a mn. Power of a Product For any number a and all integers m and n, (ab) m = a m b m. We can combine and use these properties to simplify expressions involving monomials. Example Simplify (-ab ) 3 (a ) 4. (-ab ) 3 (a ) 4 = (-ab ) 3 (a 8 ) Power of a Power = (-) 3 (a 3 )(b ) 3 (a 8 ) Power of a Product = (-) 3 (a 3 )(a 8 )(b ) 3 Group the coefficients and the variables = (-) 3 (a 11 )(b ) 3 Product of Powers = -8a 11 b 6 Power of a Power The product is -8a 11 b 6. Simplify each expression. 1. (y 5 ). (n 7 ) 4 3. (x ) 5 (x 3 ) 4. -3(ab 4 ) 3 5. (-3ab 4 ) 3 6. (4x b) 3 7. (4a ) (b 3 ) 8. (4x) (b 3 ) 9. (x y 4 ) 5 10. (a 3 b )(b 3 ) 11. (-4xy) 3 (-x ) 3 1. (-3j k 3 ) (j k) 3 13. (5 a b) 3 ( 1 5 abf ) 14. (xy) (-3x )(4y 4 ) 15. (x 3 y z ) 3 (x z) 4 16. (-n 6 y 5 )(-6n 3 y )(ny) 3 17. (-3a 3 n 4 )(-3a 3 n) 4 18. -3(x) 4 (4x 5 y) Chapter 7 90 Glencoe Algebra 1
Dividing Monomials Quotients of Monomials To divide two powers with the same base, subtract the exponents. Quotient of Powers Power of a Quotient For all integers m and n and any nonzero number a, a m a n = a m-n. For any integer m and any real numbers a and b, b 0, ( a b) m = a m b m. Example 1 Simplify a 4 b 7 ab. Assume Example Simplify ( a 3 b 5 3 b ) 3. Assume that no denominator equals zero. a 4 b 7 ab = ( a 4 ) that no denominator equals zero. a ( b 7 b ) Group powers with the same base. ( a 3 b 5 3 b ) 3 = ( a 3 b 5 ) 3 Power of a Quotient (3 b 3 ) = (a 4-1 )(b 7 - ) Quotient of Powers = a 3 b 5 = 3 ( a 3 ) 3 ( b 5 ) 3 Power of a Product Simplify. (3) 3 ( b ) 3 The quotient is a 3 b 5. = 8 a 9 b 15 Power of a Power 7 b 6 = 8 a 9 b 9 7 The quotient is 8 a 9 b 9 7. Quotient of Powers Simplify each expression. Assume that no denominator equals zero. 1. 5 5 5. m 6 m 4 4. a a 5. x 5 y 3 x 5 y 7. x y 6 y 4 x 10. ( r 5 w 3 8. ( a b r 4 w ) 4 11. ( 3 r 6 n 3 3 3. p 5 n 4 p n 6. - y 7 14 y 5 ) 3 a 9. ( 4 p 4 r 4 3 p r ) r 5 n ) 4 1. r 7 n 7 t n 3 r 3 t 3 Chapter 7 91 Glencoe Algebra 1
Multiplying Polynomials Multiply Binomials To multiply two binomials, you can apply the Distributive Property twice. A useful way to keep track of terms in the product is to use the FOIL method as illustrated in Example. Example 1 Find (x + 3)(x - 4). Example Find (x - )(x + 5) using Horizontal Method (x + 3)(x - 4) the FOIL method. (x - )(x + 5) = x(x - 4) + 3(x - 4) = (x)(x) + x(-4) + 3(x)+ 3(-4) = x - 4x + 3x - 1 = x - x - 1 Vertical Method x + 3 ( ) x - 4-4x - 1 x + 3x x - x - 1 The product is x - x - 1. First Outer Inner Last = (x)(x) + (x)(5) + (-)(x) + (-)(5) = x + 5x + (-x) - 10 = x + 3x - 10 The product is x + 3x - 10. Find each product. 1. (x + )(x + 3). (x - 4)(x + 1) 3. (x - 6)(x - ) 4. ( p - 4)( p + ) 5. ( y + 5)( y + ) 6. (x - 1)(x + 5) 7. (3n - 4)(3n - 4) 8. (8m - )(8m + ) 9. (k + 4)(5k - 1) 10. (3x + 1)(4x + 3) 11. (x - 8)(-3x + 1) 1. (5t + 4)(t - 6) 13. (5m - 3n)(4m - n) 14. (a - 3b)(a - 5b) 15. (8x - 5)(8x + 5) 16. (n - 4)(n + 5) 17. (4m - 3)(5m - 5) 18. (7g - 4)(7g + 4) Chapter 7 101 Glencoe Algebra 1
