CHAPTER Inequalities Solutions Key Are You Ready?. B. E. F. D. C 6. b - a = 6 - = 7. ab = ()(6) = 9. a + b = + 6 = 8 8. b a = 6 =. <..7 >.. % =.. <. 6x + x = 7x. -8a + a = -a 7..x +.x = 6.x 9. ( - d) = () + (-d) = - d. ( + m) = () + (m) = + m. -7x = -7x -7 = -7 x = -. h = 6 () h = ()6 h = 7. 6x = 6x 6 = 6 x = 7 9. - y = - () y = ()(-) y = -6 6. 9 x - x = -6 x 8. (x + ) = (x) + () = x + 6. (r - ) = (r) + (-) = r -. s - = 8 + + s =. y + = - - y = -9 6. t + = - - - t = - 8. r - 8 = - + 8 +8 r = - Graphing and writing inequalities Check it out!. all real numbers greater than a. b. - w - w w w - - -. x <. c. -6 - - - - -. Let d represent the amount the employee can earn per hour. 8. d 8. 6 8 think and discuss. Both graphs include the real numbers greater than. The graph of x also includes.. Inequality Graph x > x - Exercises guided practice - - - - - -. A solution of an inequality makes the inequality true when substituted for the variable.. all real numbers greater than or equal to. all real numbers greater than -. all real numbers less than. all real numbers greater than or equal to 6. 7. -6 - - - - - 8. ( - ) > m > m 8 > m m < 8 6 8 9. p ÇÇÇ 7 + 8 p ÇÇ p. a -. b > -8. c <.. d < -7. e. f 6 6. Let m represent the number of members present. m, where m is a whole number 8 9 7. Let r represent the athlete s heart rate. r <, where r is positive 8 6 Holt McDougal Algebra
practice and problem solving 8. all real numbers less than 9. all real numbers less than. all real numbers greater than. all real numbers less than or equal to. 6 7 8 9. d > ( - 8) d > (-) d > - - -9-6 -. - - - -. t - t 9 - t 6. u 7. v < - 8. w > -. 9. x > -.. y <. z 9 6. Let s represent the speed allowed in miles per hour. s, where s is nonnegative. Let y represent the number of years of experience. y 6. x is greater than 7.. h is less than -. 6. d is less than or equal to. 7. r is greater than or equal to -. 8. g > 9 7 8 9. e > 9. p 7. f > 6 7 8 9 - - -. Let t represent the temperature on Earth in F. t.9.9. Let p represent the profits. p <,,. Let h represent the height in inches. h 6 6 7 8. Let e represent the elevation in feet. e 6. Possible answer: x represents the distance in miles between two locations. 7. Possible answer: x represents the age in years of a child at a childcare center where x is positive. 8. Possible answer: 9. Possible answer: x represents the hour on x represents the number an analog clock when of millions of albums x is a natural number. sold by a popular band.. A. D. B. C. A is incorrect; it should be drawn with an empty circle. a. Let s represent the amount she can spend. s - 9 s, where s is nonnegative b. c. s - s where s is nonnegative 6. 6 7. Look for a solid or empty circle. A solid circle tells you to use or, and an empty circle tells you to use < or >. Then look at the direction of the arrow. An arrow pointing left tells you to use < or, and an arrow pointing right tells you to use > or. 8a. less than or equal to b. greater than or equal to test prep 9. D; Since - () = - = - and - -, is not a solution of the inequality - x -. 6. F; Since - = and, is not a solution of the inequality - x <. 6. C; Try t = -. Since - (-) = 6 and - 6, t = - is a solution so it should be shaded. Try t =. Since - = and -, t = is a solution so it should also be shaded. Therefore C is correct. challenge and extend 6. all real numbers 6. any nonzero numbers such that x and y are the same sign and x < y 6. any numbers such that y is less than or equal to x 6. < 66. > 67. any number between. and.7 68. yes; infinitely many 69. Draw an empty circle at. Then draw arrows going left and right from. - solving inequalities by adding or subtracting Check it out! a. s + - - s + 9 s 9 6 9 b. > - + t + + > + t t < 6 Holt McDougal Algebra
c. q -. < 7. +. +. q - < q < 9. Let m represent the number of milligrams of iron. + m - - m, where m is nonnegative Sarah can consume mg or less without exceeding the RDA.. Let p represent the number of more pounds he needs to break the school record. + p > 8 - - p > Josh needs to bench press more than additional pounds to break the school record. Think and discuss. Substitute the endpoint into the related equation. Then substitute a value in the solution region into the original inequality. d - = -6 d - > -6 (-) - -6 () - > -6-6 -6 - > -6. You can add or subtract the same number on both sides of an equation or inequality, and the statement will still be true.. Properties of Inequality Addition: a - < 9 + + a < Subtraction: a + > - - a > 6 7 6. Let d represent the additional amount you need to spend for the restaurant to deliver. 7.9 + d. -7.9-7.9 d 7. You must spend at least $7. more for the restaurant to deliver. practice and problem solving 7. a - 8.. > q -.8 + + +.8 +.8 a. > q q <. 6. 9. - + x < - + + x <. r + - - r -. + + w 6 + w -6-6 w 7 However, weight is nonnegative. - 7 The crate cannot weigh more than 7 pounds.. 7 + g -7-7 g 8 where g is nonnegative Mindy can add no more than 8 gallons to the tank.. x - > + + x >. n + 6-6 -6 n - exercises guided practice. < p + 6-6 - 6 6 < p p > 6 6 8. - + x - + + x - - - - -. w + - - w - - -. z - > - + + z > -9. + t - - t where t is nonnegative - -9-6 - - - - The temperature can increase no more than F.. r - + + r 8 6 7 8 9 7. - + q > 9 + + q > 9 9..8 p + - -.8 p p.8.8.6 - - - 6. x + - - x - - - - 8. x + < 7 - - x < 6 6 6 6 6. - x - + + x x - - - 6 7 Holt McDougal Algebra
