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Solving Two-Step Equations LESSON 6-1 Course 3 Problem of the Day There are 15 girls in a class of 27 students. Using lowest terms, what fraction of the students are girls? What fraction are boys? 5 4 , 9 9 Lesson Main Lesson 6-1 Feature Solving Two-Step Equations LESSON 6-1 Course 3 Check Skills You’ll Need (For help, go to Lesson 1-6.) 1. Vocabulary Review What does it mean to isolate the variable? Solve each equation. 2. x + 4 = – 3 3. c – 5 = 1 4. 5 + a = 35 Check Skills You’ll Need Lesson Main Lesson 6-1 Feature Solving Two-Step Equations LESSON 6-1 Course 3 Check Skills You’ll Need Solutions 1. to get the variable alone on one side of the equation 2. x+4=–3 x+4–4 =–3–4 x=–7 3. c–5= 1 c–5+5= 1+5 c=6 4. a = 30 Lesson Main Lesson 6-1 Feature Solving Two-Step Equations LESSON 6-1 Course 3 Additional Examples Solve 4p + 27 = 61.48. 4p + 27 = 61.48 4p + 27 – 27 = 61.48 – 27 4p = 34.48 34.48 4p = 4 4 Subtract 27 from each side. Simplify. Divide each side by 4. Simplify. p = 8.62 Check 4p + 27 = 61.48 61.48 Substitute 8.62 for p. 61.48 = 61.48 The solution checks. 4(8.62) + 27 Quick Check Lesson Main Lesson 6-1 Feature Solving Two-Step Equations LESSON 6-1 Course 3 Additional Examples At a recent breakfast, four friends paid for their drinks and shared the cost of a bag of doughnuts. Joe’s drink was $1.75. He paid $3.20 total for breakfast. What equation can be used to find the cost of the doughnuts? How much did the bag of doughnuts cost? Words cost of Joe’s drink (cost of bag of doughnuts 4) plus is $3.20 = 3.20 Let d = the cost of a bag of doughnuts. Equation 1.75 Lesson Main d 4 + Lesson 6-1 Feature Solving Two-Step Equations LESSON 6-1 Course 3 Additional Examples (continued) The equation is 1.75 + d =3.2. You can solve the equation to find the 4 cost. 1.75 + d = 3.2. 4 1.75 – 1.75 + d 4 d 4 (4) d 4 = 3.2 – 1.75 Subtract 1.75 from each side. = 1.45 Simplify. = (4)1.45 Multiply each side by 4. d = 5.8 Simplify. The cost of a bag of doughnuts is $5.80. Lesson Main Lesson 6-1 Quick Check Feature Solving Two-Step Equations LESSON 6-1 Course 3 Lesson Quiz Solve each equation. 1. n – 14 = 1 3 45 2. 16 – 2x = 9 3.5 Write and solve an equation. 3. Suppose you bought a $2.75 sandwich and two drinks of equal price. You spend $4.25 in all. How much does one drink cost? 2d + 2.75 = 4.25; $.75 4. Admission to a museum costs $3.75 per person. A group of friends attend. They spend a total of $30 on lunch. At the end of the day, they had spent a total of $56.25. How many friends attended the museum? 3.75x + 30 = 56.25; 7 friends Lesson Main Lesson 6-1 Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Problem of the Day Suppose you had the following set of numbers: 2, 3, 5, 7, 8, 11 Using only two numbers at a time, write as many fractions greater than 1 as possible from this set. 3 , 5 , 7 , 8 , 11 , 5 , 7 , 8 , 11 , 7 , 8 , 11 , 8 , 11 , 11 2 2 2 2 2 3 3 3 3 5 5 5 7 7 8 Lesson Main Lesson 6-2 Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Check Skills You’ll Need (For help, go to Lesson 1-5.) 1. Vocabulary Review Is the expression (5 + 3a) – 15 simplified? Explain. Simplify each expression. 2. – 8(r + 3) 4. 35(2 – t) 3. – 7(s – 5) Check Skills You’ll Need Lesson Main Lesson 6-2 Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Check Skills You’ll Need Solutions 1. No; 15 can be subtracted from 5. The simplest form is 3a – 10. 2. – 8(r) + (– 8)(3) = – 8r – 24 3. –7(s) – (– 7)(5) = – 7s + 35 4. 