Solving Equations by Lauren McCluskey

DO NOW Solve each equation. 1.3n – 7 + 2n = 8n x = 7x x x = 2(3x + 3) x + 3x = 2(2x + 5) Solve each equation. 1.3n – 7 + 2n = 8n x = 7x x x = 2(3x + 3) x + 3x = 2(2x + 5)

Equations with Variables on Both Sides: Use the Addition or Subtraction property of Equality to get the variables on one side of the equation.

Example: 4p - 10 = p + 3p -2p Combine like terms: p + 3p - 2p = 2p Use the subtraction property of equality: 4p - 10 = 2p -2p 2p -10 = 0 4p - 10 = p + 3p -2p Combine like terms: p + 3p - 2p = 2p Use the subtraction property of equality: 4p - 10 = 2p -2p 2p -10 = 0

Example: cont. 2p - 10 = 0 Undo subtraction: p = 10 Undo multiplication: 2p / 2 = 10/ 2 p = 5 2p - 10 = 0 Undo subtraction: p = 10 Undo multiplication: 2p / 2 = 10/ 2 p = 5

Try It! 1)6b + 14 = -7 - b 2) w = -8w + w 3) z = 10z - 4 1)6b + 14 = -7 - b 2) w = -8w + w 3) z = 10z - 4

Check your answers: 1) b = -3 2) w = 4 3) z = 2 1) b = -3 2) w = 4 3) z = 2

Identity or No Solution: “ An equation has no solution if no value of the variable makes the equation true.” “An equation that is true for every value of the variable is an identity.” “ An equation has no solution if no value of the variable makes the equation true.” “An equation that is true for every value of the variable is an identity.”

Examples  4x + 5 = x + 3(x + 2) – 1  2p – 6 + p = 7 + 3p  4x + 5 = x + 3(x + 2) – 1  2p – 6 + p = 7 + 3p

2.5: Defining One Variable in Terms of Another:  “Some problems involve two or more unknown quantities. To solve such problems, first decide which unknown quantity the variable will represent. Then express the other unknown quantity in terms of that variable.”

Example: “The width of a rectangle is 2 cm less than its length. The perimeter of the rectangle is 16cm. What is the length of the rectangle?” Let l = length Let l - 2 = width P= 2l + 2w So 2(l) + 2(l -2) = 16cm “The width of a rectangle is 2 cm less than its length. The perimeter of the rectangle is 16cm. What is the length of the rectangle?” Let l = length Let l - 2 = width P= 2l + 2w So 2(l) + 2(l -2) = 16cm

 The length of a rectangle is 6 in. more than its width.The perimeter of  the rectangle is 24 in.What is the length of the rectangle?  The length of a rectangle is 6 in. more than its width.The perimeter of  the rectangle is 24 in.What is the length of the rectangle?

Example: cont. 2l + 2(l-2) = 16 cm 2l + 2l - 4 = 16cm 4l - 4 = 16 cm l = 20 cm 4l / 4 = 20/ 4 so l = 5 2l + 2(l-2) = 16 cm 2l + 2l - 4 = 16cm 4l - 4 = 16 cm l = 20 cm 4l / 4 = 20/ 4 so l = 5

Try it! The length of a rectangle is 6 more than 3 times as long as the width. The perimeter is 36 m. What are the measurements?

Check your answer: l = 15m w = 3m OR: 15m x 3m l = 15m w = 3m OR: 15m x 3m

Consecutive Integers: “Consecutive integers differ by 1.” Consecutive even or consecutive odd integers differ by 2. Example: The sum or two consecutive odd integers is 84. What are the integers? “Consecutive integers differ by 1.” Consecutive even or consecutive odd integers differ by 2. Example: The sum or two consecutive odd integers is 84. What are the integers?

Consecutive Integers Let x = the 1st integer Let x + 2 = the 2nd integer x + x + 2 = 84 2x + 2 = x = 82 x = 41; x + 2 = 43 So the integers are 41 and 43. Let x = the 1st integer Let x + 2 = the 2nd integer x + x + 2 = 84 2x + 2 = x = 82 x = 41; x + 2 = 43 So the integers are 41 and 43.

Try It! The sum of three consecutive integers is 48. What are the integers?

Check your answer: The integers are: 15, 16, and 17. The integers are: 15, 16, and 17.

Distance= Rate x Time On May 18, 1990 the fastest speed of any national railroad was achieved by the French high speed train Train ´a Grande Vitess as it traveled over a distance from Cortalain to Tours, France. A commentator said that this speed was so fast that if it continued at that rate, the train would travel 6404 miles in 20 hours. How fast did the train travel on that date?

Rate* Times = Distance  When the distances covered are equal, we can set the two expressions equal to each other and solve for x.  When the distances combine to make up the total distance, we can add the expressions, set it equal to the total distance, and solve for x.  When the distances covered are equal, we can set the two expressions equal to each other and solve for x.  When the distances combine to make up the total distance, we can add the expressions, set it equal to the total distance, and solve for x.

Try It! Adapted from Prentice Hall: 1) A group of campers left the campsite in a canoe going 10km/h. Two hours later, another group left in a motor boat going 22km/h. How long did it take the second group to catch up? Adapted from Prentice Hall: 1) A group of campers left the campsite in a canoe going 10km/h. Two hours later, another group left in a motor boat going 22km/h. How long did it take the second group to catch up?

Try It! 2) On his way to work, your uncle averaged 20 mph. On his way home, he averaged 40mph. If the total time was 1 1/2hours, how long did it take him to drive to work?

Try It! 3) Sarah and John left Perryville going in opposite directions. Sarah drives 12mph faster than John. After 2 hours, they are 176 miles apart. Find Sarah’s and John’s speeds.

Check your answers:  

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