. < 6 + c -6-6 - < c c > -. y - > + + y > c. 8 + m -8-8 m 78, where m is nonnegative. 78 - - - -. x +.. -. -. x - - - a. 8 Possible answer: Washington is about 8 miles away. Daryl is willing to drive miles. So he can drive up to - 8 = miles in Washington. Therefore 78 is reasonable. - - - b. c.. Let d represent the number of days he can still wear his contact lenses. + d - - d 9 where d is nonnegative 6 9 Alex can wear his contact lenses for up to 9 more days. 6. x - + + x x ; C 8. x + 6 > 9-6 -6 x > ; A 7. 8 > x - (-) 8 > x + - - > x x < ; B 9. - x - 7 +7 + 7 x x ; D. It is a reasonable answer. If you round each number to the nearest whole number and then solve the inequality + x <, the solution is x <.. 96 + + p,6 9 + p,6-9 -9 p, where p is nonnegative There can be at most, people in the other types of seats.. Possible answer: If a scale is unbalanced and the same amount is added to or subtracted from both sides, the scale should maintain the same amount of imbalance.. When you isolate the variable in each inequality, you get x and x.. Both inequalities have all numbers greater than as solutions. x + also includes. The graph of x + > has an empty circle at, but the graph of x + has a solid circle at. a. + = 8 miles b. 8 + m test prep 6. A; If you round each number and then solve the inequality + p <, the solution is p <. 7. F; If x is the number of bottles Dave has, then x + represents the number of bottles Sam and Dave have, which is at most. Therefore, situation F represents x +. 8. A; Solving gives p < - which is graph A. 9. J; The solutions of n + 6 are n. However the solutions of n - are n. So J does not have the same soltuions of n + 6. challenge and extend. 6 9 + x 69 8 + x - 8-8 x x. 6 + m > 7 6 6 + m > 6-6 - 6 m > 6 m > - -. r - 7 r - 7 + + r r. Sometimes; when b is positive, the inequality is true.. Always; adding the same number to both sides keeps the statement true.. Always; a and c are the two greater numbers, so their sum will be greater than that of the two lesser numbers. 6. b and c are equal. Holt McDougal Algebra
- solving inequalities by multiplying or dividing Check it out! a. k > k > k > 6 c. ( 6 8 g > 7 g ) > (7) g > 6 6 7 8 b.. > -.h. -. < -.h -. -7 < h h > -7-9 -8-7 -6 - b. - q - q - q q - - - a. -x -() -(-x) - x x - - -. Let g represent the number of -ounce servings. g 8 g 8 g.8 Since only a whole number of glasses can be filled,,,,,,, 6, 7, 8, 9,,, or -ounce servings can be filled. think and discuss. They are alike because you can multiply or divide by a postive number on both sides and the inequality or equation will still be true. They are different because you have to reverse the inequality symbol if you multiply or divide by a negative number.. Solving Inequalities by Using Multiplication and Division Divide Multiply exercises guided practice. b > 7 b > 7 b > 9 By a positive number x < 8 x 8 < x < x x () () x 6 6 9 By a negative number -x > 9 -x 9 < - - x < - x - x (-) (-) -. - 8b - 8 8b b - b b - x - - - - -. d > 6 ( d ) > (6) d > 8 6 8..m..m... m.. 7. 9s > -8 9s 9 > -8 9 s > - - - -. 9. -x < - -x - > - - x >.. 6. -.n <. -.n -. >. -. n > -. -.6 -. -. -. -. d. -6 < (-6) d -6 > (-6) d > - -6 - - - - -. > t -6-6() < -6 ( t -6 ) -7 < t t > -7. d 6 d 6 d 6. 8.. ( k > 6 k ) > ( (6) k > 9 r 6 9 ) ( r 8 r r 8 6 7 ) 8 9 b - 8 - ( b - ) -(8) b -6-8 -7-6 - -. > -8g - < g g > - -8 < -8g -8 - -. -h -6 -h - -6 - h.6...6.8 6. - m -7 - ( - m ) -(-7) m 6 Holt McDougal Algebra
7. Let n represent the number of nights he can stay. 8n 8n 8 8 n 6.87 Tom can stay only a whole number of nights. So Tom can stay,,,,,, or 6 nights. practice and problem solving 8. < t < 9. t j < t ( j ) () t > j 6. -8 < 8c -8 8 < 8c 8 - < c c > -. - - - - w - ( w ) (-) w -8 - -8 -. 6y <. 6y 6 <. 6 y <.7 6..6.7.8.9 ( x x ) ( x ) 8. - > p - > p -. > p p < -. - -. - -. - 8 6. > d > d 7 > d d < 7. ( h 6 7 8 9 h 7 ) ( 7 ) h 8 7 8 7. c - c - c - - - - - - 7. 6b (6b) 6 6 ( ) b 9. b 8-8 ( b 8 ) 8(-) b -6 - - -6-8. -9a > 8-9a -9 < 8-9 a < -9 - -9-6 -. -6p >.6-6p -6 <.6-6 p < -. -. -. -. -.. - 6 f < -6 ( - 6 f ) > -6() f > - - - - - 6. -w - -w - -. w w -. 8. -.8 -.6 -. -. - < f - - ( ) > - ( f - ) - > f f < - -6 - - - - -. -.b < -7.7 -.b -. > -7.7 -. b >..... - ( < r - ) > - ( r - > r r < - - ) - - - - y - ( y - > - - ) < - ( - - - y < - ). -.t < -9 -.t -. > -9 -. t > 6 7. -z > -z - < - z < - - - - - -9 9. -k 7 -(-k) -(7) k -7-9 -8-7 -6 -. 6 - p - (6) - - p p - ( - -6 - -8 - p ). Let r represent the number of pieces of rope. 8r 8r 8 8 r Roz can only cut a whole number of pieces, so she can cut,,, or pieces of rope.. -8x < -8x -8 > -8 x > - -6 - - - - -. t t t 8 6 8 6 Holt McDougal Algebra