35(2) – (35)(t) = 70 – 35t Lesson Main Lesson 6-2 Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Additional Examples Combine like terms in the expression 8p + 13p + p. 8p + 13p + p = 8p + 13p + 1p Rewrite p as 1p. = (8 + 13 + 1)p Distributive Property = 22p Combine like terms by adding. Quick Check Lesson Main Lesson 6-2 Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Additional Examples Carlos buys 6 tubes of paint and 3 pieces of fabric to make an art project. Shauna buys 8 tubes of paint and 1 piece of fabric. Define and use variables to represent the total cost. Words Carlos 6 tubes of paint plus 3 pieces of fabric Let t = the cost of a tube of paint. Let f = the cost of a piece of fabric. Expression Words Shauna Expression Lesson Main 6t 3f + 8 tubes of paint plus + 8t Lesson 6-2 1 piece of fabric f Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Additional Examples (continued) Combined Expression (6t + 3f ) + (8t + f ) Commutative Property of Addition (6t + 3f) + (8t + f) = 6t + 3f + 8t + f = (6 + 8)t + (3 + 1)f Distributive Property = 14t + 4f Simplify. Quick Check Lesson Main Lesson 6-2 Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Additional Examples Simplify 7t – 2(t – 3). 7t – 2(t – 3) = 7t + (–2)(t – 3) Add the opposite of 2(t – 3). = 7t + [–2t – (–6)] Distributive Property = 7t + (–2t) + 6 Simplify. = [7 + (–2)]t + 6 Distributive Property = 5t + 6 Simplify. Quick Check Lesson Main Lesson 6-2 Feature Simplifying Algebraic Expressions LESSON 6-2 Course 3 Lesson Quiz Simplify each expression. 1. – 13c + c –12c 2. 4y – 7 + 8y 12y – 7 3. 1 – 6(b – 9) – 6b + 55 4. Karen buys 4 boxes of cereal and 3 bags of almonds at the grocery store. Her brother, David, buys 2 boxes of cereal. Define and use variables to represent the total cost. Let c = the cost of a box of cereal and let a = the cost of a bag of almonds. Then 6c + 3a represents the total cost. Lesson Main Lesson 6-2 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Problem of the Day Find the lowest common denominator (LCD) for each set of numbers. a. 1 , 3 , 1 2 4 3 b. 2 , 5 , 1 3 12 6 9 18 Lesson Main Lesson 6-3 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Check Skills You’ll Need (For help, go to Lesson 6-2.) 1. Vocabulary Review Identify the like terms in 3x + 2x = 8 – x. Simplify. 2. 5 – 3m + 7 – 23m 3. 4(7 – 3r) 4. (q + 1)5 + 3q Check Skills You’ll Need Lesson Main Lesson 6-3 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Check Skills You’ll Need Solutions 1. 3x, 2x, – x 2. 12 – 26m 3. 4(7) + (4)(– 3r) = 28 – 12r 4. 5(q) + 5(1) + 3q = (5 + 3)q + 5 = 8q + 5 Lesson Main Lesson 6-3 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Additional Examples Solve 2c + 2 + 3c = 12. 2c + 2 + 3c = 12 2c + 3c + 2 = 12 5c + 2 = 12 5c + 2 – 2 = 12 – 2 5c = 10 5c 10 = 5 5 c=2 Lesson Main Commutative Property Combine like terms. Subtract 2 from each side. Simplify. Divide each side by 5. Simplify. Lesson 6-3 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Additional Examples (continued) Check 2c + 2 + 3c = 12 2(2) + 2 + 3(2) 12 12 = 12 Substitute 2 for c. The solution checks. Quick Check Lesson Main Lesson 6-3 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Additional Examples Eight cheerleaders set a goal of selling 424 boxes of cards to raise money. After two weeks, each cheerleader has sold 28 boxes. How many more boxes must each cheerleader sell? Words 8 cheerleaders • (28 boxes + additional boxes = 424 boxes Let x = the number of additional boxes. Equation 8 • (28 + x) = 424 8(28 + x) = 424 Lesson Main Lesson 6-3 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Additional Examples Quick Check (continued) 224 + 8x = 424 Distributive Property 224 – 224 + 8x = 424 – 224 Subtract 224 from each side. 8x = 200 Simplify. 8x 200 = 8 8 Divide each side by 8. x = 25 Simplify. Each cheerleader must sell 25 more boxes. Check for Reasonableness Round 8 to 10 and 28 to 20. The cheerleaders sold about 10 • 20, or 200 boxes. This means each cheerleader must sell about 22 more boxes. 25 is close to 22. The answer is reasonable. Lesson Main Lesson 6-3 Feature Solving Multi-Step Equations LESSON 6-3 Course 3 Lesson Quiz Solve the following equations. 