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# Ppt for lesson on equations tws

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• 1. THREE DAYTHREE DAY Unit Topic/Title:Unit Topic/Title: EQUATIONS:EQUATIONS: AN INTRODUCTION /AN INTRODUCTION / PROBLEM SOLVINGPROBLEM SOLVING
• 2. THREE DAYTHREE DAY LESSON OBJECTIVESLESSON OBJECTIVES  To solve equations using theTo solve equations using the Addition, Subtraction, DivisionAddition, Subtraction, Division and Multiplication Properties forand Multiplication Properties for Equations.Equations.  To solve equations of the formTo solve equations of the form x +b =cx +b =c  To solve equations of the formTo solve equations of the form ax = c ; x/a =cax = c ; x/a =c  To solve equations of the formTo solve equations of the form ax + b =c ; x/a +b =cax + b =c ; x/a +b =c
• 3. THREE DAYTHREE DAY INDIANA STANDARDSINDIANA STANDARDS TECHNOLOGYTECHNOLOGY  BMS.T.5.1 ---- Technology as a Communication Tool:BMS.T.5.1 ---- Technology as a Communication Tool: Students use telecommunications to collaborate, publish,Students use telecommunications to collaborate, publish, and interact with peers, teachers, and other audiences.and interact with peers, teachers, and other audiences. Students use a variety of technologies to convey informationStudents use a variety of technologies to convey information such as e-mail, e-learning, video conferencing, andsuch as e-mail, e-learning, video conferencing, and telephonytelephony  BMS.T.6.1 ----- Technology as an information Research Tool:BMS.T.6.1 ----- Technology as an information Research Tool: Students use technology to access, review, evaluate, andStudents use technology to access, review, evaluate, and select information from multiple resources for reportingselect information from multiple resources for reporting purposes. Students write appropriate research reports.purposes. Students write appropriate research reports.  BMS.T.7.1 ---- Technology as a Problem-Solving and DataBMS.T.7.1 ---- Technology as a Problem-Solving and Data Driven Decision- Making Tool: Students use technology toDriven Decision- Making Tool: Students use technology to develop strategies for solving problems.develop strategies for solving problems.
• 4. STUDENT ACADEMICSTUDENT ACADEMIC STANDARDSSTANDARDS  A1.2.1A1.2.1 Solve linear equationsSolve linear equations  A1.2.2 Solve equations andA1.2.2 Solve equations and formulas for a specifiedformulas for a specified variablevariable  A1.2.6 Solve word problemsA1.2.6 Solve word problems that involve linear equations,that involve linear equations, formulas, and inequalities.formulas, and inequalities.
• 5. KEY DEFINITIONSKEY DEFINITIONS  ADDITION PROPERTY FOR EQUATIONS:ADDITION PROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, if a = b, then a + c = b + c.For all real numbers a, b, and c, if a = b, then a + c = b + c.  SUBTRACTION PPROPERTY FOR EQUATIONS:SUBTRACTION PPROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, if a = b, then a – c = b – c.For all real numbers a, b, and c, if a = b, then a – c = b – c.  DIVISION PROPERTY FOR EQUATIONS:DIVISION PROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, (cFor all real numbers a, b, and c, (c ≠ 0), if a = b,≠ 0), if a = b,  then a/c = b/c.then a/c = b/c.  MULTIPLICATION PROPERTY FOR EQUATIONS:MULTIPLICATION PROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, (c ≠ 0), if a = b,For all real numbers a, b, and c, (c ≠ 0), if a = b,  then c · a = c · b.then c · a = c · b.  EQUATION: A mathematical statement that contains an = sign that isEQUATION: A mathematical statement that contains an = sign that is used between two numerical or algebraic expressions. An equationused between two numerical or algebraic expressions. An equation is also a statement in which the item on the left is equal to the itemis also a statement in which the item on the left is equal to the item on the right.on the right.  Example: 15 = 15Example: 15 = 15