. x < x ) < () x < ( 7. -9p -9-9p -9-6 p p -6-8 -6 - - 9. - b > - - ( - b ) < - b < - - -. 7x 7x 7 7 x ( - ) 6. - b -6 - ( - b ) - ( -6) b 6. ( p - p ) p - ( -) - - - - 8. t > - (t) > - 6 ( - t > - 6-6 - 6. 6 >.6r 6.6 >.6r.6 6 > r r < 6. ) - 6 6 8 h -6-6 ( h -6 ) -6() h - - - - -. t () ( t ) t t 6 8. You reverse the symbol only when you mulitply or divide both sides of the inequality by the same negative number. 6. A lw l(.).l.. l 6 where l is positive The length of the rectangle is at most 6 inches. 7. -.t. -.t -.. -. t -; C 9. -. -.t -. -. -.t -. t t ; A 8. 6. t - 9 9 t ) 9(-) t -7; D t - -6-6) -6 (- ) t ; B 9 ( -6 ( t 6. Let b represent the number of bags. The shelter needs at least lb of cat chow per week. Since there are weeks in a year, the shelter needs at least = lb of cat chow per year. b b b 6 The shelter needs at least 6 bags of cat chow per year. 6. Let p represent the possible number of points a student could earn..9p 67.9p.9 67.9 p 6 where p is nonnegative A student could earn no more than 6 points. 6. Multiplying both sides of an inequality by zero makes both sides equal zero, so there is no longer an inequality to solve. 6. A is incorrect. Both sides are divided by a positive number, so the inequality symbol should not be reversed. 6. Let g represent the number of guests..g 8.g. 8. g 6 where g is nonnegative Jan can invite up to 6 guests. 66a. 7n b. 7n 7n 7 7 n. The club can reserve up to rooms. c. 6n 6n 6 6 n.8 No, they still can reserve only or fewer rooms. test prep 67. B; - y > has the solution y < -6 but y < - has the solution y < -. 68. G; The graph represents x - but the solution to -x is x -. 69. B; Since s represents the number of stamps, the total cost of the stamps is.9s and this must be less than or equal to. 7 Holt McDougal Algebra
7. Answers may vary. x > 8 x > 8 Divide both sides by. x > x > ( x ) > () Multiply both sides by. x > x > x > Divide both sides by. x > challenge and extend 7. - 6 g 6 - ( 7 ) - 6 ( - 6 g ) - g g - 7. m > 7 8 m > 7 8 8 ( 8 m ) > 8 ( 7 ) m > 7. ( x < 8. x ) < (8.) x <.7 7. f 8 f 7 8 ( 8 f ) 8 ( 7 ) f 7. Round to x. x x x So, the greatest possible integer solution is x = 76. yes; < and < < 77. no; < but ready to go on? Section A Quiz. all real numbers greater than -. all real numbers less than or equal to 8. all real numbers greater than or equal to. all real numbers greater than. - - - 7. g < ÇÇÇ 8 + g < Ç 9 g < 6 6. 8. h h 8 8 6. a <, where a is nonnegative. k + 7 - - k - - - 7. r - 8 - + 8 +8 r - -6 - - - - -. m, where m is nonnegative 6. > p - + + 7 > p p < 7 8. - + p < -6 + + p < - 6 7-6 - - - - - 9. Let b represent the number of gift baskets. 6 + b -6-6 b Allie must sell at least more baskets.. Let m represent the amount of money. 7. + m. -7. -7. m. Dante can spend $. at most.. -x < 8 -x - > 8 - x > -. - - - ( t t ) () t 6 8 6. d - ( d ) (-) d -9 - -9-6 -. 8 > -6c - < c c > - 8-6 < -6c -6 - - - - 8 9. Let r represent the number of ribbons. r 8 r 8 r.7 Since only a whole number of ribbons can be cut, Riley can cut,,,,, or strips of ribbon. 9. x -. y <. z -.. t, where t is a whole number 6 7 8 Holt McDougal Algebra
- solving two-step and multi-step inequalities Check it out! a. - x + 6-6 - 6-8 x -8 x -6 x x -6-8 -6 - - c. - n 7 ( - n ) (7) - n - - -n -n - - n - b. > x + - - ( x + - ) < -() x + < -6 - - x < - - - - - -9 a. m + > m + > - - m > m > m > 8 9 think and discuss. minimum value of v.. Mulitply both sides by, and then subtract from both sides.. Divide the left side to get x +, subtract from both sides, and then multiply both sides by.. How are they alike? To solve multi-step equations or inequalities, follow the order of operations to simplify the expressions on both sides of the equal sign or inequality symbol, and then undo each operation. Exercises guided practice. m + > - - m > m > m > 6 Solving Multi-Step Equations and Inequalities How are they different? When solving multi-step inequalities, you must reverse the inequality symbol if you multiply or divide both sides by a negative number. There are many solutions of an inequality but usually only one solution of an equation.. d + - - d - d - d - - - - -9-8 b. + (x + ) > + (x) + () > + x + 8 > x + > - - x > -8 x > -8 x > - -6 - - - - - c. 8 < x - 8 8) < 8 ( x - 8 ) 8) < 8 ( 8 x ) + 8 (- ) < x - + + 7 < x 8 ( 8 ( 7 < x < x x >. Let g represent the grade on the second test. 9 + g 9 ( 9 + g ) (9) 9 + g 8-9 -9 g 8 Jim must score at least 8 on the second test. 6 8. 6 -x + - - -x - -x - - x x - - - - + x. > - ( + x ) > (-) + x > - - - x > -6 - - -6-8 7. - x 7 ( - x ) (7) - x - - -x 8 -x - 8 - x -9-6 - - - - -. c - 7 > + 7 +7 c > c > c > 6 6. <.x -.7 +.7 +.7.7 <.x.7. <.x. 8. < x x > 8. 8. x + - - x - x - x - 6 8. 9 - - - - - -9-6 - 9 Holt McDougal Algebra
