EXAMPLE. Solving a Literal Equation and Evaluating Its Solution REFLECT

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1 Name Class Date Solving for a Variable Going Deeper Essential question: How do you solve literal equations and rewrite formulas? A literal equation is an equation in which the coefficients and constants have been replaced by letters. In the following Eplore, you will see how a literal equation can be used to represent specific equations having the same form A-REI.2.3 EXPLORE Understanding Literal Equations A For each equation given below, solve the equation by writing two equivalent equations: one where the -term is isolated and then one where is isolated = = = -1 B Identify the two properties of equality that you used in part A. List them in the order that you used them. C Each equation in part A has the general form a + b = c where a 0. Solve this literal equation for using the properties of equality that you identified in part B. D a + b = c Write the literal equation. a = - Subtract b from both sides. = - Divide both sides by a. Show that the solution of the literal equation gives the same solution of = 7 as you found in part A. Recognize that when a = 3, b = 1, and c = 7, the literal equation a + b = c gives the specific equation = 7. = - = Write the literal equation s solution. - Substitute 3 for a, 1 for b, and 7 for c. = Simplify. REFLECT 1a. Why must the restriction a 0 be placed on the literal equation a + b = c? Chapter 1 33 Lesson 6

2 1b. Choose one of the other specific equations from part A. Show that the solution of the literal equation gives the solution of the specific equation. When you solve a literal equation, you use properties of equality and other properties to isolate the variable. The result is not a number, but rather an epression involving the letters that represent the coefficients and constants. 2 A-REI.2.3 EXAMPLE Solving a Literal Equation and Evaluating Its Solution Solve the literal equation a( + b) = c where a 0. Then use the literal equation s solution to obtain the solution of the specific equation 2( + 7) = -6. A Solve a( + b) = c for. Use the properties of equality to justify your solution steps. a( + b) = c + b = Original equation Property of Equality = - Property of Equality B Obtain the solution of 2( + 7) = -6 from the literal equation s solution by letting a = 2, b = 7, and c = -6. = - Write the literal equation s solution. = - Substitute 2 for a, 7 for b, and -6 for c. = REFLECT Simplify. 2a. When solving a( + b) = c, why do you divide by a before you subtract b? 2b. Write an equation that has the form a( + b) = c. Find the solution of your equation using the literal equation s solution. Chapter 1 34 Lesson 6

3 2c. Another way to solve a( + b) = c is to start by using the Distributive Property. Show and justify the solution steps using this method. 2d. When you start solving a( + b) = c by dividing by a, you get = c_ - b. When you a start solving a( + b) = c by distributing a, you get = c - a ab. Use the fact that you can rewrite c - a ab as the difference of two fractions to show that the two solutions are equivalent. 3 A-CED.1.4 EXAMPLE Solving a Formula for a Variable Solve the formula for the given variable. Justify each step in your solution. A The formula V = lwh gives the volume of a rectangular prism with length l, width w, and height h. Solve the formula for h to find the height of a rectangular prism with a given volume, length, and width. V = lwh Original equation V = lwh h = Simplify. B The formula E = 1_ 2 k 2 gives the potential energy E of a spring with spring constant k that has been stretched by length. Solve the formula for k to find the constant of a spring with a given potential energy and stretch. E = 1 2 k 2 Original equation E = 1 2 k 2 2E = k 2 2E = k 2 k = Simplify. Chapter 1 35 Lesson 6

4 REFLECT 3a. The formula T = p + sp gives the total cost of an item with price p and sales ta s, epressed as a decimal. Describe a situation in which you would want to solve the formula for s. 3b. What is true about the restrictions on the value of a variable in a formula that might not be true of other literal equations? 4 A-CED.1.4 EXAMPLE Writing and Rearranging a Formula The flower garden at the right is made up of a square and an isosceles triangle. Write a formula for the perimeter P in terms of, and then solve for to find a formula for the side length of the square in terms of P. A Write a formula for the perimeter of each shape. Use only the sides of the square and the triangle that form the outer edges of the figure. 7 8 Perimeter of square = Perimeter of triangle = B Combine the formulas. P = C Solve the formula for. P = P = ( + ) ( ) P = ( ) ( ) ( ) P = Write the combined formula. Distributive Property Find the sum. Write the result as an improper fraction. Multiplication Property of Equality = Simplify. Chapter 1 36 Lesson 6