• 6. DAY ONEDAY ONE PROCEDURE / ACTIVITIESPROCEDURE / ACTIVITIES Students will solve equations withStudents will solve equations with variables on both sides.variables on both sides. To solve equations of the formTo solve equations of the form x = b = cx = b = c Example: Solve x - 7 = 6Example: Solve x - 7 = 6 X – 7 = 6X – 7 = 6 X – 7 + 7 = 6 + 7X – 7 + 7 = 6 + 7 X = 13X = 13
• 7. PROCEDUREPROCEDURE To solve equations of the form ax = cTo solve equations of the form ax = c Solve: 7x = -21Solve: 7x = -21 7x = -21 (solve for x)7x = -21 (solve for x) 7x7x == -21-21 (divide both sides by 7)(divide both sides by 7) 7 77 7 1x = -31x = -3 X = -3X = -3 Website to visit: yourteacher.comWebsite to visit: yourteacher.com
• 8. PROCEDUREPROCEDURE To solve equations with the form :To solve equations with the form : xx = c= c aa Solve:Solve: xx = 5= 5 -3-3 -3-3 ·· xx = -3 · 5 Multiply both sides by –3.= -3 · 5 Multiply both sides by –3. -3-3 check:check: xx = 5= 5 -15-15 = 5= 5 X = -15 -3 -3X = -15 -3 -3 5 = 55 = 5
• 9. DAY TWODAY TWO PROCEDUREPROCEDURE To solve equations of the formTo solve equations of the form ax + b = cax + b = c Solve: 2x + 6 = 14Solve: 2x + 6 = 14 2x + 6 = 142x + 6 = 14 2x + 6 – 6 = 14 – 62x + 6 – 6 = 14 – 6 2x + 0 = 82x + 0 = 8 2x2x == 88 2 22 2 x = 4x = 4
• 10. DAY THREEDAY THREE PROCEDURE / ACTIVITIESPROCEDURE / ACTIVITIES  Students will solve equations with variables on bothStudents will solve equations with variables on both sides.sides. Solve 5x – 8 = 3x + 12Solve 5x – 8 = 3x + 12 -3x +5x – 8 = -3x +3x +12-3x +5x – 8 = -3x +3x +12 ↔ Add -3x to each side.↔ Add -3x to each side. 2x – 8 = 0 + 12 ↔ Combine like terms.2x – 8 = 0 + 12 ↔ Combine like terms. 2x – 8 = 12 ↔ Now we have one variable2x – 8 = 12 ↔ Now we have one variable term.term. 2x – 8 + 8 = 12 + 8 ↔ Add 8 to each side.2x – 8 + 8 = 12 + 8 ↔ Add 8 to each side. 2x2x == 2020 ↔ Divide each side by 2↔ Divide each side by 2 2 22 2 x = 10x = 10 Therefore, the solution is 10.Therefore, the solution is 10.
• 11. DAY FOURDAY FOUR WORD PROBLEMWORD PROBLEM  Word problems can lead to equations with the variable onWord problems can lead to equations with the variable on both sides. Solve:both sides. Solve: Twenty more than 4 times Jack’s age is the same as 6Twenty more than 4 times Jack’s age is the same as 6 times his age.times his age.  Read > The problem asks for Jack’s age.Read > The problem asks for Jack’s age.  Plan > Use a variable to represent Jack’s age. Let a =Plan > Use a variable to represent Jack’s age. Let a = his age.his age. (20 more than(20 more than 4 times Jack’s age) (4 times Jack’s age) (is the same as) (6is the same as) (6 times his age).times his age). Solve > 4a + 20 = 6aSolve > 4a + 20 = 6a -4a + 4a + 20 = -4a + 6a-4a + 4a + 20 = -4a + 6a 2020 == 2a2a 2 22 2 10 = a10 = a
• 12. DAY FIVEDAY FIVE WORD PROBLEMWORD PROBLEM  20 MORE THAN 4 Times Jack’s age is the same as 620 MORE THAN 4 Times Jack’s age is the same as 6 times is age.times is age. 44 · 10 + 20 | 6 · 10· 10 + 20 | 6 · 10 40 + 20 | 6040 + 20 | 60 60 = 60 True60 = 60 True Therefore, Jack is 10 yearsTherefore, Jack is 10 years old.old. Supplemental Website:Supplemental Website:
• 13. SOLVE THESE PROBLESSOLVE THESE PROBLES 1.1. 6X + 7 = 3X + 166X + 7 = 3X + 16 2.2. 6 + 10X = 8X + 126 + 10X = 8X + 12 3. 6P + 13 = 9P – 53. 6P + 13 = 9P – 5
• 14. Answer to Problem #1Answer to Problem #1  6x + 7 + 3x + 166x + 7 + 3x + 16  6x – 6x +7 = 3x -6x +166x – 6x +7 = 3x -6x +16  0 + 7 = -3X + 160 + 7 = -3X + 16  7 – 16 = -3X + 16 – 167 – 16 = -3X + 16 – 16  -9-9 == -3X-3X -3 = -3-3 = -3 3 = X3 = X
• 15. Answer to ProblemAnswer to Problem #1(same problem)#1(same problem)  6X + 7 = 3X + 166X + 7 = 3X + 16  6X -3X + 7 = 3X – 3X +166X -3X + 7 = 3X – 3X +16  3X + 7 = 0 + 163X + 7 = 0 + 16  3X + 7 - 7 = 16 – 73X + 7 - 7 = 16 – 7  3X + 0 = 93X + 0 = 9  3X3X == 99 3 33 3 X = 3X = 3