1. 2m + 4 – 8m = 28 m = –4 2. 2(f – 1) + f = 37 f = 13 3. 4.5(4x – 12) = 144 x = 11 4. Jasmine earns a certain amount per hour for the first 40 hours worked in a week. Each hour she works over 40 in one week, she earns an additional $4.50 per hour. If Jasmine works 46 hours one week, and earned $383.50 that week, how much does she earn per hour for the first 40 hours? $7.75 per hour Lesson Main Lesson 6-3 Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Problem of the Day Audra drinks 24.75 oz of milk in 1 day. How many ounces of milk does she drink in a week? 173.25 oz Lesson Main Lesson 6-4 Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Check Skills You’ll Need (For help, go to Lesson 6-2.) 1. Vocabulary Review Operations that undo each other are called Simplify. 2. 9(t + 7) – 16 3. 12 – 6(2r – 8) 4. 2x – (5x + 7) Check Skills You’ll Need Lesson Main Lesson 6-4 Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Check Skills You’ll Need Solutions 1. inverse operations 2. 9t + 47 3. 60 – 12r 4. – 3x – 7 Lesson Main Lesson 6-4 Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Additional Examples Solve 9 + 2p = –3 – 4p. 9 + 2p = –3 – 4p 9 + 2p + 4p = –3 – 4p + 4p 9 + 6p = –3 9 – 9 + 6p = –3 – 9 Add 4p to each side. Combine like terms. Subtract 9 from each side. 6p = –12 Simplify. 6p –12 = 6 6 Divide each side by 6. p = –2 Lesson Main Simplify. Lesson 6-4 Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Additional Examples (continued) Check 9 + 2p = –3 – 4p 9 + 2(–2) –3 – 4(–2) 5=5 Substitute –2 for p. The solution checks. Quick Check Lesson Main Lesson 6-4 Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Additional Examples Each week you set aside $18 for a stereo and put the remainder in a savings account. After 7 weeks, the amount you place in the savings account is 4.2 times your total weekly pay. How much do you make each week? Words 7 (weekly amount – 18) = 4.2 Weekly amount Let x = the amount earned weekly. 7 Lesson Main Lesson 6-4 (x - 18) = 4.2 x Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Additional Examples (continued) 7(x – 18) = 4.2x 7x – 126 = 4.2x 7x – 7x – 126 = 4.2x – 7x Distributive Property Subtract 7x from each side. – 126 = – 2.8x Combine like terms. – 126 – 2.8x = – 2.8 – 2.8 45 = x Divide each side by – 2.8. Simplify. Quick Check Lesson Main Lesson 6-4 Feature Solving Equations With Variables on Both Sides LESSON 6-4 Course 3 Lesson Quiz Solve each equation. 1. 4(3u – 1) = 20. u=2 2. 5t – 4 = t – 8. t = –1 3. 3(k – 8) = – k. k=6 Lesson Main Lesson 6-4 Feature Solving Inequalities by Adding or Subtracting LESSON 6-5 Course 3 Problem of the Day If 1.5 truckloads of dirt cost $39.00, what is the cost of 1 truckload? $26.00 Lesson Main Lesson 6-5 Feature Solving Inequalities by Adding or Subtracting LESSON 6-5 Course 3 Check Skills You’ll Need (For help, go to Lesson 1-6.) 1. Vocabulary Review A(n) ? is a mathematical sentence with an equal sign. Solve each equation. 2. x + 15 = – 3 3. y + 22 = 9 4. a – 28 = – 4 Check Skills You’ll Need Lesson Main Lesson 6-5 Feature Solving Inequalities by Adding or Subtracting LESSON 6-5 Course 3 Check Skills You’ll Need Solutions 1. equation 2. x + 15 = – 3 x + 15 – 15 = – 3 – 15 x = – 18 4. a – 28 = – 4 a – 28 + 28 = – 4 + 28 a = 24 Lesson Main 3. Lesson 6-5 y + 22 = 9 y + 22 – 22 = 9 –22 y = – 13 Feature Solving Inequalities by Adding or Subtracting LESSON 6-5 Course 3 Additional Examples Solve p – 3 < –5. p – 3 < –5 p – 3 + 3 < –5 + 3 p < –2 Notice that 3 is subtracted from p. Add 3 to each side. Simplify. Check Step 1 See if –2 is a solution to the related equation. p – 3 = –5 –2 – 3 –5 –5 = –5 Lesson Main Write the related equation. Substitute –2 for p. Simplify. Lesson 6-5 Feature Solving Inequalities by Adding or