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Quiz Evaluate each expression for the given values of the variables. 1. 6x + 9 for x = 2 2. 4x + 3y for x = 4, y = 2 3. If n is the amount of money in a savings account, then the expression n + 0.05n can be used to find the amount in the account after it has earned interest for one year. Find the total in the account after one year if $100 is the initial amount. Write an algebraic expression for each word phrase. 4. 8 less than a number k 5. 7 times the sum of n and 2 Write an algebraic expression and used it to solve the problem. 6. Tony buys n raffle tickets for $0.75 each. If he buys 12 of them, how much will they cost?
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Pre-Algebra Solving Addition and Subtraction Equations 1-3
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Write an algebraic expression for each word phrase. 1. a number x decreased by 9 2. 5 times the sum of p and 6 3. 2 plus the product of 8 and n 4. the quotient of 4 and a number c x 9 5(p + 6) 2 + 8n 4 c Warm Up
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Learn to solve equations using addition and subtraction.
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equation solve solution inverse operation isolate the variable Addition Property of Equality Subtraction Property of Equality Vocabulary
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An equation uses an equal sign to show that two expressions are equal. All of these are equations. 3 + 8 = 11r + 6 = 1424 = x – 7 100 2 = 50 To solve an equation, find the value of the variable that makes the equation true. This value of the variable is called the solution of the equation.
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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. 13= 15 ? So 5 is not solution. x + 8 = 15 ? 5 + 8 = 15 ? Determining Whether a Number is a Solution of an Equation
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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. 15= 15 ? So 7 is a solution. x + 8 = 15 ? 7 + 8 = 15 ? Example Continued
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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. 31= 15 ? So 23 is not a solution. x + 8 = 15 ? 23 + 8 = 15 ? Example Continued
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Determine which value of x is a solution of the equation. x – 4 = 13; x = 9, 17, or 27 Substitute each value for x in the equation. Substitute 9 for x. 5 = 13 ? So 9 is not a solution. x – 4 = 13 ? 9 – 4 = 13 ? Try This
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Determine which value of x is a solution of the equation. x – 4 = 13; x = 9, 17, or 27 Substitute each value for x in the equation. Substitute 17 for x. 13 = 13 ? So 17 is a solution. x – 4 = 13 ? 17 – 4 = 13 ? Try This Continued
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Determine which value of x is a solution of the equation. x – 4 = 13; x = 9, 17, or 27 Substitute each value for x in the equation. Substitute 27 for x. 23 = 13 ? So 27 is not a solution. x – 4 = 13 ? 27 – 4 = 13 ? Try This Continued
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Addition and subtraction are inverse operations, 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.
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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 + 4 2 + 7 = 9 x = y + z ADDITION PROPERTY OF EQUALITY WordsNumbersAlgebra
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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 3 4 + 4 = 8 x = y z z z z SUBTRACTION PROPERTY OF EQUALITY WordsNumbersAlgebra
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Solve. Subtract 10 from both sides. A. 10 + n = 18 10 + n = 18 –10 0 + n = 8 n = 8 Identity Property of Zero: 0 + n = n. Check 10 + n = 18 ? 10 + 8 = 18 18 = 18 ? Solving Equations Using Addition and Subtraction Properties
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Solve. Add 8 to both sides. B. p – 8 = 9 p – 8 = 9 + 8 p + 0 = 17 p = 17 Identity Property of Zero: p + 0 = p. Check p – 8 = 9 ? 17 – 8 = 9 9 = 9 ? Solving Equations Using Addition and Subtraction Properties
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Solve. Add 11 to both sides. C. 22 = y – 11 22 = y – 11 + 11 33 = y + 0 33 = y Identity Property of Zero: y + 0 = 0. Check 22 = y – 11 ? 22 = 33 – 11 22 = 22 ? Solving Equations Using Addition and Subtraction Properties
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+ = + = 34 16,550 x + 0 = 16,516 A. Jan took a 34-mile trip in her car, and the odometer showed 16,550 miles at the end of the trip. What was the original odometer reading? Subtract 34 from both sides. x + 34 = 16,550 The original odometer reading was 16,516 miles. odometer reading at the beginning of the trip miles traveled x –34 x = 16,516 Solve: odometer reading at the end of the trip
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+ = + = n 1125 0 + n = 230 B. From 1980 to 2000, the population of a town increased from 895 residents to 1125 residents. What was the increase in population during that 20-year period? Subtract 895 from both sides. 895 + n = 1125 The increase in population was 230. initial population increase in population 895 –895 n = 230 Solve: population after increase
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+ = + = 27 535 x + 0 = 508 A. Isabelle earned $27 interest and now has a balance of $535 in the bank. What was her balance before interest was added? Subtract 27 from both sides. x + 27 = 535 Isabelle had a balance of $508 before interest was added. balance before interest interest earned x –27 x = 508 Solve: balance after interest
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+ = + = n 502 0 + n = 30 B. From June to July, the water level in a lake has increased from 472 feet to 502 feet. What was the increase in water level during that 1-month period? Subtract 472 from both sides. 472 + n = 502 The increase in water level was 30 feet. initial water level increase in water level 472 –472 n = 30 Solve: water level after increase
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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 Lesson Quiz
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