• 16. HOMEWORKHOMEWORK WILL BEWILL BE ASSIGNEDASSIGNED
• 18. POST - TESTPOST - TEST EQUATIONSEQUATIONS
• 19. THREE DAYTHREE DAY Unit Topic/Title:Unit Topic/Title: EQUATIONS:EQUATIONS: AN INTRODUCTION /AN INTRODUCTION / PROBLEM SOLVINGPROBLEM SOLVING
• 20. THREE DAYTHREE DAY LESSON OBJECTIVESLESSON OBJECTIVES  To solve equations using theTo solve equations using the Addition, Subtraction, DivisionAddition, Subtraction, Division and Multiplication Properties forand Multiplication Properties for Equations.Equations.  To solve equations of the formTo solve equations of the form x +b =cx +b =c  To solve equations of the formTo solve equations of the form ax = c ; x/a =cax = c ; x/a =c  To solve equations of the formTo solve equations of the form ax + b =c ; x/a +b =cax + b =c ; x/a +b =c
• 21. THREE DAYTHREE DAY INDIANA STANDARDSINDIANA STANDARDS TECHNOLOGYTECHNOLOGY  BMS.T.5.1 ---- Technology as a Communication Tool:BMS.T.5.1 ---- Technology as a Communication Tool: Students use telecommunications to collaborate, publish,Students use telecommunications to collaborate, publish, and interact with peers, teachers, and other audiences.and interact with peers, teachers, and other audiences. Students use a variety of technologies to convey informationStudents use a variety of technologies to convey information such as e-mail, e-learning, video conferencing, andsuch as e-mail, e-learning, video conferencing, and telephonytelephony  BMS.T.6.1 ----- Technology as an information Research Tool:BMS.T.6.1 ----- Technology as an information Research Tool: Students use technology to access, review, evaluate, andStudents use technology to access, review, evaluate, and select information from multiple resources for reportingselect information from multiple resources for reporting purposes. Students write appropriate research reports.purposes. Students write appropriate research reports.  BMS.T.7.1 ---- Technology as a Problem-Solving and DataBMS.T.7.1 ---- Technology as a Problem-Solving and Data Driven Decision- Making Tool: Students use technology toDriven Decision- Making Tool: Students use technology to develop strategies for solving problems.develop strategies for solving problems.
• 22. STUDENT ACADEMICSTUDENT ACADEMIC STANDARDSSTANDARDS  A1.2.1A1.2.1 Solve linear equationsSolve linear equations  A1.2.2 Solve equations andA1.2.2 Solve equations and formulas for a specifiedformulas for a specified variablevariable  A1.2.6 Solve word problemsA1.2.6 Solve word problems that involve linear equations,that involve linear equations, formulas, and inequalities.formulas, and inequalities.
• 23. KEY DEFINITIONSKEY DEFINITIONS  ADDITION PROPERTY FOR EQUATIONS:ADDITION PROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, if a = b, then a + c = b + c.For all real numbers a, b, and c, if a = b, then a + c = b + c.  SUBTRACTION PPROPERTY FOR EQUATIONS:SUBTRACTION PPROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, if a = b, then a – c = b – c.For all real numbers a, b, and c, if a = b, then a – c = b – c.  DIVISION PROPERTY FOR EQUATIONS:DIVISION PROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, (cFor all real numbers a, b, and c, (c ≠ 0), if a = b,≠ 0), if a = b,  then a/c = b/c.then a/c = b/c.  MULTIPLICATION PROPERTY FOR EQUATIONS:MULTIPLICATION PROPERTY FOR EQUATIONS:  For all real numbers a, b, and c, (c ≠ 0), if a = b,For all real numbers a, b, and c, (c ≠ 0), if a = b,  then c · a = c · b.then c · a = c · b.  EQUATION: A mathematical statement that contains an = sign that isEQUATION: A mathematical statement that contains an = sign that is used between two numerical or algebraic expressions. An equationused between two numerical or algebraic expressions. An equation is also a statement in which the item on the left is equal to the itemis also a statement in which the item on the left is equal to the item on the right.on the right.  Example: 15 = 15Example: 15 = 15