Using the Distributive Property Solve Equations by Factoring The following property, along with factoring, can be used to solve certain equations. Zero Product Property For any real numbers a and b, if ab = 0, then either a = 0, b = 0, or both a and b equal 0. Example Solve 9x + x = 0. Then check the solutions. Write the equation so that it is of the form ab = 0. 9x + x = 0 Original equation x(9x + 1) = 0 Factor the GCF of 9x + x, which is x. x = 0 or 9x + 1 = 0 x = 0 x = - 1 9 The solution set is {0, - 1 9}. Zero Product Property Solve each equation. Check Substitute 0 and - 1 for x in the original equation. 9 9x + x = 0 9x + x = 0 9(0) + 0 0 9 (- 1 9) + (- 1 0 9) 1 0 = 0 9 + (- 1 0 9) 0 = 0 Solve each equation. Check your solutions. 1. x(x + 3) = 0. 3m(m - 4) = 0 3. (r - 3)(r + ) = 0 4. 3x(x - 1) = 0 5. (4m + 8)(m - 3) = 0 6. 5t = 5t 7. (4c + )(c - 7) = 0 8. 5p - 15p = 0 9. 4y = 8y 10. 1x = -6x 11. (4a + 3)(8a + 7) = 0 1. 8y = 1y 13. x = -x 14. (6y - 4)(y + 3) = 0 15. 4m = 4m 16. 1x = 3x 17. 1a = -3a 18. (1a + 4)(3a - 1) = 0 Chapter 8 108 Glencoe Algebra 1
Quadratic Equations: x + bx + c = 0 Factor x + bx + c To factor a trinomial of the form x + bx + c, find two integers, m and p, whose sum is equal to b and whose product is equal to c. Factoring x + bx + c x + bx + c = (x + m)(x + p), where m + p = b and mp = c. Example 1 Factor each polynomial. Example Factor x + 6x - 16. a. x + 7x + 10 In this trinomial, b = 7 and c = 10. Factors of 10 Sum of Factors 1, 10 11, 5 7 Since + 5 = 7 and 5 = 10, let m = and p = 5. x + 7x + 10 = (x + 5)(x + ) b. x - 8x + 7 In this trinomial, b = -8 and c = 7. Notice that m + p is negative and mp is positive, so m and p are both negative. Since -7 + (-1) = -8 and (-7)(-1) = 7, m = -7 and p = -1. x - 8x + 7 = (x - 7)(x - 1) Factor each polynomial. In this trinomial, b = 6 and c = -16. This means m + p is positive and mp is negative. Make a list of the factors of -16, where one factor of each pair is positive. Factors of -16 Sum of Factors 1, -16-15 -1, 16 15, -8-6 -, 8 6 Therefore, m = - and p = 8. x + 6x - 16 = (x - )(x + 8) 1. x + 4x + 3. m + 1m + 3 3. r - 3r + 4. x - x - 6 5. x - 4x - 1 6. x - x + 11 7. t - 4t - 1 8. p - 16p + 64 9. 9-10x + x 10. x + 6x + 5 11. a + 8a - 9 1. y - 7y - 8 13. x - x - 3 14. y + 14y + 13 15. m + 9m + 0 16. x + 1x + 0 17. a - 14a + 4 18. 18 + 11y + y 19. x + xy + y 0. a - 4ab + 4b 1. x + 6xy - 7y Chapter 8 109 Glencoe Algebra 1
Name: Factoring Quadratics with a=1 Date: Pd: Directions: Factor each polynomial completely (you may need to factor the GCF first). All work should be neat and organized on a separate sheet of paper. Circle / box your final answer. 1. u - 1u. 6x + x 3. z + 10z + 1 4. x + 14x + 48 5. q 9q + 18 6. k 4k 3 7. b + 18b + 81 8. m + 6m 7 9. p 5p + 6 10. n 7n 44 11. j 1j + 36 1. h + 8h + 1 13. b + b 4 14. a 3a 4 15. y 3y 88 16. r + 14r + 49 17. a 49 18. x 8x + 15 19. x + 7x + 10 0. 5t + 15t 1. w 8w + 1. c 5c 14 3. 10r 35r 4. z z 4 5. t + 1t + 36 6. a 36 7. a 6a + 9 8. 4p 16 9. w 5/4 30. t 6t 8 31. x + 4 3. x + 5x 6 33. c + 6c + 9 34. b 9b + 16 35. y -11y 1 36. x x + 8 37. p + 19p + 48 38. h 30 39. g 1g 8 40. 5 t 41. j 10j + 5 4. x x + 11 BONUS: 43. 3z + 4z 4 44. y 5y - 1
Quadratic Equations: x + bx + c = 0 Solve Equations by Factoring Factoring and the Zero Product Property can be used to solve many equations of the form x + bx + c = 0. Example 1 Solve x + 6x = 7. Check your solutions. x + 6x = 7 Original equation x + 6x - 7 = 0 Rewrite equation so that one side equals 0. (x - 1)(x + 7) = 0 Factor. x - 1 = 0 or x + 7 = 0 Zero Product Property x = 1 x = -7 Solve each equation. The solution set is {1, -7}. Since 1 + 6 = 7 and (-7) + 6(-7) = 7, the solutions check. Example ROCKET LAUNCH A rocket is fired with an initial velocity of 88 feet per second. How many seconds will it take for the rocket to hit the ground? The formula h = vt - 16t gives the height h of the rocket after t seconds when the initial velocity v is given in feet per second. h = vt - 16t Formula 0 = 88t - 16t Substitute. 