ALGEBRA 1 LESSON 10-2 (For help, go to Lessons 1-4, 1-5, and 2-1.) Complete each statement with. 1.–3 + 4 – –3 – – – Solve each equation. 4.x – 4 = 55.n – 3 = –5 6.t + 4 = –57.k + = Solving Inequalities Using Addition and Subtraction 10-2

ALGEBRA 1 LESSON –3 + 4 – > –1 2. –3 – 6 4 – 6 –9 < –2 3. – – –1.4 > – x – 4 = 5 x = 9 5.n – 3 = –5 n = –2 6.t + 4 = –5 t = –9 Solutions Solving Inequalities Using Addition and Subtraction k + = k = – = – =

ALGEBRA 1 LESSON 10-2 Solve p – 4 < 1. Graph the solutions. p – < 1 + 4Add 4 to each side. p < 5Simplify. Solving Inequalities Using Addition and Subtraction 10-2

ALGEBRA 1 LESSON 10-2 Solve 8 d – 2. Graph and check your solution. > d – 2 + 2Add 2 to each side. > 10 d, or d 10Simplify. >< Check: 8 = d – 2Check the computation – 2Substitute 10 for d. 8 = 8 8 d – 2Check the direction of the inequality. 8 9 – 2Substitute 9 for d. 8 7 > > > Solving Inequalities Using Addition and Subtraction 10-2

ALGEBRA 1 LESSON 10-2 Solve c + 4 > 7. Graph the solutions. c + 4 – 4 > 7 – 4Subtract 4 from each side. c > 3Simplify. Solving Inequalities Using Addition and Subtraction 10-2

ALGEBRA 1 LESSON 10-2 In order to receive a B in your literature class, you must earn more than 350 points of reading credits. Last week you earned 120 points. This week you earned 90 points. How many more points must you earn to receive a B? Solving Inequalities Using Addition and Subtraction 10-2 points earnedrequired points needed Relate:plus is more than Define:Let = the number of points needed.p Write: p >

ALGEBRA 1 LESSON 10-2 (continued) You must earn 141 more points p > 350Combine like terms p > p – 210 > 350 – 210Subtract 210 from each side. p > 140Simplify. Solving Inequalities Using Addition and Subtraction 10-2

ALGEBRA 1 LESSON 10-2 Solve each inequality. Graph the solutions. 1.p – 7 –52.w – 3 < –9 3.x + 6 > h > > p 2 > w < –6 x > –2 < 4 h, or h 4 > Solving Inequalities Using Addition and Subtraction 10-2

(For help, go to Lessons 2-1 and 3-1.) ALGEBRA 1 LESSON 10-2 Solve each equation. 1.8 = t2.14 = –21x3. = –1 4.5d = 325. x = – n = 9 Write an inequality for each graph x6x Solving Inequalities Using Multiplication and Division 10-2

ALGEBRA 1 LESSON 10-2 Solutions 1. 8 = t t = 8 t = 8(2) = = –21x –21x = 14 x = – = – = –1 x = –1(6) = –6 x6x6 4. 5d = 32 d = = x = –12 x = –12 x = – n = 9 = n = n x –1 8.x > 3 < Solving Inequalities Using Multiplication and Division 10-2

ALGEBRA 1 LESSON 10-2 Solve > –2. Graph and check the solutions. z3z3 z > –6Simplify each side. 3 > 3(–2)Multiply each side by 3. Do not reverse the inequality symbol. z3z3 ( ) z3z3 > –2Check the direction of the inequality. Check: = –2 Check the computation. z3z3 – = –2Substitute –6 for z –2 = –2Simplify. – > –2Substitute –3 for z –1 > –2Simplify. Solving Inequalities Using Multiplication and Division 10-2

ALGEBRA 1 LESSON 10-2 Solve 3 – x. Graph and check the solutions. < 3535 ( ) 5353 – 5353 – (3) > ( ) 3535 x Multiply each side by the reciprocal of –, which is –, and reverse the inequality symbol –5 x, or x –5Simplify. <> Solving Inequalities Using Multiplication and Division 10-2

ALGEBRA 1 LESSON 10-2 (continued) Check: 3 = – xCheck the computation. 3 = – (–5) Substitute –5 for x. 3 = 3 3 – xCheck the direction of the inequality. 3 – (–10)Substitute –10 for x < < < Solving Inequalities Using Multiplication and Division 10-2

ALGEBRA 1 LESSON 10-2 Solve –4c < 24. Graph the solutions. c > –6Simplify. Divide each side by –4. Reverse the inequality symbol. –4c –4 > 24 –4 Solving Inequalities Using Multiplication and Division 10-2

