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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 2 Equations, Inequalities and Problem Solving.

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Presentation on theme: "Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 2 Equations, Inequalities and Problem Solving."— Presentation transcript:

1 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Chapter 2 Equations, Inequalities and Problem Solving

2 22 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

3 33 1. Solve P = a + b + c for b. 2. Solve V = ⅓Ah for A. 3. Solve I = PRT for R. 4. Solve A = ½(B + b)h for h. Bellwork: b = P – a – c A = 3V / h R = I / PT h = 2A / (B + b)

4 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. 2.6 Formulas and Problem Solving

5 55 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Objectives:  Use formulas to solve problems  Solve a formula or equation for one of its variables

6 66 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Formulas A formula is an equation that states a known relationship among multiple quantities (has more than one variable in it) **Literal Equations** A = lw (Area of a rectangle = length · width) I = PRT (Simple Interest = Principal · Rate · Time) P = a + b + c (Perimeter of a triangle = side a + side b + side c) d = rt (distance = rate · time) V = lwh (Volume of a rectangular solid = length · width · height)

7 77 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Example 1 Charles Pecot can afford enough fencing to enclose a rectangular garden with a perimeter of 140 feet. If the width of the garden is to be 30 feet, find the length. l = length of the garden w = width of the garden = 30 feet P= perimeter of the garden = 140 feet continued

8 88 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. continued Charles Pecot can afford enough fencing to enclose a rectangular garden with a perimeter of 140 feet. If the width of the garden is to be 30 feet, find the length. continued

9 99 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. continued Substitute 40 for l and 30 for w in the perimeter formula and check to see that the perimeter is 140 feet. The length of the rectangular garden is 40 feet.

10 10 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Example 2 A flower bed is in the shape of a triangle with one side twice the length of the shortest side, and the third side is 30 feet more than the length of the shortest side. Find the dimensions if the perimeter is 102 feet. The formula for the perimeter of a triangle is P = a + b + c. If we let x = the length of the shortest side, then 2x = the length of the second side, and x + 30 = the length of the third side continued

11 11 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Formula: P = a + b + c Substitute: 102 = x + 2x + x + 30 102 = x + 2x + x + 30 102 = 4x + 30 102 – 30 = 4x + 30 – 30 72 = 4x 18 = x continued x ft 2x ft x+30 ft

12 12 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Check: If the shortest side of the triangle is 18 feet, then the second side is 2(18) = 36 feet, and the third side is 18 + 30 = 48 feet. This gives a perimeter of P = 18 + 36 + 48 = 102 feet, the correct perimeter. continued State: The three sides of the triangle have a length of 18 feet, 36 feet, and 48 feet.

13 13 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Solving a Formula for a Variable It is often necessary to rewrite a formula so that it is solved for one of the variables. To solve a formula or an equation for a specified variable, we use the same steps as for solving a linear equation except that we treat the specified variable as the only variable in the equation.

14 14 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Step 1:Multiply on both sides to clear the equation of fractions if they appear. Step 2:Use the distributive property to remove parentheses if they appear. Step 3:Simplify each side of the equation by combining like terms. Step 4:Get all terms containing the specified variable on one side and all other terms on the other side by using the addition property of equality. Step 5:Get the specified variable alone by using the multiplication property of equality. Solving Equations for a Specified Varia ble

15 15 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Example 3 Solve for P. Factor out P on the right side. Divide both sides by 1 + RT. Simplify.

16 16 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Example 4 Solve T = mnr for n.

17 17 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Example 5 Solve A = P + PRT for T.

18 18 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Closure: What is a formula? What are the steps for rewriting a formula?


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