1. Gale Bach Intermediate Algebra Math 155 Fall 2013 Santa Rosa Junior College Mathematics

2. http://online.wsj.com/public/resources/documents/st_BESTJOBS0104_20110105.html Why Math? http://online.santarosa.edu/homepage/gbach/

7. Example Check: Solution The solution is x = – 5.

8. Types of Equations An identity is an equation that is true for all replacements that can be used on both sides of the equation. A contradiction is an equation that is never true. A conditional equation is sometimes true and sometimes false, depending on what the replacement is.

9. Example The equation is true regardless of the choice for x, so all real numbers are solutions. The equation is an identity equation. Solution Using the Distributive Law Combining like terms 0 = 0

10. Example The equation is false for any choice of x, so there is no solution for this equation. The equation is a contradiction. Solution Using the Distributive Law Combining like terms

11. Example There is one solution, 7. For all other choices of x, the equation is false. The equation is a conditional equation. Solution Using the Distributive Law Combining like terms

12. Solve How about technology!

15. Solving Formulas A formula is an equation that uses variables to represent a relationship between two or more quantities. Example r The area of a circle of radius r is given by the formula:

16. Example Solution Divide by l and h Simplify We want this variable alone.

17. Example Solution Subtract x/aMultiply by b Can you think of another strategy for solving for y?

18. Example Solution

19. To Solve a Formula for a Specified Variable 1.Get all the terms with the variable being solved for on one side of the equation and all the other terms on the other side, using the Addition Principle. To do this may require removing parentheses. To remove parentheses, either divide both sides by the multiplier in front of the parentheses or use the Distributive Law.

20. Example Solution Subtracting

21. To Solve a Formula for a Specified Variable 1.Get all the terms with the variable being solved for on one side of the equation and all the other terms on the other side, using the Addition Principle. To do this may require removing parentheses. To remove parentheses, either divide both sides by the multiplier in front of the parentheses or use the Distributive Law. 2.When all the terms with the specified variable are on the same side, factor (if necessary) so that the variable is written only once.

22. Example Solution Subtracting Factoring

23. To Solve a Formula for a Specified Variable 1.Get all the terms with the variable being solved for on one side of the equation and all the other terms on the other side, using the Addition Principle. To do this may require removing parentheses. To remove parentheses, either divide both sides by the multiplier in front of the parentheses or use the Distributive Law. 2.When all the terms with the specified variable are on the same side, factor (if necessary) so that the variable is written only once. 3.Solve for the variable in question by dividing both sides by the multiplier of that variable.

24. Example Solution Subtracting Factoring Dividing Simplifying

25. The Three-Step Strategy

27. Three Steps for Problem Solving with Algebra 1. Introduction 2. Body 3. Conclusion The 3 amigos use this format.

28. The First Step in Problem Solving Introduction: 1.If the problem is written, read it carefully. Then read it again, perhaps aloud. Verbalize the problems to yourself. 2.List the information given and restate the question being asked. 3. Select a variable or variables to represent any unknown(s) and clearly state what each variable represents. Be descriptive! For example, let t = the flight time, in hours; let p = Paul’s weight, in pounds; and so on. Let the variable represent the value that you know least about. 4.Find additional information. Look up formulas or definitions with which you are not familiar. Geometric formulas appear on the inside back cover of this text; important words appear in the index.

29. The First Step in Problem Solving (continued) Introduction: 5. Create a table, using variables, in which both known and unknown information is listed. Look for patterns. 6. Make and label a drawing.

30. The Second Step in Problem Solving Body: 1. Translate the problem to mathematical language. In some cases, translation can be done by writing an algebraic expression, but most problems in this course are best solved by translating to an equation. 2. Solve the equation.

31. The Third Step in Problem Solving Conclusion: 1.State your results in sentence form. 2. Make sure your answer makes sense.

34. Translating to Algebraic Expressions per of less than more than ratio twicedecreased byincreased by quotient of times minus plus divided by product ofdifference of sum of divide multiply subtract add DivisionMultiplicationSubtractionAddition Key Words

35. Example Translate to an algebraic expression: “Eight more than twice the product of 5 and a number.” Solution Eight more than the product of 5 and a number. twice

36. Example The faculty discount at a bookstore is 15%. If a sweatshirt after the discount was $32.30. What was the original price of the sweatshirt? Solution 1. Introduction. The original price minus the discount is equal to sale price. Let S = the sweatshirt’s original price. (US $)

37. 2. Body. Solve the equation. the sweatshirt price minus the discount is the sale price

38. Check. Note that 15% of $38 would be (0.15)($38) = $5.70. When this is subtracted from $38 we have Thus $38 checks with the original problem. 3. Conclusion. The original price of the sweatshirt was $38.

39. Example One angle of a triangle has the same measure as a second angle. The third angle measures 10 o more than three times the measure of the first angle. Find the measures of the angles. Solution 1. Introduction. We need to find three angles. Let x = first angle, in degrees. First The first anglex Second The same as the firstx Third10 o more than three times the first 3x + 10

40. 1.Introduction (continued). First Angle + Second Angle + Third Angle = 180 o First xx Second The same as the firstx Third10 o more than three times the first 3x + 10 2.Body. First Angle + Second Angle + Third Angle = 180 o

41. 2. Solve the equation. first anglesecond anglethird angle

42. Check. 3. Conclusion. The three angles are 34 o, 34 o, and 112 o.