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Solving Equations by 2-1 Adding or Subtracting Warm Up
Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz
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Warm Up Evaluate. 1. + 4 2. 0.51 + (0.29)
1. 2. (0.29) Give the opposite of each number. Evaluate each expression for a = 3 and b = 2. 5. a b 2 3 1 3 3 2 0.8 2 3 2 3 –8 8 14
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An equation is a mathematical statement that two expressions are equal.
A solution of an equation is a value of the variable that makes the equation true. To find solutions, isolate the variable. A variable is isolated when it appears by itself on one side of an equation, and not at all on the other side.
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Inverse Operations Operation Inverse Operation
Isolate a variable by using inverse operations which "undo" operations on the variable. An equation is like a balanced scale. To keep the balance, perform the same operation on both sides. Inverse Operations Operation Inverse Operation Addition Subtraction Subtraction Addition
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Example 1A: Solving Equations by Using Addition
Solve the equation. Check your answer. y – 8 = 24 Since 8 is subtracted from y, add 8 to both sides to undo the subtraction. y = 32 Check y – 8 = 24 To check your solution, substitute 32 for y in the original equation. 32 –
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Check It Out! Example 1b Solve the equation. Check your answer. –6 = k – 6 Since 6 is subtracted from k, add 6 to both sides to undo the subtraction. 0 = k Check –6 = k – 6 To check your solution, substitute 0 for k in the original equation. – – 6 –6 –6
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Example 2A: Solving Equations by Using Subtraction
Solve the equation. Check your answer. m + 17 = 33 – 17 –17 Since 17 is added to m, subtract 17 from both sides to undo the addition. m = 16 Check m + 17 = 33 To check your solution, substitute 16 for m in the original equation.
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Solve the equation. Check your answer.
Check It Out! Example 2a Solve the equation. Check your answer. 1 2 d = 1 Since is added to d, subtract from both sides to undo the addition. 1 2 – 1 2 d = 1 2 Check d = 1 1 2 To check your solution, substitute for d in the original equation. 1 2 1 2 1 1
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Check It Out! Example 2b Solve the equation. Check your answer. –5 = k + 5 – – 5 Since 5 is added to k, subtract 5 from both sides to undo the subtraction. –10 = k Check –5 = k + 5 To check your solution, substitute –10 for k in the original equation. –5 –10 + 5 –5 –5
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Inverse Operations Operation Inverse Operation
Solving an equation that contains multiplication or division is similar to solving an equation that contains addition or subtraction. Use inverse operations to undo the operations on the variable. Inverse Operations Operation Inverse Operation Multiplication Division Division Multiplication
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Example 1A: Solving Equations by Using Multiplication
Solve the equation. –8 = j 3 Since j is divided by 3, multiply both sides by 3 to undo the division. –24 = j Check –8 = j 3 –8 –24 3 To check your solution, substitute –24 for j in the original equation. –8 –8
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Check It Out! Example 1a Solve the equation. Check your answer. = 10 p 5 Since p is divided by 5, multiply both sides by 5 to undo the division. p = 50 Check = 10 p 5 10 50 5 To check your solution, substitute 50 for p in the original equation.
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Check It Out! Example 1b Solve the equation. Check your answer. –13 = y 3 Since y is divided by 3, multiply both sides by 3 to undo the division. –39 = y y Check –13 = 3 –13 –39 3 To check your solution, substitute –39 for y in the original equation. –13 –13
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Check It Out! Example 2a Solve the equation. Check your answer. 16 = 4c Since c is multiplied by 4, divide both sides by 4 to undo the multiplication. 4 = c Check = 4c 16 4(4) To check your solution, substitute 4 for c in the original equation.
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Check It Out! Example 2b Solve the equation. Check your answer. 0.5y = –10 Since y is multiplied by 0.5, divide both sides by 0.5 to undo the multiplication. y = –20 Check 0.5y = –10 0.5(–20) –10 To check your solution, substitute –20 for y in the original equation. –10 –10
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Check It Out! Example 2c Solve the equation. Check your answer. 15k = 75 Since k is multiplied by 15, divide both sides by 15 to undo the multiplication. k = 5 Check 15k = 75 To check your solution, substitute 5 for k in the original equation. 15(5) 75
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