Subtracting LESSON 6-5 Course 3 Additional Examples (continued) Step 2 Check the inequality symbol. Choose any number less than –2 and substitute it into the original inequality. In this case, try –3. p – 3 < –5 –3 – 3 < –5 Substitute –3 for p. –6 < –5 Steps 1 and 2 both check, so p < –2 is the solution of p – 3 < –5. Quick Check Lesson Main Lesson 6-5 Feature Solving Inequalities by Adding or Subtracting LESSON 6-5 Course 3 Additional Examples After the hairdresser cut 3 in. from Rapunzel’s hair, her hair was at least 15 in. long. How long was her hair before she had it cut? Words length before cut – length cut is at least 15 in. Let l = the length before cut. Inequality l–3 > – 15 l–3+3 > – 15 + 3 l > – 18 l – 3 > – 15 Notice that 3 is subtracted from l. Add 3 to each side. Simplify. Rapunzel’s hair was at least 18 in. long before she had it cut. Lesson Main Lesson 6-5 Quick Check Feature Solving Inequalities by Adding or Subtracting LESSON 6-5 Course 3 Lesson Quiz 1. Solve r – 9 > 8. r > 17 2. Solve and graph the inequality: x + 6 > 4. x > –2 3. Solve and graph the inequality: x – 14 < – 13. x<1 4. A lamp can use lightbulbs of up to 75 watts. The lamp is using a 45-watt bulb. At most, how many watts are available for brighter light? 30 watts Lesson Main Lesson 6-5 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Problem of the Day What is the greatest common factor (GCF) of 60 and 72? 12 Lesson Main Lesson 6-6 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Check Skills You’ll Need (For help, go to Lesson 1-7.) 1. Vocabulary Review Is the definition “Negative numbers are numbers less than or equal to zero” correct? Explain. Solve each equation. 2. 4x = – 16 3. – 8p = 808 4. – 2u = – 12.4 t 5. 1 = –8 Check Skills You’ll Need Lesson Main Lesson 6-6 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Check Skills You’ll Need Solutions 1. No; zero is not a negative number. 2. 4x = –16 4x = –16 4 4 x= –4 5. 1= 3. –8p = 808 –8p = 808 8 –8 p = –101 4. –2u = –12.4 –2u –12.4 = –2 –2 u = 6.2 t –8 (–8) · 1 = (–8) · ( t ) –8 –8 = t Lesson Main Lesson 6-6 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Additional Examples A small business sells each CD of its game software for $12. How many CDs must they sell to meet the goal of at least $84,000? Words number of CDs • $12 is at least $84,000 Let c = the number of CDs. Inequality c • 12 > – 84,000 12c > – 84,000 12c > 84,000 12 – 12 c > – 7,000 Lesson Main Divide each side by 12. Simplify. Lesson 6-6 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Additional Examples (continued) The business must sell at least 7,000 CDs. Check for Reasonableness The answer makes sense because 7,000 • 12 is 84,000, and any number over 7,000 multiplied by 12 is a number greater than 84,000. Quick Check Lesson Main Lesson 6-6 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Additional Examples Solve y > 3. Graph the solution. –4 y >3 –4 –4• y –4 <–4•3 y < –12 Multiply each side by – 4. Reverse the direction of the inequality. Simplify. Quick Check Lesson Main Lesson 6-6 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Additional Examples Solve –5b < – 15. Graph the solution. –5b < – 15 –5b > 15 –5 – –5 b > – –3 Divide each side by –5. Reverse the direction of the inequality. Simplify. Quick Check Lesson Main Lesson 6-6 Feature Solving Inequalities by Multiplying or Dividing LESSON 6-6 Course 3 Lesson Quiz Solve each inequality. 2. – 7r ≥ 14 1. 9c ≤ –36 c –4 3. r –2 a <–1 –8 a>8 4. Leroy’s class is having a cookie sale to raise money for a class trip. They will sell homemade cookies for $1.50 each. If they earn at least $200, a local business will match the amount they earn. At least how many cookies do they need to sell to get the local company to match their cookie-sale earnings? 134 cookies Lesson Main Lesson 6-6 Feature