9. (x + ) > 6 (x) + () > 6 x + 8 > 6-8 -8 x > - x > - x > - - - - -. - x + 6 - x + 6 - x - - -x -9 -(-x) -(-9) x 9 7 8 9..(x - ) > -.8.(x) +.(-) > -.8.x - > -.8 + +.x >..x. >.. x >. x + < - - x < ( x ) < ( ) x <. - x > ( - ) - x > () - x > 6 - - -x > -(-x) < -() x < - - - - practice and problem solving 6. r - 9 > 7 + 9 +9 r > 6 r > 6 r > w + 8. > 6 ( w + ) > (6) w + > - - w > 9 6 6 9. 9 v + - - 6 v (6) ( v ) v v 7. - x - - - -x - - -x - x x - - - 9. w + 99 < 77-99 -99 w < - w < - w < - - - -. -x - 8 > 6 + 8 +8 -x > -x - < - x < -6-8 -6 - - - - -. (j + ) (j) + () j + - - j -88 j -88 j -9-9 - -9-8. Let x represent the amount of sales. +.x > - -.x > 9.x. > 9. x > 9 The sales representative will make more money with the first plan with sales of more than $9. 8 6. 8 - z -8-8 - z -6 - ( - z ) - ( -6) z 9. n - 8 ( n - 8 ) () n - 8 + 8 +8 n 8 n 8 n 6 6 9 6 8. f + < - - - f < -. - -. - -. -. - > - - w + + > -w - < -w - < w w > - - - Holt McDougal Algebra
6. > - p - p () > ( > - p - - > -p - < -p - - < p p > - ) -6 - - - - - 8. (x + ) > - (x) + () > - x + > - - - x > -6 x > -6 x > -9 - -9-6 -. -8 - h - - - -h - - -h - 7 h h 7 6 7 8 9. - > 7x - (x - ) - > 7x - (x) - (-) - > 7x - x + 8 - > x + 8-8 - 8 - > x - > x - > x x < - - - -. 9 - (9 ) > x - x 9-8 > x - x -7 > 9x -7 9 > 9x 9-8 > x x < -8-6 - -8-7. v + > - - v > (v) > v > ( ) 9. > x - (x + ) > x - (x) - () > x - x - 6 > -x - 6 +6 + 6 > -x - < -x - - < x x > - -6 - - - - -. a - (- ) a - 9 + 9 +9 a a a 9. 6 - x + > 7 - x > -7-7 - x > - 9 - ( - x ) < - 9 (- ) x < 9 7 8 9. (x - ) + x > 6 (x) + (-) + x > 6 x - 6 + x > 6 x - 6 > 6 + 6 +6 x > x > 6 x > 6. 9 + (q - 8) 9 + (q) + (-8) 9 + q - 6 q - 7 +7 + 7 q q 6 q q 6 8 6 7. Let x represent the number of minutes. 9.99 < 9.99 +.x -9.99-9.99. <.x.. <.x. 8.6 < x x > 8.6 The second company s plan costs more starting at 9 minutes. 8. - > -x - 8 +8 + 8 - > -x - - < -x - < x x > - - - 9. x + - - x x x - - - Holt McDougal Algebra
. x - > 7 + + x > ( x ) > () x > 8 6 7 8 9. - x - > - x > - - -x > -(-x) < -() x <. - - - 6 - m - 6-6 - m m ) - - ( - m - 8 ( ). x - x > - -x > -8 -x - < -8 - x < 6. < x - (x + ) < x - (x) - () < x - x - < -x - + + 8 < -x 8 - > -x - -6 > x x < -6-8 -6 - - 8. - - n + n > 7-9. - - n + n > 7 - n - > + + n > 7 n > 7 n >... ( p - ) 6 - ( p - ) ( ( p - ) ) () p - 8 + + p 8 6 8 a. - b. - c. - d. e. - f.. x + 9 <. 6 - x - 9-9 - - -x x < ( x - ) -x - < () - x x < 8 x - 6 8 6. (x + ) 6 (x) + () 6 x + 8 6-8 -8 x - x - x -8 - - -. x + x < 7 x < 6 6 7 ( 7 6 x ) < 6 7 () x < 8 6 - - 8 - - 8. - (r - ) > - - (r) - (-) > - - r + > - 6 - r > - -6-6 -r > -8 -(-r) < -(-8) r < 8 8 6.. 6.. -.n +. 7. 6 > (x + ). -.n. 6 > (x) + () -. -. 6 > x + 8 -.n -.9 -.n -. 6 > x + 8 -.9 -. -8-8 n.8 8 > x.8 8 > x 7 > x x < 7-6 - -8 -. x - 9 7 + 9 +9 x 6 x 6 x ; B 7. -x - 6 - + -x - 6 - + 6 +6 -x -x - - x -; A 6. -6 (x - ) -6 (x) + (-) -6 x - 6 +6 + 6 x x x x ; D 8. - + ) - x - x - - - x - ( - x ( x ) - ( ) x - ; C 6 7 8 9 Holt McDougal Algebra
9. Let m represent the number of months. + < 7 + m 6 < 7 + m -7-7 < m < m. < m m >. A consumer will pay less for the machine at Easy Electronics for months or more. 6a. ()(x + ) < b. (x + ) < (x) + () < x + < - - x < 7. x < 7. x < 9. c. h = x + h (9.) + h 9 + h The height of the triangle is less than or equal to in. 6a. Number Process Cost b. C = + n + + () 6 + () 9 + () 8 n + (n) + (n) c. C + n - - n n n They can make or fewer CDs. 6. r -.9 >.9 +.9 +.9 r > 9.8 r > 9.8 r >.96 r = 6.. Divide both sides by, and then subtract from both sides.. Distribute on the left side, subtract 6 from both sides, then divide both sides by. test prep 6. A; Substituting 6 for y gives the inequality 8 > x +. Solving for x gives 7 > x or x < 7. 6. G; If x represents the number of tickets, the price of all the tickets is.x. The total cost is + 6 +.x and this must be less than or equal to. 66. B; Since the > sign is used, and since only situation B involves less than, B must be correct. 67. 9 Let x represent the number of points scored in the second game. (x - 8) + x + > x + > - - x > 6 x > 6 x > 8 The team must have scored at least 9 points. challenge and extend 68. (x + ) - 6x + 6 (x) + () - 6x + 6 x + 6-6x + 6 - x - - -x - -x - - - x 6 69. -8 > -(x + 9) - + x -8 > -(x) - (9) - + x -8 > -x - 9 - + x -8 > -x - + + - > -x -(-) < -(-x) < x x > 6 + x 7. - (x - ) > + x - (x) - (-) > + x - x + > - - - - x > - - - x > - x ) - ( - < -(-) x < Holt McDougal Algebra