0 = 16t(143 - t) Factor. 16t = 0 or 143 - t = 0 Zero Product Property t = 0 t = 143 Solve each equation. The value t = 0 represents the time at launch. The rocket returns to the ground in 143 seconds, or a little less than.5 minutes after launch. Solve each equation. Check the solutions. 1. x - 4x + 3 = 0. y - 5y + 4 = 0 3. m + 10m + 9 = 0 4. x = x + 5. x - 4x = 5 6. x - 1x + 36 = 0 7. t - 8 = -7t 8. p = 9p - 14 9. -9-8x + x = 0 10. x + 6 = 5x 11. a = 11a - 18 1. y - 8y + 15 = 0 13. x = 4-10x 14. a - 18a = -7 15. b = 10b - 16 Use the formula h = vt - 16t to solve each problem. 16. FOOTBALL A punter can kick a football with an initial velocity of 48 feet per second. How many seconds will it take for the ball to return to the ground? 17. BASEBALL A ball is thrown up with an initial velocity of 3 feet per second. How many seconds will it take for the ball to return to the ground? 18. ROCKET LAUNCH If a rocket is launched with an initial velocity of 1600 feet per second, when will the rocket be 14,400 feet high? Chapter 8 110 Glencoe Algebra 1
Solving Quadratic Equations by Using the Quadratic Formula Quadratic Formula To solve the standard form of the quadratic equation, ax + bx + c = 0, use the Quadratic Formula. Quadratic Formula -b ± the formula x = b - 4ac a that gives the solutions of ax + bx + c = 0, where a 0 Example 1 Solve x + x = 3 by Example Solve x - 6x - = 0 by using the Quadratic Formula. Rewrite the equation in standard form. using the Quadratic Formula. Round to the nearest tenth if necessary. x + x = 3 Original equation For this equation a = 1, b = -6, and c = -. x + x - 3 = 3-3 Subtract 3 from each side. x = -b ± b - 4ac x + x - 3 = 0 Simplify. a Now let a = 1, b =, and c = -3 in the Quadratic Formula. x = -b ± b - 4ac a a = - ± () - 4(1)(-3) (1) - + = 16 x = - + 4 or x = - + 4 = 1 = -3 The solution set is {-3, 1}. = 6 ± (-6) - 4(1)(-) (1) = 6 ± 44 x = 6 ± 44 or x = 6-44 x 6.3-0.3 The solution set is {-0.3, 6.3}. Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary. 1. x - 3x + = 0. x - 8x = -16 3. 16x - 8x = -1 4. x + 5x = 6 5. 3x + x = 8 6. 8x - 8x - 5 = 0 7. -4x + 19x = 1 8. x + 6x = 5 9. 48x + x - 15 = 0 10. 8x - 4x = 4 11. x + 5x = 8 1. 8x + 9x - 4 = 0 13. x + 9x + 4 = 0 14. 8x + 17x + = 0 Chapter 9 15 Glencoe Algebra 1
Simplifying Radical Expressions Product Property of Square Roots The Product Property of Square Roots and prime factorization can be used to simplify expressions involving irrational square roots. When you simplify radical expressions with variables, use absolute value to ensure nonnegative results. Product Property of Square Roots For any numbers a and b, where a 0 and b 0, ab = a b. Example 1 Simplify 180. 180 = 3 3 5 Prime factorization of 180 = 3 5 Product Property of Square Roots = 3 5 Simplify. = 6 5 Simplify. Example 10a b 5 c 4 Simplify 10a b 5 c 4. = 3 3 5 a b 5 c 4 = 3 5 a b 4 b c 4 = 3 5 a b b c = a b c 30b Simplify each expression. 1. 8. 68 3. 60 4. 75 5. 16 6. 3 6 7. 5 8. 5 10 9. 4a 10. 9x 4 13. 4 10 3 6 14. 3x 3 3x 4 17. 4a 4 b 18. 81x 4 y 11. 300a 4 15. 0a b 4 19. 150a b c 1. 18c 6 16. 100x 3 y 0. 7a 6 b 3 c 1. 45x y 5 z 8. 98x 4 y 6 z Chapter 10 137 Glencoe Algebra 1