Course Addition and Subtraction Equations 1-11 Addition and Subtraction Equations Course 2 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

Course Addition and Subtraction Equations Warm Up Determine if the given numbers are solutions to the given equations. 1. x = 2 for 4x = 9 2. x = 5 for 8x + 2 = x = 4 for 3(x – 2) = 10 no yes no

Course Addition and Subtraction Equations Problem of the Day Four couples have dinner together. The wives are Ginny, Helen, Sarah, and Bridget. The husbands are Mark, Alex, Stephen, and Henry. Who is married to whom? Sarah is Mark’s sister. Sarah introduced Henry to his wife. Bridget has 2 brothers, but her husband is an only child. Ginny is married to Stephen. Ginny and Stephen, Helen and Mark, Sarah and Alex, Bridget and Henry

Course Addition and Subtraction Equations Learn to solve one-step equations by using addition or subtraction.

Course Addition and Subtraction Equations Vocabulary Addition Property of Equality inverse operations Subtraction Property of Equality

Course Addition and Subtraction Equations To solve an equation means to find a solution to the equation. To do this, isolate the variable— that is, get the variable alone on one side of the equal sign. x = 8 – 5 x + 5 = 8 7 – 3 = y 7 = 3 + y The variables are isolated.The variables are not isolated. Recall that an equation is like a balanced scale. If you increase or decrease the weights by the same amount on both sides, the scale will remain balanced.

Course Addition and Subtraction Equations You can add the same amount to both sides of an equation, and the statement will still be true = = 9 ADDITION PROPERTY OF EQUALITY WordsNumbersAlgebra x = y + z = + z x + z = y+ z

Course Addition and Subtraction Equations Use inverse operations when isolating a variable. Addition and subtraction are inverse operations, which means that they “undo” each other.

Course Addition and Subtraction Equations Solve the equation b – 7 = 24. Check your answer. Additional Example 1: Solving an Equation by Addition b – 7 = b = 31 Think: 7 is subtracted from b, so add 7 to both sides to isolate b. Check b – 7 = – 7 = 24 ? 24 = 24 ? Substitute 31 for b. 31 is a solution.

Course Addition and Subtraction Equations Check It Out: Example 1 Solve the equation y – 3 = 21. Check your answer. y – 3 = y = 24 Think: 3 is subtracted from y, so add 3 to both sides to isolate y. Check y – 3 = – 3 = 21 ? 21 = 21 ? Substitute 24 for y. 24 is a solution.

Course Addition and Subtraction Equations You can subtract the same amount from both sides of an equation, and the statement will still be true = 11 –3 – = 8 SUBTRACTION PROPERTY OF EQUALITY WordsNumbersAlgebra x = y – z = – z x – z = y – z

Course Addition and Subtraction Equations Solve the equation t + 14 = 29. Check your answer. Additional Example 2: Solving an Equation by Subtraction t + 14 = 29 – 14 t = 15 Check t + 14 = = 29 ? 29 = 29 ? Think: 14 is added to t, so subtract 14 from both sides to isolate t. Substitute 15 for t. 15 is a solution.

Course Addition and Subtraction Equations Check It Out: Example 2 Solve the equation x + 11 = 36. Check your answer. x + 11 = 36 – 11 x = 25 Check Think: 11 is added to x, so subtract 11 from both sides to isolate x. x + 11 = = 36 ? 36 = 36 ? Substitute 25 for x. 25 is a solution.

Course Addition and Subtraction Equations The Giants scored 13 points in a game against Dallas. They scored 7 points for a touchdown and the rest of their points for field goals. How many points did they score on field goals? Additional Example 3: Sports Application Let f represent the field goal points. 7 points + field goal points = points scored 7 + f = 13 – 7 f = 6 They scored 6 points on field goals. Subtract 7 from both sides to isolate f.

Course Addition and Subtraction Equations Check It Out: Example 3 A basketball player scored 23 points during a game. Of those points, 3 were from 3-point goals and the remainder were 2 point goals. How many points did he score with 2 point goals? Let x equal the points scored by 2 point goals. 3 point goals + 2 point goals = points scored 3 + x = 23 – 3 x = 20 He scored 20 points from 2 point goals. Subtract 3 from both sides to isolate x.

