6:10 26 GUIDELINES FOR MODELING WITH EQUATIONS 1. Identify the Variable. Identify the quantity that the problem asks you to find. This quantity can usually be determined by a careful reading of the question that is posed at the end of the problem. Then introduce notation for the variable (call it \( x \) or some other letter). 2. Translate from Words to Algebra. Read each sentence in the problem again, and express all the quantities mentioned in the problem in terms of the variable you defined in Step 1. To organize this information, it is sometimes helpful to draw a diagram or make a table. 3. Set Up the Model. Find the crucial fact in the problem that gives a relationship between the expressions you listed in Step 2. Set up an equation (or model) that expresses this relationship. 4. Solve the Equation and Check Your Answer. Solve the equation, check your answer, and express it as a sentence that answers the question posed in the problem. 1) Express the following quantity in terms of \( s \). The average of three test scores, if the first two scores are 66 and 84 and the third score is \( s \). 2) A cell phone company charges a monthly fee of \( \$ 10 \) for the first 1000 text messages and 10 cents for every additional text message. Miriam's bill for text messages for the month of June is \( \$ 38.50 \). Write an equation that can be used to find the number of additional text messages sent. Problems About Interest \[ \mathrm{I}=\text { Prt } \] I = interest (\$) \( \mathrm{P}= \) Principal (\$) \( r= \) rate (as a decimal) \( t= \) time (in years) 3) Phyllis invested \( \$ 12,000 \), a portion earning a simple interest rate of \( 4 \frac{1}{2} \% \) per year and the rest earning a rate of \( 4 \% \) per year. After 1 year the total interest earned on these investments was \( \$ 525 \). How much money did she invest at each rate? 4) Suzie inherits \( \$ 100,000 \) and invests it in two certificates of deposit. One certificate pays \( 6 \% \) and the other pays \( 4 \frac{1}{2} \% \) simple interest annually. If Mary's total interest is \( \$ 5,775 \) per year, how much money is invested at each rate? Problems About Area or Length -fleet02-xythos.content.blackboardcdn.com - Private

Let's work through each of the provided problems step-by-step using the **GUIDELINES FOR MODELING WITH EQUATIONS**. --- ## **1) Express the Average of Three Test Scores in Terms of \( s \)** **Problem Statement:** The average of three test scores, where the first two scores are 66 and 84, and the third score is \( s \). ### **Step 1: Identify the Variable** - The quantity we need to find is the **average of the three test scores**. - Let’s denote the average by \( A \). ### **Step 2: Translate from Words to Algebra** - The first two scores are **66** and **84**. - The third score is represented by \( s \). - The average of the three scores is calculated by summing all three scores and dividing by **3**. ### **Step 3: Set Up the Model** \[ A = \frac{66 + 84 + s}{3} \] ### **Step 4: Solve the Equation and Check Your Answer** - Since the problem only asks to **express the average in terms of \( s \)**, we can leave the equation as is. - **Final Expression:** \[ A = \frac{150 + s}{3} \] (Here, \( 66 + 84 = 150 \), simplifying the expression.) --- ## **2) Determine the Number of Additional Text Messages Sent by Miriam** **Problem Statement:** A cell phone company charges a monthly fee of \$10 for the first 1000 text messages and 10 cents for every additional text message. Miriam's bill for text messages for the month of June is \$38.50. Write an equation that can be used to find the number of additional text messages sent. ### **Step 1: Identify the Variable** - Let’s denote the **number of additional text messages** Miriam sent beyond the initial 1000 as \( x \). ### **Step 2: Translate from Words to Algebra** - **Base Fee:** \$10 covers the first 1000 text messages. - **Additional Cost:** 10 cents (which is \$0.10) per additional text message. - **Total Bill:** \$38.50. ### **Step 3: Set Up the Model** \[ \text{Total Bill} = \text{Base Fee} + (\text{Cost per Additional Message} \times \text{Number of Additional Messages}) \] Plugging in the known values: \[ 38.50 = 10 + 0.10x \] ### **Step 4: Solve the Equation and Check Your Answer** - **Solving for \( x \):** \[ 38.50 - 10 = 0.10x \\ 28.50 = 0.10x \\ x = \frac{28.50}{0.10} \\ x = 285 \] - **Interpretation:** Miriam sent **285 additional text messages** beyond the first 1000 in June. --- ## **3) Phyllis's Investment Allocation** **Problem Statement:** Phyllis invested \$12,000, a portion earning a simple interest rate of \( 4 \frac{1}{2} \% \) per year and the rest earning a rate of \( 4\% \) per year. After 1 year, the total interest earned on these investments was \$525. How much money did she invest at each rate? ### **Step 1: Identify the Variable** - Let \( x \) be the amount invested at \( 4.5\% \). - Then, \( 12,000 - x \) is the amount invested at \( 4\% \). ### **Step 2: Translate from Words to Algebra** - **Interest from \( 4.5\% \) investment:** \( 0.045x \) - **Interest from \( 4\% \) investment:** \( 0.04(12,000 - x) \) - **Total Interest:** \$525. ### **Step 3: Set Up the Model** \[ 0.045x + 0.04(12,000 - x) = 525 \] ### **Step 4: Solve the Equation** - **Expand the equation:** \[ 0.045x + 480 - 0.04x = 525 \] - **Combine like terms:** \[ 0.005x + 480 = 525 \] - **Subtract 480 from both sides:** \[ 0.005x = 45 \] - **Divide both sides by 0.005:** \[ x = \frac{45}{0.005} \\ x = 9,000 \] - **Interpretation:** - **\$9,000** was invested at **4.5%**. - **\$3,000** was invested at **4%**. --- ## **4) Suzie's Certificates of Deposit Investment** **Problem Statement:** Suzie inherits \$100,000 and invests it in two certificates of deposit. One certificate pays \( 6\% \) and the other pays \( 4 \frac{1}{2} \% \) simple interest annually. If Suzie's total interest is \$5,775 per year, how much money is invested at each rate? ### **Step 1: Identify the Variable** - Let \( x \) be the amount invested at \( 6\% \). - Then, \( 100,000 - x \) is the amount invested at \( 4.5\% \). ### **Step 2: Translate from Words to Algebra** - **Interest from \( 6\% \) investment:** \( 0.06x \) - **Interest from \( 4.5\% \) investment:** \( 0.045(100,000 - x) \) - **Total Interest:** \$5,775. ### **Step 3: Set Up the Model** \[ 0.06x + 0.045(100,000 - x) = 5,775 \] ### **Step 4: Solve the Equation** - **Expand the equation:** \[ 0.06x + 4,500 - 0.045x = 5,775 \] - **Combine like terms:** \[ 0.015x + 4,500 = 5,775 \] - **Subtract 4,500 from both sides:** \[ 0.015x = 1,275 \] - **Divide both sides by 0.015:** \[ x = \frac{1,275}{0.015} \\ x = 85,000 \] - **Interpretation:** - **\$85,000** was invested at **6%**. - **\$15,000** was invested at **4.5%**. --- By following these systematic steps—**Identifying Variables**, **Translating to Algebra**, **Setting Up Models**, and **Solving Equations**—you can effectively tackle and solve various algebraic problems.