1. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Over 20 years, the population of a town decreased by 275 people to a population of 850. Write and solve an equation to find the original population. Example 4: Application Write an equation to represent the relationship. + 275 p =1125 p – d = c original population minus current population decrease in population is p – 275 = 850 Since 275 is subtracted from p, add 275 to both sides to undo the subtraction. p – d = c The original population was 1125 people.

2. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting A person's maximum heart rate is the highest rate, in beats per minute, that the person's heart should reach. One method to estimate maximum heart rate states that your age added to your maximum heart rate is 220. Using this method, write and solve an equation to find a person's age if the person's maximum heart rate is 185 beats per minute. Check It Out! Example 4

3. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Check It Out! Example 4 Continued a + r = 220 age added to 220 maximum heart rate is Write an equation to represent the relationship. – 185 a = 35 a + 185 = 220 Substitute 185 for r. Since 185 is added to a, subtract 185 from both sides to undo the addition. a + r = 220 A person whose maximum heart rate is 185 beats per minute would be 35 years old.

4. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 4: Application Write an equation to represent the relationship. Ciro puts of the money he earns from mowing lawns into a college education fund. This year Ciro added $285 to his college education fund. Write and solve an equation to find how much money Ciro earned mowing lawns this year. 1 4 one-fourth times earnings equals college fund m = $1140 Substitute 285 for c. Since m is divided by 4, multiply both sides by 4 to undo the division. Ciro earned $1140 mowing lawns.

5. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Check it Out! Example 4 Write an equation to represent the relationship. The distance in miles from the airport that a plane should begin descending, divided by 3, equals the plane's height above the ground in thousands of feet. A plane began descending 45 miles from the airport. Use the equation to find how high the plane was flying when the descent began. Distance divided by 3 equals height in thousands of feet 15 = h Substitute 45 for d. The plane was flying at 15,000 ft when the descent began.

6. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Lesson Quiz: Part 2 7. A person's weight on Venus is about his or her weight on Earth. Write and solve an equation to find how much a person weighs on Earth if he or she weighs 108 pounds on Venus. 9 10

7. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Jan joined the dining club at the local café for a fee of $29.95. Being a member entitles her to save $2.50 every time she buys lunch. So far, Jan calculates that she has saved a total of $12.55 by joining the club. Write and solve an equation to find how many time Jan has eaten lunch at the café. Example 4: Application

8. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 4: Application Continued 1 Understand the Problem The answer will be the number of times Jan has eaten lunch at the café. List the important information: Jan paid a $29.95 dining club fee. Jan saves $2.50 on every lunch meal. After one year, Jan has saved $12.55.

9. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 4: Application Continued Let m represent the number of meals that Jan has paid for at the café. That means that Jan has saved $2.50m. However, Jan must also add the amount she spent to join the dining club. 2 Make a Plan total amount saved dining club fee amount saved on each meal =– 12.55 = 2.50m – 29.95

10. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 4: Application Continued 12.55 = 2.50m – 29.95 Solve 3 + 29.95 42.50 = 2.50m 2.50 17 = m Since 29.95 is subtracted from 2.50m, add 29.95 to both sides to undo the subtraction. Since m is multiplied by 2.50, divide both sides by 2.50 to undo the multiplication.

11. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 4: Application Continued Look Back4 Check that the answer is reasonable. Jan saves $2.50 every time she buys lunch, so if she has lunch 17 times at the café, the amount saved is 17(2.50) = 42.50. Subtract the cost of the dining club fee, which is about $30. So the total saved is about $12.50, which is close to the amount given in the problem, $12.55.

12. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Sara paid $15.95 to become a member at a gym. She then paid a monthly membership fee. Her total cost for 12 months was $735.95. How much was the monthly fee? Check It Out! Example 4

13. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Check It Out! Example 4 Continued 1 Understand the Problem The answer will the monthly membership fee. List the important information: Sara paid $15.95 to become a gym member. Sara pays a monthly membership fee. Her total cost for 12 months was $735.95.

14. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Check It Out! Example 4 Continued Let m represent the monthly membership fee that Sara must pay. That means that Sara must pay 12m. However, Sara must also add the amount she spent to become a gym member. 2 Make a Plan total cost initial membership monthly fee = + 735.95 = 12m + 15.95

15. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Check It Out! Example 4 Continued 735.95 = 12m + 15.95 Solve 3 – 15.95 – 15.95 720 = 12m 12 60 = m Since 15.95 is added to 12m, subtract 15.95 from both sides to undo the addition. Since m is multiplied by 12, divide both sides by 12 to undo the multiplication.

16. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Check It Out! Example 4 Continued Look Back4 Check that the answer is reasonable. Sara pays $60 a month, so after 12 months Sara has paid 12(60) = 720. Add the cost of the initial membership fee, which is about $16. So the total paid is about $736, which is close to the amount given in the problem, $735.95.

17. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be? Example 11: Application PersonBulbs Jon60 bulbs plus 44 bulbs per hour Sara96 bulbs plus 32 bulbs per hour

18. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 11: Application Continued Let b represent bulbs, and write expressions for the number of bulbs planted. 60 bulbs plus 44 bulbs each hour the same as 96 bulbs plus 32 bulbs each hour When is ? 60 + 44b = 96 + 32b – 32b To collect the variable terms on one side, subtract 32b from both sides. 60 + 12b = 96

19. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 11: Application Continued Since 60 is added to 12b, subtract 60 from both sides. 60 + 12b = 96 –60 – 60 12b = 36 Since b is multiplied by 12, divide both sides by 12 to undo the multiplication. b = 3

20. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 11: Application Continued After 3 hours, Jon and Sara will have planted the same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either expression for b = 3: 60 + 44b = 60 + 44(3) = 60 + 132 = 192 96 + 32b = 96 + 32(3) = 96 + 96 = 192 After 3 hours, Jon and Sara will each have planted 192 bulbs.

21. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Example 12 Let g represent Greg's age, and write expressions for his age. four times Greg's age decreased by 3 is equal to three times Greg's age increased by 7. 4g – 3 = 3g + 7

22. Holt Algebra 1 2-1 Solving Equations by Adding or Subtracting Example 12 Continued 4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides. g – 3 = 7 –3g Since 3 is subtracted from g, add 3 to both sides. + 3 + 3 g = 10 Greg is 10 years old.