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Solving Equations by Adding or Subtracting. 1-7. Warm Up. Problem of the Day. Lesson Presentation. Course 3. Solving Equations by Adding or Subtracting. 1-7. Course 3. Learn to solve equations using addition and subtraction. Learn to solve equations using multiplication and division.
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Solving Equations by Adding or Subtracting 1-7 Warm Up Problem of the Day Lesson Presentation Course 3
Solving Equations by Adding or Subtracting 1-7 Course 3 Learn to solve equations using addition and subtraction. Learn to solve equations using multiplication and division.
Solving Equations by Adding or Subtracting 1-7 Course 3 Vocabulary equation inverse operation
Solving Equations by Adding or Subtracting 1-7 100 = 50 2 Course 3 An equation is a mathematical sentence that uses an equal sign to show that two expressions have the same value. All of these are equations. 3 + 8 = 11 r + 6 = 14 24 = x – 7 To solve an equation that contains a variable, find the value of the variable that makes the equation true. This value of the variable is called the solution of the equation. Course 3
Solving Equations by Adding or Subtracting 1-7 ? ? x + 8 = 15 5 + 8 = 15 ? 13= 15 Course 3 Additional Example 1: Determining Whether a Number is a Solution of an Equation Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. Substitute 5 for x. So 5 is not solution.
Solving Equations by Adding or Subtracting 1-7 ? ? x + 8 = 15 7 + 8 = 15 ? 15= 15 Course 3 Additional Example 1 Continued Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. Substitute 7 for x. So 7 is a solution.
Solving Equations by Adding or Subtracting 1-7 ? ? x + 8 = 15 23 + 8 = 15 ? 31= 15 Course 3 Additional Example 1 Continued Determine which value of x is a solution of the equation. x + 8 = 15; x = 5, 7, or 23 Substitute each value for x in the equation. Substitute 23 for x. So 23 is not a solution.
Solving Equations by Adding or Subtracting 1-7 Course 3 Addition and subtraction are inverseoperations, which means they “undo” each other. To solve an equation, use inverse operations to isolate the variable. This means getting the variable alone on one side of the equal sign. Course 3
Solving Equations by Adding or Subtracting 1-7 Words Numbers Algebra ADDITION PROPERTY OF EQUALITY + 4 + 4 + z + z Course 3 To solve a subtraction equation, like y 15 = 7, you would use the Addition Property of Equality. You can add the same number to both sides of an equation, and the statement will still be true. 2 + 3 = 5 x = y x = y 2 + 7 = 9 Course 3
Solving Equations by Adding or Subtracting 1-7 SUBTRACTION PROPERTY OF EQUALITY Words Numbers Algebra 3 3 z z Course 3 There is a similar property for solving addition equations, like x+ 9 = 11. It is called the Subtraction Property of Equality. You can subtract the same number from both sides of an equation, and the statement will still be true. 4 + 7 = 11 x = y x = y 4 + 4 = 8 Course 3
Solving Equations by Adding or Subtracting 1-7 ? 10 + 8 = 18 ? 18 = 18 Course 3 Additional Example 2A: Solving Equations Using Addition and Subtraction Properties Solve. 10 + n = 18 10 + n = 18 –10 –10 Subtract 10 from both sides. 0 + n = 8 Identity Property of Zero: 0 + n = n. n = 8 Check 10 + n = 18
Solving Equations by Adding or Subtracting 1-7 ? 17 – 8 = 9 ? 9 = 9 Course 3 Additional Example 2B: Solving Equations Using Addition and Subtraction Properties Solve. p – 8 = 9 p – 8 = 9 + 8 + 8 Add 8 to both sides. p + 0= 17 Identity Property of Zero: p + 0 = p. p = 17 Check p – 8 = 9
Solving Equations by Adding or Subtracting 1-7 ? 22 = 33 – 11 ? 22 = 22 Course 3 Additional Example 2C: Solving Equations Using Addition and Subtraction Properties Solve. 22 = y – 11 22 = y – 11 + 11 + 11 Add 11 to both sides. 33 = y + 0 Identity Property of Zero: y + 0 = 0. 33 = y Check 22 = y – 11
Solving Equations by Multiplying or Dividing 1-8 MULTIPLICATION PROPERTY OF EQUALITY Words Numbers Algebra 4 • 4 • z z Course 3 You can solve a division equation using the Multiplication Property of Equality. You can multiply both sides of an equation by the same number, and the statement will still be true. 2 • 3 = 6 x = y 2 • 3 = 6 x = y 8 • 3 = 24 Course 3
Solving Equations by Multiplying or Dividing 1-8 Words Numbers Algebra DIVISION PROPERTY OF EQUALITY 12 = 6 2 Course 3 You can solve a multiplication equation using the Division Property of Equality. You can divide both sides of an equation by the same nonzero number, and the equation will still be true. 4 • 3 = 12 x = y 4 • 3 = 12 x = y 2 2 z z Course 3
Solving Equations by Multiplying or Dividing 1-8 ? 6(8) = 48 ? 48 = 48 Course 3 Additional Example 1A: Solving Equations Using Division Solve 6x = 48. 6x = 48 6x = 48 Divide both sides by 6. 6 6 1 • x = x 1x = 6 x = 8 Check 6x = 48 Substitute 8 for x.
Solving Equations by Multiplying or Dividing 1-8 ? –9(–5) = 45 ? 45 = 45 Course 3 Additional Example 1B: Solving Equations Using Division Solve –9y = 45. –9y = 45 –9y = 45 Divide both sides by –9. –9 –9 1 • y = y 1y = –5 y = –5 Check –9y = 45 Substitute –5 for y.
Solving Equations by Multiplying or Dividing 1-8 b b b –4 –4 –4 = 5 = 5 –20 ? = 5 –4 ? 5 = 5 Course 3 Additional Example 2: Solving Equations Using Multiplication Solve = 5. –4 • –4 • Multiply both sides by –4. b = –20 Check Substitute –20 for b.
Solving Equations by Multiplying or Dividing 1-8 c c c –3 –4 –3 = 5 = 5 –15 ? = 5 –3 ? 5 = 5 Course 3 Check It Out: Example 2 Solve = 5. –3 • –3 • Multiply both sides by –3. c = –15 Check Substitute –15 for c.
Solving Equations by Adding or Subtracting 1-7 Course 3 Lesson Quiz Determine which value of x is a solution of the equation. 1.x + 9 = 17; x = 6, 8, or 26 2.x – 3 = 18; x = 15, 18, or 21 Solve. 3.a + 4 = 22 4.n – 6= 39 5. The price of your favorite cereal is now $4.25. In prior weeks the price was $3.69. Write and solve an equation to find n, the increase in the price of the cereal. 8 21 a = 18 n = 45 3.69 + n = 4.25;$0.56
Solving Equations by Multiplying or Dividing 1-8 x –4 Course 3 Lesson Quiz Solve. 1.3t = 9 2. –15 = 3b 3. = –7 4.z ÷ 4 = 22 5. A roller coaster descends a hill at a rate of 80 feet per second. The bottom of the hill is 400 feet from the top. How long will it take the coaster rides to reach the bottom? t = 3 b = –5 x = 28 z = 88 5 seconds