5 REFLECT 4a. What are the restrictions on the values of P and? Eplain. 4b. How could you write a formula for the area of the square in terms of P? PRACTICE 1. Show and justify the steps for solving + a = b. Then use the literal equation s solution to obtain the solution of + 2 = Show and justify the steps for solving a = b where a 0. Then use the literal equation s solution to obtain the solution of 3 = Show and justify the steps for solving a = b + c where a b. Then use the literal equation s solution to obtain the solution of 2 = + 7. Chapter 1 37 Lesson 6

6 Solve each formula for the indicated variable. 4. Formula for the surface area of a rectangular prism: SA = 2(lw + hw + hl), for w 5. Formula for the area of a trapezoid: A = 1 (a + b)h, for b 2 6. An electrician sent Bonnie an invoice in the amount of a dollars for 6 hours of work that was done on Saturday. The electrician charges a weekend fee f in addition to an hourly rate r. Bonnie knows what the weekend fee is. Write a formula Bonnie can use to find r, the rate the electrician charges per hour. 7. The swimming pool below is made up of a square and two semicircles. Write a formula for the perimeter P in terms of, and then solve for to find a formula for the side length of the square in terms of P. 1 2 Chapter 1 38 Lesson 6

7 Name Class Date 1-6 Additional Practice Answer each of the following. 1. The formula C = 2 r relates the radius r of a circle to its circumference C. Solve the formula for r. 2. The formula y = m + b is called the slope-intercept form of a line. Solve this formula for m. Solve for the indicated variable. 3. 4c = d for c 4. n 6m = 8 for n 5. 2p + 5r = q for p = y + z for 7. a b = c for b 8. h 4 j = k for j Answer each of the following. 9. The formula c = 5p relates c, the total cost in dollars of hosting a birthday party at a skating rink, to p, the number of people attending. a. Solve the formula c = 5p for p. b. If Allie s parents are willing to spend $300 for a party, how many people can attend? 10. The formula for the area of a triangle is A = 1 2 bh, where b represents the length of the base and h represents the height. a. Solve the formula A = 1 bh for b. 2 b. If a triangle has an area of 192 mm 2, and the height measures 12 mm, what is the measure of the base? Chapter 1 39 Lesson 6

8 Problem Solving Use the table below, which shows some track and field gold medal winners, to answer questions 1 4. Round all answers to the nearest tenth. 1. Solve the formula d = rt for r Summer Olympics 2. Find Johnson s average speed in meters per second. 3. Find Garcia s average speed in meters per second. Gold Medal Winner M. Greene, USA K. Kenteris, Greece M. Johnson, USA Race Time (s) 100 m m m A. Garcia, 110 m hurdles The world record of seconds in Cuba the 200-meter race was set by Michael Johnson in Find the difference between Johnson s average speed and Kenteris average speed. Select the best answer. 5. The cost to mail a letter in the United States in 2008 was $0.41 for the first ounce and $0.26 for each additional ounce. Solve C = (z 1) for z. A z = B z = C C C C z = 0.26 D z = C Degrees Celsius and degrees Fahrenheit are related by the equation C = 5 (F 32). Solve for F. 9 A F = 9C + 27 C F = 5 9 C + 32 B F = 9 5 C D F = 9 5 C The formula V = Bh shows how to find 3 the volume of a pyramid. Solve for B. F B = 3V H B = 3Vh h G B = 3V h J B = 3V + h 8. The cost of operating an electrical device is given by the formula C = Wtc 1000 where W is the power in watts, t is the time in hours, and c is the cost in cents per kilowatt-hour. Solve for W. F W = 1000C tc G W = Ctc 1000 H W = 1000C + tc J W = 1000C tc Chapter 1 40 Lesson 6