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Let c be any real number. If a = b, then a + c = b + c If a = b, then a ( c )= b ( c ) x + 7 – 7 = 22 – 7 x = 15 If a = b, then a – c = b – c If a = b, then a /c = b /c –7=–7 x = 15 Vertical steps = less writing. 2 + 12 = + 12 x 41 5 5 x 9 2 x 26 “UNDO” the operations on the x term.. Direction: “SOLVE.” this means, WORK BACKWARDS Rules to follow: SIMPLIFY to 1 variable term. You may need to combine like terms, distributive property to remove ( )’s, and Clear Fractions by multiplying both sides by the LCD (least common denominator). 3x 5 17 –5=–5 3x 12 3 3 x4 45 x 13 – 45 = – 45 x 32 -1 -1 x 32 4 x 7 1 3 +7 = +7 2 4 x 8 3 3 4 3 4 x6 16.3 7.2x 8.18 – 16.3 = – 16.3 7.2x 24.48 -7.2 -7.2 x 3.4 Sect 2.2 Solving using Order of Operations Combining Like Terms in the simplifying stage. 3x 4x 14 7 x 14 7 7 x 2 Always move the smallest variable term! x 5 8x 6 + 8x = + 8x 7x 5 6 –5=–5 7x 1 7 7 1 x 7 6x 5 7 x 10 4x 7 5 1x 17 4x + 4x = + 4x 5 3x 17 –5 =–5 3x 12 3 3 x4 Distributive Property in the simplifying stage. 2 5x 5 3x 2 1 2 5x 25 3x 6 1 5x 23 3x 7 + 5x =+ 5x 23 8x 7 +7= +7 16 8x 8 8 2 x To clear fractions, multiply both sides by the LCD. LCD = 6 2 2 16 6 x 2x 6 3 6 4x 1 12 x – 4x = – 4x 1 8x 8 8 1 x 8 Two approaches. 2 3x 2 8 5 6 4 5 5 x 8 5 5 5 6x 4 40 –4=–4 6x 36 6 6 x6 5 2 4 2 3x 2 8 5 5 2 3x 2 20 –2=–2 3x 18 3 3 x6 To clear fractions, multiply both sides by the LCD. 6 12 6 12 3 12 LCD = 12 1 3 3 2 4 x 12 2 2 4 3 6x 18 9 8 1005.3 1.2x 1.94100 530 120x 194 6x 9 8 – 530 = – 530 120x 336 –9=–9 6x 1 To clear decimals, multiply both sides by 10. Every multiplication of 10 moves the decimal one place to the right. We will multiply by 100. x 1 6 120 = 120 x 2.8 6 6 All the above examples are Conditional equations…they have a solution. Contradiction Equation. (No solution) Identity Equations. (Have infinite sols.) 2 x 7 7x 1 5x 3x 5 2x 2 x 3x 5 2x 4 x 2 x 7 7 x 7 5x 3x 5 3x 4 2x 7 2x 7 5 4 77 False Statement No Solution Contradiction Equation True Statement Infinite Solutions Identity Equation 1 A bh 2 1 2 A bh 2 2 Solve for b. Isolate the b by undoing the operations taking place on the variable b. 2 A bh h h 2A b h Solve for W. Remove the terms with no W 1st. Next isolate the W by undoing the operations taking place on the variable W. P 2L 2W – 2L = – 2L P 2L 2W 2 2 P 2L W 2 Solve for a. K 917 6w h a – 917 = – 917 Move the terms and factors that are outside the ( ) ‘s to the other side. Since we have – a, add the a to the left and subtract the fraction to the right. a will be isolated. K 917 6w h a 6 6 K 917 wha 6 a K 917 K 917 a 6 6 a wh K 917 6 Solve for h. Remove the terms with no h 1st. Next isolate the h by undoing the operations taking place on the variable h. A 2rh 2r 2 2r 2 2r 2 A 2r 2 2rh 2r 2r A 2r 2 h 2r Sect 2.4 Solving Percentage Problems n 1 n n% = n0.01 100 100 Convert decimals (fractions) to percent and percent to decimal. Fraction to a percent n % d 100 Convert to a percent. 3 x 5 100 Cross multiply and products are equal. 5x 300 5 x 300 5 5 x 60 60% Decimal to a percent. Multiply by 100 Convert to a percent. 0.359 100 35.9% Notice we moved the decimal point 2 places to the right. Percent to a decimal. Drop % and divide by 100 Convert to a decimal. 19.9% 19.9 100 0.199 Notice we moved the decimal point 2 places to the left. The values that associates with the words “is” and “of” We will work both techniques…pick your favorite. x 11 0.01 49 x 5.39 Not to bad… just like Sect 1.1 and 1.2. What is 11% of 49? x 11 is 100 49 of 100x 11 49 100 100 x 5.39 Notice we will always cross multiply by the numbers on the diagonal and divide by the 3rd number! 3 16.01 x 3 0.16 * x 0.16 0.16 18.75 x is 16 3 100 x of Cross multiply the numbers on the diagonal and divide by the 3rd number! Calculator !!! x 100 3 16 18.75 x .01 50 34 Here is where problems may occur! x really needs to multiplied by 0.01 because of the word percent! 