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Foundations of College Algebra Chapter 2 Linear Equations Ratios and Proportions

Foundations of College Algebra Chapter 2 Linear Equations Ratios and Proportions. Lesson 2.6 Objective Day 1 I will learn to words into a ratio, solve a proportion and an equation using. Definition ratio – a comparison of two quantities with the of measure using division.

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Foundations of College Algebra Chapter 2 Linear Equations Ratios and Proportions

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  1. Foundations of College Algebra Chapter 2 Linear EquationsRatios and Proportions Lesson 2.6 Objective Day 1 I will learn to words into a ratio, solve a proportion and an equation using .

  2. Definition ratio – a comparison of two quantities with the of measure using division. If the units are not the same to the units of the numerator Ways to write a ratio of number a to number b: 1. a to b 2. a :b 3. ab

  3. Translating words into ratiosStep 1 – place first number in . place second number in . Step 2 – check that denominator units are as numerator. If not CONVERT! 1. The ratio of 5 hr to 3 hr: 2. The ratio of 6 hrs to 3 days: 3. The ratio of 3 days to 2 weeks 4. The ratio of 12 hours to 4 days

  4. Definition:proportion – two that are equal. it is a special type of . if a proportion is then the cross products MUST be EQUAL. a=cproportion, b and d ≠ 0b d ad = bc cross products are equal Be Careful:The cross product property is true when there is more than one term on one side. x + 2 = 3 or x + 1 = 5 2 3 4 7

  5. Determine if the following proportions are true or false. a. 3 =15 b. 6 =30 4 20 7 32 c. 21= 62 d. 13= 91 15 45 17 119

  6. Cross product property can be used to solve for a missing value of a proportion. Step 1: the original proportion down. Step 2: the cross products. Step 3: the linear equation for the variable. Solve the proportion: 5 = x 9 63 Step 1: Write 5 = x 9 63 Step 2: Calculate 5 · 63 = 9xStep 3: Solve 315 = 9x315 = 9x ; x = 35 9 9

  7. Cross product property can be used to solve for a missing value of a proportion. Step 1: Write the original proportion down. Step 2: Calculate the cross products. Step 3: Solve the linear equation for the variable. Solve the proportion: x = 35 6 42

  8. Cross product property can be used to solve for a missing value of a proportion. Step 1: Write the original proportion down. Step 2: Calculate the cross products. Distribute if necessary Step 3: Solve the linear equation for the variable. Solve the proportion: x – 2 = x + 1 5 3

  9. Cross product property can be used to solve for a missing value of a proportion. Step 1: Write the original proportion down. Step 2: Calculate the cross products. Distribute if necessary Step 3: Solve the linear equation for the variable. Solve the proportion: x + 6 = 2 2 5

  10. HomeworkPage 146 Page 146 Day 1: 3 – 12; 22 – 26; 29, 31, 32, 34, 35, 38, 41, 42, 43 Update Vocabulary

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