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Equations and Solutions - Warm Up Problem of the Day Lesson

In this lesson, students determine if given numbers are solutions to given equations. They also solve a problem of the day involving couples and their relationships.

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Equations and Solutions - Warm Up Problem of the Day Lesson

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  1. Warm Up Problem of the Day Lesson Presentation Lesson Quizzes

  2. 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 = 42 3.x = 4 for 3(x – 2) = 10 no yes no

  3. 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

  4. Sunshine State Standards MA.7.A.3.4 Use the properties of equality to represent an equation in a different way and to show that two different equations are equivalent in a given context. Also MA.7.A.3.3

  5. Vocabulary Addition Property of Equality Inverse operations Subtraction Property of Equality

  6. To solvean 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 = y7 = 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.

  7. 2 + 3 = 5 x = y + 4 + 4 + z = + z 2 + 7 = 9 x + z = y+ z You can add the same amount to both sides of an equation, and the statement will still be true.

  8. Use inverse operations when isolating a variable. Addition and subtraction are inverse operations, which means that they “undo” each other. 2 7 +5 2 5 7 = =

  9. ? 31 – 7 = 24 ? 24 = 24 Additional Example 1: Using the Addition Property of Equality Solve the equation b – 7 = 24. Check your answer. b – 7 = 24 Think: 7 is subtracted from b, so add 7 to both sides to isolate b. + 7 +7 b = 31 Check b – 7 = 24 Substitute 31 for b.  31 is a solution.

  10. ? 24 – 3 = 21 ? 21 = 21 Check It Out: Example 1 Solve the equation y – 3 = 21. Check your answer. y – 3 = 21 Think: 3 is subtracted from y, so add 3 to both sides to isolate y. + 3 +3 y = 24 Check y – 3 = 21 Substitute 24 for y. 24 is a solution. 

  11. 4 + 7 = 11 x = y –3 – 3 – z = – z 4 + 4 = 8 x – z = y – z You can subtract the same amount from both sides of an equation, and the statement will still be true.

  12. ? 15 + 14 = 29 ? 29 = 29 Additional Example 2: Using the Subtraction Property of Equality Solve the equation t + 14 = 29. Check your answer. t + 14 = 29 Think: 14 is added to t, so subtract 14 from both sides to isolate t. – 14 – 14 t = 15 Check t + 14 = 29 Substitute 15 for t.  15 is a solution.

  13. ? 25 + 11 = 36 ? 36 = 36 Check It Out: Example 2 Solve the equation x + 11 = 36. Check your answer. x + 11 = 36 Think: 11 is added to x, so subtract 11 from both sides to isolate x. – 11 – 11 x = 25 Check x + 11 = 36 Substitute 25 for x.  25 is a solution.

  14. Additional Example 3: Sports Application 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? Let f represent the field goal points. 7 points + field goal points = points scored 7 + f = 13 7 + f = 13 Subtract 7 from both sides to isolate f. – 7 – 7 f = 6 They scored 6 points on field goals.

  15. 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 = 23 Subtract 3 from both sides to isolate x. – 3 – 3 x = 20 He scored 20 points from 2 point goals.

  16. Lesson Quizzes Standard Lesson Quiz Lesson Quiz for Student Response Systems

  17. Lesson Quiz Solve each equation. Check your answer. 1.x – 9 = 4 2.y + 6 = 72 3. 21 = n – 41 4. 127 = w + 31 5. 81 = x – 102 x = 13; 13 – 9 = 4 y = 66; 66 + 6 = 72 n = 62; 21 = 62 – 41 w = 96; 127 = 96 + 31 x = 183; 81 = 183 – 102 6. 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

  18. Lesson Quiz for Student Response Systems 1. Solve the equation x – 11 = 5. Check your answer. A.x = 16; 16 – 11 = 5 B.x = 6; 5 + 6 = 11 C.x = 55; 55 ÷ 11 = 5 D.x = 21; 21 – 11 = 5

  19. Lesson Quiz for Student Response Systems 2. Solve the equation 36 = n – 22. Check your answer. A.n = 58; 36 = 58 – 22 B.n = 12; 12 + 22 = 36 C.n = 14; 22 = 36 – 14 D.n = 60; 36 = 60 –22

  20. Lesson Quiz for Student Response Systems 3. Solve the equation z + 9 = 15. Check your answer. A.z = 24; 24 – 9 = 15 B.z = 135; 9(15) = 135 C.z = 6; 6 + 9 = 15 D.z = 9; 9 + 9 = 18

  21. Lesson Quiz for Student Response Systems 4. Solve the equation 143 = w + 33. Check your answer. A.w = 100; 143 = 100 + 33 B.w = 110; 143 = 110 + 33 C.w = 166; 143 = 166 – 33 D.w = 176; 143 = 33 – 176

  22. Lesson Quiz for Student Response Systems 5. A shop owner sold 27 books on Friday. This was 9 more than the number of books he sold on Thursday. Identify an equation and its solution that shows the number of books he sold on Thursday. A.x + 27 = 36; 9 books B. 9x = 27; 3 books C.x + 9 = 27; 18 books D.x – 9 = 27; 36 books

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