7. x > - - - 7. x - - - 7. -x > -x - < - x < 7. x < 7. x - - - 76. -x > -(-x) < -() x < - b..x -. +.9x <.x + 6.x -. <.x + 6 -.x -.x.x -. < 6 +. +..x < 6..x. < 6.. x < 6 - - - - - - - solving inequalities with variables on both sides check it out! a. x 7x + 6 -x -x x + 6-6 - 6-6 x -6 x - x x - - - - b. t + < -t - 6 +t +t 7t + < -6 - - 7t < -7 7t 7 < -7 7 t < - - - -. Let f represent the number of flyers. +.f <.f -.f -.f <.f. <.f. 6 < f f > 6 The cost at A-Plus Advertising is less than the cost at Print and More if more than 6 flyers are printed. a. ( - r) (r - ) () + (-r) (r) + (-) - r r - 6 + r +r 8r - 6 +6 + 6 6 8r 6 8 8r 8 r r - - - a. (y - ) y + (y) + (-) y + y - y + -y -y - 7 no solutions b. x - < x + -x -x - < all real numbers think and discuss. Subtract c from both sides of the inequality so that all variable terms are on the right side. Then subtract from both sides so that all constant terms are on the left side.. Solutions of Inequalities with Variables on Both Sides All real numbers x + < x + Exercises guided practice. x > x - 6 -x -x -x > -6 -x - < -6 - x < No solutions x + > x + 9 6. 7x + > 8x - 9-7x -7x > x - 9 +9 + 9 6 > x 6 > x > x x <. 7y + y - -y -y 6y + - - - 6y -6 6y 6-6 6 y - - - - - - - Holt McDougal Algebra
. -r < - r +r + r -r < -r - > - r > - -6 - - - - - 6..x -.8.x -. -.x -.x.x -.8 -. +.8 +.8.x..x... x. c - > 8c + -c -c - > c + - - -6 > c -6 > c - > c c < - - - -. (6 - x) < x (6) + (-x) < x - x < x + x +x < 6x 6 < 6x 6 < x x > - - -. f + f - f - f - f - ( ) - (- f ) - f f -. x > (7 - x) x > (7) + (-x) x > - x +x + x 7x > 7x 7 > 7 x > 6 7. Let p represent the number of pizzas. + p < 7p - p -p < p < p. < p p >. They will have to sell at least pizzas to make a profit. 8. ( + x) ( + x) () + (x) () + (x) + x 6 + x - x - x + x 6 - - x - x - x -7-9 -8-7 -6-9. -( - p) > (p + ) -() - (-p) > (p) + () - + p > p + - p -p - > p + - - -7 > p p < -7-9 -8-7 -6 - -6 - - - - -. -6.7 +.6t <.t -.6t -.6t -6.7 < -.t -6.7 -. > -.t -. 6.8 > t t < 6.8 6 6. 6. 6.6 6.8. (x - ) -( - x) (x) + (-) -() - (-x) x - - + x -x - x - - all real numbers. (y + ) < y + 6. v + < v - 7 (y) + () < y + -v -v y + < y + < -7 7 -y -y no solutions < 7 no solutions 7. b - b - 6 -b -b - -6 all real numbers 9. k + 7 (k + ) k + 7 (k) + () k + 7 k + 8 -k -k 7 8 7 no solutions 8. (x - ) > x (x) + (-) > x x - > x -x -x - > 7 no solutions Holt McDougal Algebra
practice and problem solving. x x + 8 -x -x -x 8 -x - 8 - x - -6 - - - - -..x -. <.x -.8 -.x -.x -. <.6x -.8 +.8 +.8.6 <.6x.6.6 <.6x.6 < x x > - - -. 7 + b b - b -b 7 + b -7-7 b -7-9 -8-7 -6 -..8m -. >.8m +.8 -.8m -.8m.m -. >.8 +. +..m >..m. >.. m > 6 6. lw > bh (x + ) > ( x + 6)() (x + ) > (x + 6) (x) + () > (x) + (6) x + > x + 8 -x -x 7x + > 8 - - 7x > 6 7x 7 > 6 7 x > 8. 9y + > y - 7 -y -y y + > -7 - - y > - y > - y > - - - -. 7 - t < t - + t +t 7 < 9t - + + 9 < 9t 9 - - - 9 < 9t 9 < t t > 7. ( - x) (x - ) () + (-x) (x) + (-) 8 - x x - + x +x 8 9x - + + 8 9x 8 9 9x 9 x x - - - 8. -(n + ) < 6( - n) -(n) - () < 6() + 6(-n) -n - < 6-6n +6n +6n n - < 6 + + n < 8 n < 8 n < 6 6 8 9. 9(w + ) w 9(w) + 9() w 9w + 8 w -9w -9w 8 w 8 w 6 w w 6. 6 8 r + r - r - r -6 ( - 6 r - 6 r ) - r r - ) -6 ( -6 - - - - -. ( - x) < -(x - ) () + (-x) < -(x) - (-) 6 - x < -x + + x +x 6 < 7 no solutions. 7 - y > - y + y + y 7 > all real numbers.. +.t >.8t - -.t -.t. >.t - + + 7. >.t 7.. >.t. > t t < 6. ( - n) < n - 7 () + (-n) < n - 7 8 - n < n - 7 + n +n 8 < n - 7 +7 + 7 < n < n < n n > 6 6 Holt McDougal Algebra
. ( + z) z + 6 () + (z) z + 6 + z z + 6 - z -z 6 all real numbers 6. -(k - ) ( - k) -(k) - (-) () + (-k) -k + - k +k + k 7 no solutions 7. (x - ) x (x) + (-) x x - x -x -x - all real numbers 9. t - > t + -t -t - > t + - - - > t - > t -7 > t t < -7-9 -8-7 -6 -. -(y + ) - 6 < y + -(y) - () - 6 < y + -y - - 6 < y + -y - < y + +y +y - < 6y + - - - < 6y - 6 < 6y 6 - < y y > - -6 - - - - -. x + 9 - x < x 9 - x < x + x +x 9 < x 9 < x < x x > 6 8. (v - 9) + v (v) + (-9) + v v - 7 + v -v - v -7 7 no solutions. 8 + 9p > p - - 9p -9p 8 > p - + + 9 > p 9 > p 6 > p p < 6. (x - ) < -x (x) + (-) < -x x - < -x -x -x - < -x - - > -x - > x x < - - -. - x - x + x + x x ( x ) ( x. -(x - 7) - - x < 8x + -(x) - (-7) - - x < 8x + -x + - - x < 8x + - x < 8x + + x +x < x + - - - < x - < x - < x x > - - - - 6. -(r - ) ( - r) -(r) - (-) () + (-r) -6r + - 6r +6r + 6r all real numbers -6 - - - - - 7. -7x - + x (x + ) + 8-7x - + x (x) + () + 8-7x - + x x + + 8 -x - x + +x +x - x + - - - x - x -6 x x -6-8 -6 - - ) 6 6 6 6 7 7 Holt McDougal Algebra
