1. WARM UP Simplify: 1. 1.10(35.5x) = 3 355x

3. OBJECTIVES  Solve equations containing fractions or decimals  Solve equations containing parenthesis  Use the principal of zero products  Translate problems to equations

4. VOCABULARY  Least common denominator (LCD)  If and only if

5. CLEARING FRACTIONS OR DECIMALS  When an equation contains fractions or decimals, we can use the multiplication property to eliminate them.  The process is called clearing the equation of fractions or decimals.  Example:Solve We multiply both sides of the equations by the least common denominator, in this case 4.

6. EXAMPLE Multiplying by 4 Using the distributive property Simplify The solution is

7. MORE EXAMPLES We multiply both sides of the equation by a power of ten. Multiplying by 100 The solution is 3 Solve: 12.4 – 3.6x = 1.48 100(12.4 – 3.64x) = 100  1.48 1240 – 364x = 148 – 364x = 148 + (-1240) – 364x = -1092 x = 3

8. TRY THIS…. 1.2. 6.3x – 9.6 = 3 10(6.3x) – 9.6 = (3)10 63x – 96 = 30 63x = 126 x = 2

9. EQUATIONS WITH PARENTHESIS  When an equation contains parenthesis, we first use the distributive property to remove them. Then we proceed as before. Example: Solve 3(7 – 2x) = 14 – 8(x – 1) 21 – 6x = 14 – 8x + 8 Using the addition principle Using the distributive property 21 – 6x = 22 – 8x 8x – 6x = 22 + (-21) 2x = 1 Collecting like terms x = ½

10. GUIDELINES FOR SOLVING EQUATIONS 1. Use the distributive property to remove parenthesis, if necessary. 2 Clear fractions or decimals, if necessary 3. Collect like terms on both sides of the equation. 4. Use the addition and multiplication properties to solve for the variable.

11. TRY THIS…. 3(y – 1) – 1 = 2 – 3(y + 5) 3y – 3 – 1 = 2 – 3y – 15 3y – 4 = -13 – 3y 3y + 3y = -13 + 4 6y = -9

12. THE PRINCIPLE OF ZERO PRODUCTS  When we multiply two numbers, the product will be zero if one of the factors is zero.  Furthermore, if a product is zero, then at least one of the factors must be zero.  A statement that A and B must be both true or both false can be abbreviated as A if and only if B. This is equivalent to if A then B, and if B then A.  The Principle of Zero Products:  For any real numbers a and b, ab = 0, if and only if a = 0 or b = 0

13. EXAMPLES Using the principle of zero products Solve each equation separately The solution set is {-4, 2}. Solve: (x + 4)(x – 2) = 0 x + 4 = 0 x – 2 = 0 x = -4 There are two solutions, -4 and 2. We can show the solutions as a set by listing them inside brackets. x = 2 or

14. MORE EXAMPLES Using the principle of zero products Solve each equation separately The solution set is {0, -½}. Solve: 7x(4x + 2) = 0 7x = 0 or 4x + 2 = 0 7x = 0 The solutions are 0 and -½. x = -½ or

15. MORE EXAMPLES Using the principle of zero products Solve each equation separately The solution set is. Solve: (-2x + 5)(5x + 1) = 0 -2x + 5 = 0 or 5x + 1 = 0 -2x = -5 5x = -1 or

16. TRY THIS…. 1. 2.x(3x – 17) = 0 3. (9x + 2)(-6x + 3) = 0 (x – 19)(x + 5) = 0

17. USING EQUATIONS  Many problems can be solved by translating them into mathematical language. Often this means translating a problem into an equation. When appropriate, drawing a diagram usually helps us to understand the problem.  PROBLEM SOLVING GUIDELINES UNDERSTAND the problem Develop and carry out a PLAN Find the ANSWER and CHECK.

18. EXAMPLE A 28 ft. rope is cut into two pieces. One piece is 3 ft. longer than the other. How long are the pieces? Question: What are the lengths of the two pieces of rope? UNDERSTAND the problem: Data: The total length of the rope is 28 ft. One piece of rope is 3 ft. longer than the other piece. A diagram can help translate the problem to an equation. 28 ft. one piece3 ft. longer piece

19. EXAMPLE CONTINUED Here is one way to translate: Length of 1 piece of rope plus length of other is 28 x + (x + 3) = 28 x + x + 3 = 28 Develop and carry out a PLAN: 2x+ 3 = 28 2x = 25

20. EXAMPLE CONTINUED One piece is 12½ ft. long and the other piece is 3 ft. longer, or 15½ ft. long. The sum of the lengths is 28 ft. Find the ANSWER and CHECK.

21. TRY THIS…. 1. A 23 ft. cable is cut into two pieces, one three times as long as the other. How long are the pieces?

22. EXAMPLE Five plus a number is seven times the number. What is the number? 5 + 2x = 7  x 5 plus twice the number is seven times the number We have used x to represent the unknown number. 5 + 2x = 7x 5 = 7x – 2x 5 = 5x x = 1 Twice 1 is 2. If we add 5, we get 7. Seven times 1 is also 7. This checks, so the answer to the problem is 1.

23. TRY THIS…. 1. If seven times a certain number is subtracted from 6, the result is five times the number. What is the number? Chris bought a baseball card in 1975. By 1985, the value of the card had tripled. Since 1985, the value has increased by another $30. The card is now worth 5 times its original value. How much was the card worth when Chris bought it?2.

24. EXAMPLE The price of a mobile home was cut 11% to a new price of $48,950. What was the original price? x - 11%  x = $48,950 Original price minus 11% of original price is new price We have used x to represent the unknown number. x – 0.11  x = 48,950 1x – 0.11x = 48,950 (1-0.11)x = 48,950 0.89x = 48,950 The solution of the equation is $55,000. Check: 11% of 55,000 is 6,050. Subtracting 6,050 from 55,000 is 48, 950. x = 55,000

25. TRY THIS…. 1. A clothing store drops the price of a suit 25% to a sale price of $93. What was the original price? An investment is made at 12% simple interest. It grows to $812 at the end of 1 year. How much was invested originally? (Hint: Recall the expression P + Prt regarding the return on a principal of P dollars.)2.

26. EXAMPLE The sum of two consecutive integers is 35. What are the integers. x + (x + 1) = 35 First integer plus second integer = 35 Since the integers are consecutive, we know that one of them is 1 greater than the other. We call one of them x and the other x + 1. x + (x + 1) = 35 2x + 1 = 35 2x = 34 Since x is 17, then x + 1, the second integer is 18. The answers 17 and 18. These are both integers and consecutive. Their sum is 35, so the answers check in the problem. x = 17

27. GUIDELINES FOR WRITING NUMBERS IN ALGEBRAIC PROBLEMS 1. If two numbers are consecutive, call one x and the other x + 1. 2. If two numbers are consecutive even numbers, call one 2x and the other 2x + 2. 3. If two numbers are are consecutive odd numbers, call one 2x + 1 and the other 2x + 3. 4. If a number x is increased by by n%, the new number can be written

28. TRY THIS…. 1. The sum of two consecutive odd integers is 36. What are the integers? Field goals in basketball count two points each. In one game, Jose had one less field goal than Gary. Together they scored 46 points on field goals. a)How many points did each player score on field goals? b)How many field goals did each player make?2.

29. HOMEWORK  Textbook pg. 64-65 #2, 4, 12, 30 & pg. 69 #2, 4, 6, & 12  Journal Question #1 (Answer the following question in a minimum of two paragraphs: “Find a newspaper ad that gives a sale price and the percent off. Show how to find the original price. List the steps.