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1.3 Solving Equations

1.3 Solving Equations. Algebra 2 Mrs. Spitz Fall 2006. 1.2 Even Answers. 48. 8c – 13d. Objective. Solve equations using the properties of equality. Application.

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1.3 Solving Equations

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  1. 1.3 Solving Equations Algebra 2 Mrs. Spitz Fall 2006

  2. 1.2 Even Answers 48. 8c – 13d

  3. Objective • Solve equations using the properties of equality

  4. Application • Janet Graves needed to buy some stamps for her graduation announcements. She bought some 25-cent stamps and three times as many 29-cent stamps. She paid a total of $22.40. How many of each type of stamp did she buy? • Hint: Let s = the number of 25-cent stamps and 3s = the number of 29-cent stamps.

  5. HMMMM. . . Let s = the number of 25-cent stamps and 3s = the number of 29-cent stamps. .25s + .29(3s) = 22.40 .25s + .87s = 22.40 1.12s = 22.40 s = 22.40 ÷ 1.12 s = 20 So, she bought 20 25-cent stamps. Now if I multiply 20 times 3, I’ll have the number of 29-cent stamps. 3(20) = 60 Janet bought 60 29-cent stamps and 20 25-cent stamps. Application

  6. Introduction • This problem can be solved by writing and then solving an open sentence. Sentences with variables to be replaced, such as 3x – 7 = 21 and 3x + 4 > 9, are called open sentences. When you solve an open sentence, you find replacements for the variables that will make the sentence true. Each of these replacements is called a solution of the open sentence.

  7. Introduction • An equation states that two mathematical expressions are equal. Solving an equation is like solving an open sentence. You find values that you can put in place of the variables so that the equation is true.

  8. Introduction • Real numbers have certain properties that we can use when we solve equations or open sentences. Some of those are listed below.

  9. Ex. 1: Name the property of equality illustrated in each statement. • If 36 · 2 = 72, then 72 = 36 · 2. • 21.4 = 21.4 • If 8 = 6 + 2 and 6 + 2 = 5 + 3, then 8 = 5 + 3 • symmetric property • reflexive property • transitive property

  10. Substitution Property • Some equations can be solved by making a substitution. The substitution property allows you to replace an expression with another equivalent expression.

  11. y = 8(0.3) + 1.2 y = 2.4 + 1.2 y = 3.6 Write equation down Substitute 2.4 for 8(0.3) Addition property Ex. 2: Solve y = 8(0.3) + 1.2

  12. x + 28.3 = 56.0 x + 28.3 + (- 28.3) = 56 + (-28.3) x = 27.7 Write equation down Additive inverse Solve Ex. 3: Solve x + 28.3 = 56.0 This equation could also be solved by subtracting 28.3 from each side. Some equations may be solved by multiplying or dividing each side by the same number.

  13. 8x = 48 ⅛ · 8x = ⅛ · 48 x = 6 Write equation down Multiplicative Inverse or reciprocal Solve Ex. 4: Solve 8x = 48 This equation could also be solved by dividing each side by 8. Check: 8x = 48 8(6) = 48 48 = 48

  14. -⅔k = 14 (-3/2)(-⅔)k = (-3/2)(14) k = - 21 Write equation down Multiplicative Inverse or reciprocal Solve a b = c c Ex. 5: Solve -⅔k = 14 Check: -⅔(-21) ? 14 42/3 ? 14 14 = 14

  15. NOTE: • In order to solve some equations, it may be necessary to apply more than one property. The following examples illustrate the use of several properties in solving equations.

  16. 0.75(8a + 20) – 2(a – 1) = 3 6a + 15 – 2a + 2 = 3 4a + 17 = 3 4a = -14 a = -3.5 Write down the problem correctly. Distributive and substitution properties Commutative, distributive and substitution properties Subtraction and substitution properties Division and substitution properties Ex. 6: Solve 0.75(8a + 20) – 2(a – 1) = 3

  17. 68x + 373 = 802 68x = 429 x = 6.309 Write down the problem correctly. Subtraction and substitution properties Division and substitution properties Ex. 7: Use a calculator to solve 68x + 373 = 802

  18. Assignment • pg. 21 #4-30 all

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