8. - ( n + 8) + ( n) - (8) + - n - n n - n - n - 8 + n - n 8 - - n - - 7 8 - -n - ( - ) -(-n) n n 9. Let s represent the number of seconds. + 8s > 8 + t - s - t + s > 8 - - s > 8 s > 8 s > 6.67 The red kite will be higher than the blue kite in 7 s. a. decreased by b. 9 - (x - ) c. increase by d. 9 + (x - ) e. 9 - (x - ) < 9 + (x - ) 9 - x + < 9 + x - -9 + -9 + 7- x + < x 9 - x < x +x +x 9 < x 9 < x.9 < x, where x is a whole number 6 a. +.n b. n c. +.n <.n -.n -.n < 7.n 7. < 7.n 7. < n n > They must sell CDs or more to make a profit.. x + > x -x -x > - x - () < - ( - x ) - < x x > -. x + < x - -x -x < x - + + < x < x 7 < x x > 7. x - < 6x - 8 -x -x - < x - 8 +8 + 8 - < x x > -. - ( - x x - -x -x - x - x ) -(-) x 6. Let v represent the number of videos..99v 9.99 +.99v -.99v -.99v v 9.99 You would need to rent videos or more for Video View to be less expensive than Movie Place. 7. No; the sum of the measures x and x - would need to be greater than x. x + (x - ) > x. x - > x has no solutions. 8. The steps are identical except when you multiply or divide both sides of an inequality by a negative number, when you must reverse the inequality symbol. 9. x can never be greater than itself plus. 6. B is incorrect. The student should have added x to both sides to undo the subtraction. test prep 6. D; Isolating a and b gives b <, so b must be negative. 6. J; Multiplying both sides by - causes the inequality sign to change and gives a > -b. 6. A; Since 7 ( - (-)) = 7() = 8 > -6 = (-) = ((-)-), - is a solution. 6. J; Dividing both sides by - gives the inequality x > which is shown in graph J. 6. Possible answer: Parking lot A starts with cars and 7 more cars park each hour. Parking lot B starts with cars and more park each hour. The inequality helps you find that parking lot A has more cars than parking lot B after hours. 8 Holt McDougal Algebra
challenge and Extend 66. + x + x - x - x + x - - - x (-) ( x ) -6 x x -6 67..6x -.7 > 6.x - (-.x).6x -.7 > 6.x +.x.6x -.7 > 8.x -.6x -.6x -.7 > 6.9x -.7 6.9 > 6.9x 6.9 - > x x < - 68..x - 7.x < 8.x - 9. -6.x < 8.x - 9. -8.x -8.x -.7x < -9. -.7x -.7 > -9. -.7 x > 69. -w + -8-7 7 - + w 9 9 -w + - 9 7 9 + w -w - 8 + w +w +w - 8 + 7w -8-8 - 7w - 7 7w 7-6 7 w w - 6 7 7. Check students work: the number in the square should be less than the number in the circle. 7. Check students work: the number in the square should be greater than the number in the circle. 7. Check students work: if the same number is added to both sides and x is negative, the inequality is an identity. -6 solving compound inequalities check it out!. Let c be the free chlorine in a pool.. c. 6 a. -9 < x - < - + + + < x < 6 a. + r < OR r + > 9 - - - - r < OR r > 6 8 6 b. 7x OR x < - 7x 7 7 x < - x OR x < - - - - b. - n + < - - - -9 9 n < 6 - n < 6 - n < - - - a. -9 < y < - b. x - OR x think and discuss. y > can be written as < y. Combine the ineqalities < y and y to write < y.. A 6 B x > 7 x < 8 x > AND x < x > OR x < Exercises guided practice 6 7 8 6 7 8. intersection. Let t represent the temperature. 7 t 9 7 7 8 8 9 9 9 Holt McDougal Algebra
. - < x + < 7 - - - - < x < - -. x + - - - x x x 8. - < x - < + + + < x < 6 9. < x - < 7 + + + 6 < x < 8 6 < x < 8 < x < 6. < x + < - - - < x < 6 7. x + < -6 OR x + > 6 - - - - x < -8 OR x > 6 6. < x + < - - - 8 < x < 8 8 < x < 8 < x < 9 6 7 8 9. x - < -7 OR x + > + + - - x < - OR x > - - - -. x + < OR x + > 8 - - - - x < OR x > x < x < OR x > 6-8 -6 - - 8. r - < OR r - > + + + + r < OR r > 6 9. n + < OR n + > 7 - - - - n < OR n > 6. x - < - OR x - > - + + + + x < OR x > 6. - a -. b - OR b >. c < OR c 9. d < 8 practice and problem solving. Let k be the distance from the surface of the Earth. 6 k 6 6. - < x + < - - - - < x < - - - 7. n - 7 + + + 6 n 6 n n 6 6-6 - 6. x + < OR x + > 8 - - - - x < OR x > 6. x + < OR x - > - - + + x < - OR x > - - -. p < OR p >. q < OR q 6. -6 < r 7. - < s < 8. Let f represent the frequencies in Hz. 8. f 69. 8. 69. 9a. + 8n gives the cost of the studio and technicians. They will spend between $ and $. b. + 8n - - - - 8n - 8 8n 8 8 -. n.6 n cannot be negative since time cannot be negative. c. Right now they can use the studio for.6 hours. 6 h -.6 h =.97 h.97 h $8 h = $ They need an additional $ to use the studio for 6 hours.. -6 < x < 6. x -6 - - 6 - - - 6 Holt McDougal Algebra