• 24. DAY ONEDAY ONE PROCEDURE / ACTIVITIESPROCEDURE / ACTIVITIES Students will solve equations withStudents will solve equations with variables on both sides.variables on both sides. To solve equations of the formTo solve equations of the form x = b = cx = b = c Example: Solve x - 7 = 6Example: Solve x - 7 = 6 X – 7 = 6X – 7 = 6 X – 7 + 7 = 6 + 7X – 7 + 7 = 6 + 7 X = 13X = 13
• 25. PROCEDUREPROCEDURE To solve equations of the form ax = cTo solve equations of the form ax = c Solve: 7x = -21Solve: 7x = -21 7x = -21 (solve for x)7x = -21 (solve for x) 7x7x == -21-21 (divide both sides by 7)(divide both sides by 7) 7 77 7 1x = -31x = -3 X = -3X = -3 Website to visit: yourteacher.comWebsite to visit: yourteacher.com
• 26. PROCEDUREPROCEDURE To solve equations with the form :To solve equations with the form : xx = c= c aa Solve:Solve: xx = 5= 5 -3-3 -3-3 ·· xx = -3 · 5 Multiply both sides by –3.= -3 · 5 Multiply both sides by –3. -3-3 check:check: xx = 5= 5 -15-15 = 5= 5 X = -15 -3 -3X = -15 -3 -3 5 = 55 = 5
• 27. DAY TWODAY TWO PROCEDUREPROCEDURE To solve equations of the formTo solve equations of the form ax + b = cax + b = c Solve: 2x + 6 = 14Solve: 2x + 6 = 14 2x + 6 = 142x + 6 = 14 2x + 6 – 6 = 14 – 62x + 6 – 6 = 14 – 6 2x + 0 = 82x + 0 = 8 2x2x == 88 2 22 2 x = 4x = 4
• 28. DAY THREEDAY THREE PROCEDURE / ACTIVITIESPROCEDURE / ACTIVITIES  Students will solve equations with variables on bothStudents will solve equations with variables on both sides.sides. Solve 5x – 8 = 3x + 12Solve 5x – 8 = 3x + 12 -3x +5x – 8 = -3x +3x +12-3x +5x – 8 = -3x +3x +12 ↔ Add -3x to each side.↔ Add -3x to each side. 2x – 8 = 0 + 12 ↔ Combine like terms.2x – 8 = 0 + 12 ↔ Combine like terms. 2x – 8 = 12 ↔ Now we have one variable2x – 8 = 12 ↔ Now we have one variable term.term. 2x – 8 + 8 = 12 + 8 ↔ Add 8 to each side.2x – 8 + 8 = 12 + 8 ↔ Add 8 to each side. 2x2x == 2020 ↔ Divide each side by 2↔ Divide each side by 2 2 22 2 x = 10x = 10 Therefore, the solution is 10.Therefore, the solution is 10.
• 29. DAY FOURDAY FOUR WORD PROBLEMWORD PROBLEM  Word problems can lead to equations with the variable onWord problems can lead to equations with the variable on both sides. Solve:both sides. Solve: Twenty more than 4 times Jack’s age is the same as 6Twenty more than 4 times Jack’s age is the same as 6 times his age.times his age.  Read > The problem asks for Jack’s age.Read > The problem asks for Jack’s age.  Plan > Use a variable to represent Jack’s age. Let a =Plan > Use a variable to represent Jack’s age. Let a = his age.his age. (20 more than(20 more than 4 times Jack’s age) (4 times Jack’s age) (is the same as) (6is the same as) (6 times his age).times his age). Solve > 4a + 20 = 6aSolve > 4a + 20 = 6a -4a + 4a + 20 = -4a + 6a-4a + 4a + 20 = -4a + 6a 2020 == 2a2a 2 22 2 10 = a10 = a
• 30. DAY FIVEDAY FIVE WORD PROBLEMWORD PROBLEM  20 MORE THAN 4 Times Jack’s age is the same as 620 MORE THAN 4 Times Jack’s age is the same as 6 times is age.times is age. 44 · 10 + 20 | 6 · 10· 10 + 20 | 6 · 10 40 + 20 | 6040 + 20 | 60 60 = 60 True60 = 60 True Therefore, Jack is 10 yearsTherefore, Jack is 10 years old.old. Supplemental Website:Supplemental Website:
• 31. SOLVE THESE PROBLESSOLVE THESE PROBLES 1.1. 6X + 7 = 3X + 166X + 7 = 3X + 16 2.2. 6 + 10X = 8X + 126 + 10X = 8X + 12 3. 6P + 13 = 9P – 53. 6P + 13 = 9P – 5
• 32. Answer to Problem #1Answer to Problem #1  6x + 7 + 3x + 166x + 7 + 3x + 16  6x – 6x +7 = 3x -6x +166x – 6x +7 = 3x -6x +16  0 + 7 = -3X + 160 + 7 = -3X + 16  7 – 16 = -3X + 16 – 167 – 16 = -3X + 16 – 16  -9-9 == -3X-3X -3 = -3-3 = -3 3 = X3 = X
• 33. Answer to ProblemAnswer to Problem #1(same problem)#1(same problem)  6X + 7 = 3X + 166X + 7 = 3X + 16  6X -3X + 7 = 3X – 3X +166X -3X + 7 = 3X – 3X +16  3X + 7 = 0 + 163X + 7 = 0 + 16  3X + 7 - 7 = 16 – 73X + 7 - 7 = 16 – 7  3X + 0 = 93X + 0 = 9  3X3X == 99 3 33 3 X = 3X = 3
• 34. HOMEWORKHOMEWORK WILL BEWILL BE ASSIGNEDASSIGNED