Course Addition and Subtraction Equations Lesson Quiz Solve each equation. Check your answer. 1. x – 9 = 4 2. y + 6 = = n – = w = x – 102 y = 66; = 72 x = 13; 13 – 9 = 4 n = 62; 21 = 62 – 41 w = 96; 127 = x = 183; 81 = 183 – Tamika has sold 16 dozen cookies this week. This was 7 dozen more than she sold last week. Write and solve an equation to find how many dozen cookies she sold last week. x + 7 = 16; 9 dozen

Course Addition and Subtraction Equations 1-12 Multiplication and Division Equations Course 2 Warm Up Warm Up Problem of the Day Problem of the Day Lesson Presentation Lesson Presentation

Course Addition and Subtraction Equations Warm Up Solve. 1. x + 5 = 9 2. x – 34 = = x – 39 x = 4 x = 106 x = 163

Course Addition and Subtraction Equations Problem of the Day What 4-digit number am I? I am greater than 4,000 and less than 5,000. The sum of my hundreds digit and my ones digit is 9. Twice my tens number is 2 more than my thousands digit. The product of my hundreds digit and my ones digit is 0. I am not an even number. 4,039

Course Addition and Subtraction Equations Learn to solve one-step equations by using multiplication or division.

Course Addition and Subtraction Equations Vocabulary Multiplication Property of Equality Division Property of Equality

Course Addition and Subtraction Equations Like addition and subtraction, multiplication and division are inverse operations. They “undo” each other. ÷

Course Addition and Subtraction Equations If a variable is divided by a number, you can often use multiplication to isolate the variable. Multiply both sides of the equation by the number.

Course Addition and Subtraction Equations Solve the equation = 13. Check your answer. Additional Example 1: Solving an Equation by Multiplication h2h2 h2h2 = 13 h2h2 = 13(2) (2) h = 26 Check h2h2 = = 13 ? Think: h is divided by 2, so multiply both sides by 2 to isolate h. Substitute 26 for h. 26 is a solution ? = 13

Course Addition and Subtraction Equations Check It Out: Example 1 Solve the equation = 30. Check your answer. x5x5 x5x5 = 30 x5x5 = 30(5) (5) x = 150 Check x5x5 = 30 Think: x is divided by 5, so multiply both sides by 5 to isolate x. 30 = 30 ? Substitute 150 for x. 150 is a solution ? = 30

Course Addition and Subtraction Equations You cannot divide by 0. Remember!

Course Addition and Subtraction Equations If a variable is multiplied by a number, you can often use division to isolate the variable. Divide both sides of the equation by the number.

Course Addition and Subtraction Equations Solve the equation 51 = 17x. Check your answer. Additional Example 2: Solving an Equation by Division 51 = 17x 17 3 = x Check 51 = 17x Think: x is multiplied by 17, so divide both sides by 17 to isolate x. Substitute 3 for x. 3 is a solution. 51 = 17(3) ? 51 = 51 ?

Course Addition and Subtraction Equations Check It Out: Example 2 Solve the equation 76 = 19y. Check your answer. 76 = 19y 19 4 = y Check 76 = 19y Think: y is multiplied by 19, so divide both sides by 19 to isolate y. Substitute 4 for y. 4 is a solution. 76 = 19(4) ? 76 = 76 ?

Course Addition and Subtraction Equations Trevor’s heart rate is 78 beats per minute. How many times does his heart beat in 10 seconds? Additional Example 3: Health Application Use the given information to write an equation, where b is the number of heart beats in 10 seconds. If you count your heart beats for 10 seconds and then multiply that by 6, you can find your heart rate in beats per minute.

Course Addition and Subtraction Equations Beats in 10s times 6 = beats per minutes Additional Example 3 Continued b · 6 = 78 6b = 78 6 b = 13 Trevor’s heart beats 13 times 10 seconds. Think: b is multiplied by 6, so divide both sides by 6 to isolate b.

Course Addition and Subtraction Equations Check It Out: Example 3 During a stock car race, one driver is able to complete 68 laps in 1 hour. How many laps would he finish in 15 minutes? If you count the number of laps in 15 minutes and multiply by 4, you can find the number of laps completed in 1 hour. Use the given information to write an equation, where n is the number of laps completed in 15 minutes.

Course Addition and Subtraction Equations Check It Out: Example 3 Continued Laps in 15 min times 4 = Laps in 1 hour n · 4 = 68 4n = 68 4 n = 17 The driver would complete 17 laps in 15 minutes. Think: L is multiplied by 4, so divide both sides by 4 to isolate n.

Course Addition and Subtraction Equations Lesson Quiz: Part I Solve the equation. Check your answer = 4x 2. 18z = = = 12y 5. z = 5; 18  5 = 90 x = 3; 12 = 4  3 y = 70; 840 = 12  70 x4x4 h 22 = 9 x = 48; 12 = x4x4 h = 198; =

Course Addition and Subtraction Equations Lesson Quiz: Part II 6. The cost of each ticket at the carnival was $0.25. Li bought $7.50 worth of tickets. How many tickets did she buy? 30