0.50x 34 0.50 0.50 x 68 68% x 34 is 100 of 50 Cross multiply the numbers on the diagonal and divide by the 3rd number! Calculator !!! x 100 34 50 68 68% In 2006, there were 300 million people in the United States, and 62.2% of them lived within 5 mi. of a Wal-Mart store. How many people lived with 5mi. of a Wal-Mart store? Any number that represents a TOTAL associates with 100% and is across from 100 in the proportion or associates with “of.” 62.2 is x 100 300 of x 62.2 300 100 186.6 million About 1.6 million students who graduated high school went to college. This was 66% of all high school graduates. How many total high school graduates are there? TOTAL is across from 100 in the proportion or associates with “of.” 66 100 1is.6 of x x 100 1.6 66 2.42 million A car dealer lowered the sticker price of a car from $20,830 to $18,955. What percent of the regular price does the sale price represent? What is the percent discount? Any number that represents a ORIGINAL PRICE associates with 100% and is across from 100 in the proportion or associates with “of.” x 18955 is x 100 18955 20830 90.563784 100 20830 of 90.6% What is the percent discount? 100% 90.6% 9.4% The total bill was $47.70 that included 6% sales tax. How much was the merchandise before tax? We have TOTAL and ORIGINAL price…which is the “of”? 106 6 100 47 is.70 ofx ORIGINAL price is the “of”? The $47.70 represent the ORIGINAL price + 6% sales tax WAIT a minute! That means our PERCENTAGE is not 6%, 106%! x 100 47.70 106 45 $45 Sect 2.5 Word Problems Sect 2.5 Word Problems Sect 2.5 Word Problems 10x 2 78 2 2 10 x 80 10 10 53 2 x 70 15 10x 70 15 15 x 5.5 10 x 55 10 10 24 3x 34 8 6x 34 8 x 8 8 6x 26 x 26 13 6 3 6 6 Sect 2.5 Word Problems 260km 3 times the unknown distance = 3x P How far he biked = 3x 3 65 195km How far he had left to go = x x 65km 3x Unknown distance = x S 3x x 260 4x 260 x 65 x F Sect 2.5 Word Problems Mile x Mile x+1 1 mile First marker = x x 279 Next consecutive marker = x + 1 279 1 280 x + x + 1 = 559 2x 1 559 2x 558 x 279 Sect 2.5 Word Problems Length P 2L 2W 288 2L 2W 288 2L 2W Width W 50 ft Width 288 2W 44 2W 288 2W 88 2W L W 44 Length L W 44 50 44 94 ft 288 4W 88 200 4W 50 W W 94 ft by 50 ft Sect 2.5 Word Problems B = Number of brochures to print Starting Cost 2($300) + The number of Brochures + B(the cost of each Brochures) Change cents to dollars! $600 $3000 + B($0.215) $3000 $3000 0.215B 2400 B 11162.8 Can print 11,162 brochures and not go over budget. Sect 2.5 Word Problems Back side is the unknown = x 40 2 40 80 Front = x + 20 40 20 60 Peak = 2x The sum of three angles of a triangle is 180 degrees. x + 2x + x + 20 = 180 4x + 20 = 180 4x = 160 x = 40 40 80 60 180 Sect 2.5 Word Problems H = the hammer price of final bid. $1150 + 8%(of final bid) = final bid 1150 0.08H 1H 1150 0.92H 1250 H The final bid must be $1250 or higher. Proportion Style. Jared has to pay the Auctioneer 8% of the final bid, so he gets 92% of the final bid. is 92 1150 100 ofH H 100 1150 92 1250 Let A > B and C is a non-zero positive constant. AB C C AC B C AB C C AC B C REMEMBER!!! AB C C AC B C AB C C AC B C A B C C How to write a solution set with set builder notation, interval notation and Graph. Let x > 3 be our solution. Graph. Set builder notation “The set of all x, such that x > 3.” ( 0 {x|x>3} Interval notation 3 Author. 3 Mr. Fitz 3, Let x < 3 be our solution. Author. x | x 3 ] Mr. Fitz Interval notation ,3 0 3 REMEMBER! 3 When x is on the left side the inequality symbol points in the direction of the graph. Compound inequalities Let x > – 2 and x < 3 be our solution. We want all x values between -2 and 3. ( ] 2 x x 3 x | 2 x 3 2 0 3 Author. Interval notation 2 2,3 3 Mr. Fitz Solve the inequalities. Graph the solution and write in interval notation. 3x 1 2x 5 – 2x – 2x x 1 5 +1 +1 x 4 4 , 4 4 1 x7 4 4 2 x 18 –2 –2 x 9 x 28 28 , 28 Flip inequality symbol 9 9, 6 5x 7 –6 –6 5x 1 –5 –5 1 x 5 2x 9 7 x 1 – 2x – 2x If you move the smallest variable term, it will stay positive. 9 5x 1 –1 1 5 , 15 3x 27 1 2 5x 30 3x 28 5x 28 –1 10 5x 5 x 2 2 + 5x + 5x 8x 28 28 + 28 5 2 x 3x 9 1 2 5x 6 Switch the inequality around so x is on the left side. 2, + 28 8x 0 8 8 x0 0 , 0 Let p = how many people $50 + $15(per person) 50 15 p 450 – 50 15 p 400 – 50 15 15 < $450 p 26.6 The party cannot exceed 26 people! Let h = how many hours worked in the 4th week Average of 4 wks > 16 hours/week 4 20 12 14 h 16 4 4 46 h 64 – 46 – 46 h 18 Dina has to work at least 18 hours in the 4th week.