. < x <. - s + s 8 6 8 6. b - 9 + + + 8 b 8 b b. - x - -. t < OR t > 7. - < x - < + + + - < x < - 6 test prep 7. D; Solving both inequalities gives x < - OR x >, which represnts all real numbers greater than or less than -. 8. H; Solving gives x < AND x > -, which is graph H. 9. B; The open circle at and arrow pointing left means that < is used. Therefore, the answer must be B or D. The closed circle at and the arrow pointing right means that must be used so the answer is B.. H; Since + = AND + =, x = is a solution to both inequalities and therefore a solution to the compound statement. 6 8. r + < - OR r - > - - + + r < - OR r > -8-8 9. a - < - OR a - > + + + + a < OR a > a < a > a < OR a > - - -. x - AND x - + + + + x 9 AND x 9 7 8 9. n - < - OR n + > 6 + + - - n < OR n > 6. Let w represent the original weight of the ball. w +. 6 -. -. -.. w.. Let m represent the additional miles needed. 7 + m 6 - - - 7 m 6. Possible answer: Margaret is expecting between and guests; g where g is a natural number.. Both graphs include numbers less than. The graph of x < AND x < 7 does not include numbers greater than or equal to. The graph of x < OR x < 7 also includes all numbers less than 7. 6. An inequality with OR always has solutions because the two simple inequalities do not have to be true at the same time. A compound inequality with no solutions must involve AND. Challenge and Extend. c - < - c < 7c c - < - c AND - c < 7c +c + c + c +c c - < AND < c + + c < AND < c c < < c c < AND. < c. < c <.... p - < p + 6 < p p - < p + 6 AND p + 6 < p -p -p -p -p p - < 6 AND 6 < p + + p < 6 AND 6 < p p < 6 6 < p p < AND < p < p < 6. s 8 - s OR s s + 6 +s + s -s -s s 8 OR s 6 s 8 s 6 s 6 OR s 9 6 9. 9 - x x OR - x 7 + x +x - - 9 6x OR -x - 9 6 6x 6 -x - - - x OR x x. - x 6 Holt McDougal Algebra
6. Solve for x. x + a AND x - 7 b - - + 7 +7 x a - AND x b + 7 a - x b + 7 For x = to be the only solution, it must be true that x. Therefore: a - = AND b + 7 = + + - 7-7 a = AND b = -6 So for x = to be the only solution, a must be equal to and b must be equal to -6. -7 solving absolute-value inequalities CHECK IT OUT! a. x 6 x 6 x - x - - - - - b. x + -. 7. +. +. x + x + AND x + - + - - - x 9 AND x - - x 9 - - - - a. x + - - x x - OR x - b. x + - - + - - x + x + - - - 9 OR x + - - x -6 OR x -8-6 - -. Let acceptable pressure be represented by p. Then, p - 7 p - 7 AND p - -7 + + + + p AND p p The range of acceptable pressures is p 7 7 a. x - 9 - + 9 + 9 x - This is true for all real numbers; all real numbers are solutions. b. x -. -8 x -. -8 x -. - The is false for all real numbers; there are no solutions. think and discuss. The former has solution x (all real numbers to the left of and including ), whereas the latter has solution x < (all real numbers to the left of, but not including ).. Absolute-Value Inequalities AND x < - < x < excercises guided practice. x - - + + x x - AND x - x - - - OR x > x > or x < -. x + - 7.8 < 6. + 7.8 +7.8 x + <. x + > - AND x + < - - - - x > - AND x < - < x < - 6 Holt McDougal Algebra
. x + < 8 - - x < 6 x > -6 AND x < 6 x > -6 x < 6 x > - AND x < - < x < - - -. x - + < - - x - < x - < AND x - > - + + + + x < 6 AND x > < x < 6 6 7 8 6. x + - + + x +. x x + AND x + - - - - - x AND x - - x - - - - - - 7. x - 6 > 6 + 6 +6 x > x < - OR x > - - 9. x 8 x 8 x x - OR x - - - - - x x - x - - - - - 8. x +.9 > 8.6 -.9 -.9 x >.7 x < -.7 OR x >.7 -.7 - -. x + > 7 x + < -7 OR x + > 7 - - - - x < -9 OR x >.7. x - + - - x - x - OR x - - + + + + x OR x 6. x + - 7 + + x + x + - OR x + - - - - x -7 OR x 7-7 7 - -. Let x represent an acceptable level of fat intake (in grams) per day. Then, x - x - AND x - - + + + + x 8 AND x x 8 Therefore an acceptable range of fat intake is grams x 8 grams. 6 7 8. x + 8-8 - 8 x -6 False for all real numbers; there are no solutions.. x + < - False for all real numbers; there are no solutions. 6. x + -8 True for all real numbers; therefore solution set is all real numbers. 7. x - + > - - - x - > - True for all real numbers; therefore solution set is all real numbers. 8. x + 7 > - 7-7 x > - True for all real numbers; therefore solution set is all real numbers. 9. x - 7 +. -. -. x - 7 -. False for all real numbers; there are no solutions. -9 - - 6 Holt McDougal Algebra
Practice and problem solving. x + 6-6 - 6 x x AND x - - x - - - - -. x - < x - < AND x - > - + + + + x < AND x > < x < 6. x - - 8 - + 8 + 8 x - x - AND x - - + + + + x 7 AND x - - x 7 - - - 6 7. x < x < AND x > - x < AND x > - x < AND x > - - < x < - - - - -. x -. + 6. - - x -.. x -.. AND x -. -. +. +. +. +. x.8 AND x x.8.8. + x + < 7 - - x + < x + < AND x + > - - - - - x < AND x > -6-6 < x < 6. x - > x - > OR x - < - + + + + x > OR x < - - - - - - 7. 6 x 6 6 x 6 6 6 x x - OR x - - 8. x - + > 8 - - x - > x - > OR x - < - + + + + x > 9 OR x < - - 9 - - 9. x+ 6 x + 6 x + 8 x + 8 OR x + -8 - - - - x 6 OR x - 6 - -. + x - > - - x - > x - > OR x - < - + + + + x > OR x < 6 7. x - + 9 > - 9-9 x - > x - > OR x - < - + + + + x > OR x < - - - - - - -8-6 - - 6 Holt McDougal Algebra