ALGEBRA 1 LESSON 10-2 Your family budgets $160 to spend on fuel for a trip. How many times can they fill the car’s gas tank if it cost $25 each time? Your family can fill the car’s tank at most 6 times. cost per total fuel tankbudget Relate:times number of tanks is at most Define:Let = the number of tanks of gas. t Write:25160t < 25t 160 < Solving Inequalities Using Multiplication and Division 10-2 Divide each side by 25. < 25t t 6.4Simplify. <

ALGEBRA 1 LESSON 10-2 Solve each inequality. Graph the solution. 1. –3 2.– < –1 3.6x < –12h y2y2 > p3p3 > y –6 > p > 3 x < 5 > < –4 h, or h –4 Solving Inequalities Using Multiplication and Division 10-2

ALGEBRA 1 LESSON 10-2 (For help, go to Lessons 2-2 and 2-3.) Solve each equation, if possible. If the equation is an identity or if it has no solution, write identity or no solution. 1.3(c + 4) = 62.3t + 6 = 3(t – 2) 3.5p + 9 = 2p – 14.7n + 4 – 5n = 2(n + 2) 5. k – + k = 6.2t – 32 = 5t + 1 Find the missing dimension of each rectangle Solving Multi-Step Inequalities 10-2

ALGEBRA 1 LESSON (c + 4) = 62.3t + 6 = 3(t – 2) c + 4 = 23t + 6 = 3t – 6 c = –2 6 = –6 no solution 3. 5p + 9 = 2p – 14.7n + 4 – 5n = 2(n + 2) 3p = –10 2n + 4 = 2n + 4 p = –3identity 1313 Solving Multi-Step Inequalities Solutions 10-2

ALGEBRA 1 LESSON k – + k = 6.2t – 32 = 5t + 1 k = –3t = 33 k = = 1 t = –11 7. P = 2( + w)8. P = 2( + w) 110 = 2( + 15)78 = 2(26 + w) 55 = = 26 + w 40 = 13 = w length = 40 cmwidth = 13 in Solutions Solving Multi-Step Inequalities 10-2

ALGEBRA 1 LESSON 10-2 Solve 5 + 4b < b – 5 < 21 – 5Subtract 5 from each side. 4b < 16Simplify. b < 4Simplify. < Divide each side by 4. 4b44b Check: 5 + 4b = 21Check the computation b < 21Check the direction of the inequality (3) < 21Substitute 3 for b (4) 21Substitute 4 for b. 21 = < 21 Solving Multi-Step Inequalities 10-2

ALGEBRA 1 LESSON 10-2 The band is making a rectangular banner that is 20 feet long with trim around the edges. What are the possible widths the banner can be if there is no more than 48 feet of trim? twice the the length lengthof trim Relate:plus twice the width can be no more than Write:2(20) + 2w 48 < Solving Multi-Step Inequalities 10-2

ALGEBRA 1 LESSON 10-2 (continued) The banner’s width must be 4 feet or less. 2(20) + 2w 48 < w 48Simplify 2(20). < w – – 40Subtract 40 from each side. < 2w 8Simplify. < w 4Simplify. < Divide each side by 2. < 2w22w Solving Multi-Step Inequalities 10-2

ALGEBRA 1 LESSON 10-2 Solve 3x + 4(6 – x) < 2. 3x + 24 – 4x < 2Use the Distributive Property. –x + 24 < 2Combine like terms. –x + 24 – 24 < 2 – 24Subtract 24 from each side. –x < –22Simplify. x > 22Simplify. > Divide each side by –1. Reverse the inequality symbol. –x –1 –22 –1 Solving Multi-Step Inequalities 10-2

Solve 8z – 6 < 3z ALGEBRA 1 LESSON z – 6 – 3z < 3z + 12 – 3zSubtract 3z from each side. 5z – 6 < 12Combine like terms. 5z – < Add 6 to each side. 5z < 18Simplify. < Divide each side by 5. 5z55z z < 3Simplify Solving Multi-Step Inequalities 10-2

ALGEBRA 1 LESSON 10-2 Solve 5(–3 + d) 3(3d – 2). < –15 – 4d + 15 –6 + 15Add 15 to each side. < –4d 9Simplify. < –15 + 5d – 9d 9d – 6 – 9dSubtract 9d from each side. < –15 – 4d –6Combine like terms. < –15 + 5d 9d – 6Use the Distributive Property. < Divide each side by –4. Reverse the inequality symbol. –4d –4 9 –4 > d –2Simplify. > 1414 Solving Multi-Step Inequalities 10-2

ALGEBRA 1 LESSON 10-2 Solve each inequality a 232.– p < p – 6 3.3(x – 4) > 4x (3c + 2) 2(3c – 2) > < a 3 > p > x < –19 < c – Solving Multi-Step Inequalities 10-2