. Let the actual temperature be represented by t, then t - 7 t - 7 AND t - 7 - + 7 + 7 + 7 + 7 t 87 AND t 6 6 t 87 The range of possible temperatures is 6 to 87. 6 87 6 6 7 7 8 8 9. + x - - x - False for all real numbers; there are no solutions.. x + - > - + + x + > - True for all real numbers; therefore solution set is the set of real numbers.. x + + - - x + - True for all real numbers; therefore solution set is the set of real numbers. 6. x - < -6 + + x < - False for all real numbers; there are no solutions. 7. x - -9 x - -9 x - - False for all real numbers; there are no solutions. 8. x + 9 9-9 - 9 x True for all real numbers; therefore solution set is the set of real numbers. 9. Always true since the absolute value is always greater than or equal to zero, which is greater than all negative numbers.. Never true since always positive and is therefore always greater than zero.. This is sometimes true as there are times when the solution may be Ø or some other set such as x >.. x x - AND x - x - -. x - x - - AND x - + + + + x - AND x - x -. x - 8 x - 8 - OR x - 8 + 8 +8 + 8 +8 x 6 OR x 6 7 8 9. a 6. b > 7. c 6 8. d < 7 9a. Middle frequency = + = = Hz. 9b. An inequality representing the range of frequencies is: ƒ or ƒ - 999 where ƒ represents frequency.. Let t represent temperature. Then SEA BASS RAINBOW TROUT EUROPEAN EEL t t 6 t - 9 t + 9-9 t - 9 - t + - t - - t + - t - t - 9 t - t -. Let n represent the number, then n > + OR n < - - - - - - + n > OR - + n < - n - > a. This is given by the difference between the values p and 8.7 or p - 8.7. b. We require this value to be less than or equal to $., therefore p - 8.7.. c. p - 8.7. p - 8.7. OR p - 8.7 -. + 8.7 + 8.7 + 8.7 + 8.7 p. OR p 7. $7. p $.. x + < k - - x < k - For no solutions, k - must be less than or equal to zero, that is, k - k. 6 Holt McDougal Algebra
. For the values within a range, use less than. The distance from the number to 6 must be less than units. So the absolute value of the difference of a number and 6 is less than. x - (-6) <. x + 6 <. Test prep. + x + < 6 - - x + < x + < AND x + > - - - - - x < - AND x > -7-7 < x < - Therefore the best answer is B. 6. x 7 + AND x 7 - - 7-7 - 7-7 x - 7 AND x - 7 - x - 7 Therefore the best answer is F. 7. w - 6 w - 6 AND w - 6 - + 6 + 6 + 6 + 6 w 9 AND w w 9 Clearly, only B is not true. challenge and extend 8. x. AND x -. x.9 +. AND x.9 -. -.9 -.9 -.9 -.9 x -.9. AND x -.9 -. x -.9. 9. x > OR x < - x > + OR x < - - - - - x - > OR x - < - x - > 6. Statements Reason. x - 6 + 7 Given. x - 6 Subtraction Property of Inequality. x - 6 - AND x - 6 Definition of absolute value. x AND x 8 Addition Property of Inequality. x AND x Division Property of Inequality ready to go on? Section B Quiz. x + < 9 - - x < 6 x < 6 x < 6. 7-6r - - 6-6r 6-6 -6r -6 - r r - - - -. a + - > - a > 6 a > 6 a >. t - > + + t > t > t > 6. (x - ) > - (x) + (-) > - x - 6 > - + 6 +6 x > x > x >. 6. < (m - 7) < (m) + (-7) < m - + + < m < m < m m > 7. + (-6) >.8p - >.8p -.8 >.8p.8 - > p p < - 8. Let s represent her mark on the second test. s + 88 9 ( s + 88 ) (9) s + 88 8-88 -88 s 96 Mindy must make a score of 96 or higher. 9. x < x + 8. 6p - > 9p -x -x -6p -6p x < 8 - > p x < 8 - > p x < 6 - - - - > p p < - 66 Holt McDougal Algebra
. r - 8 r - -r -r -8 r - + + r r r r - - -. (y + 6) > (y + ) (y) + (6) > (y) + () y + 8 > y + 8 -y -y y + 8 > 8-8 -8 y > -. ( - g) g () + (-g) g - g g + g +g g g g g. ( - x) -(x + ) () + (-x) -(x) - () - x -x - 6 + x +x -6 all real numbers. x < (x - ) x < (x) + (-) x < x - -x -x < - 7 no solutions 6. Let m represent the number of months. + 8m < + m - m - m + m < - - m < m < m < Gil will have a larger bank balance than Phillip for months. 7. - x + < 9 - - - - x < 6 - -6 - - 6 8. m + < - OR m - > 6 - - + + m < - OR m > 8 - - - 6 8 9. - x - > + + + - x > x - AND x > No solution.. - > r + OR r + < - - - - - > r OR r < r < - - -. Let t represent the temperature of the medicine. < t < 7. x + 9-9 - 9 x - x -. x + 7 - < 6 + + x + 7 < x + 7 < AND x + 7 > - - 7-7 - 7-7 x < AND x > -8-8 < x < -8.. x.. x... x 7 x -7 OR x 7-7. x - x - AND x - - + + + + x 6 AND x - - x 6 7 6. x - 9. < -.7 + 9. + 9. x <. -. < x <. 7. + x > - - - 9 x > - x 9 > - 9 x > - 9 x is always greater than -. So x is all real numbers. 67 Holt McDougal Algebra
8. 7 + x > - 7-7 x > 6 x > 6 x > x < - OR x > 9. d - d - AND d - - + + + + d AND d 88 88 d 88 study guide: review. inequality. union. compound inequality. intersection. solution of an inequality - 6. 8. Graphing and writing inequalities -6 - - - -. ( - ) < k (-) < k - < k k > - 6-6 - - - - - 7. 9... a <. k -.. q < -. Let t represent the temperature. t 7 8 7 96 6 9 9.. 6 6. Let s represent the number of students present. s, where s is a natural number 6 9 8 7. Let m represent the number of minutes to complete the lab. - m < solving inequalities by adding or subtracting 8. t + < - - t < 7 9. k - 7 - + 7 +7 k. - < m + - - - < m m > - -6 - - - - -. w - < 6. + + w < 9. 8 8. 9 9.. h - < + + h < - - -. x +. 6.8 -. -. x... 6. > a - + + > a a < 6. > 7 + v -7-7 - > v v < - - - - 6. Let m represent the number of miles left to run.. + m -. -. m. Tammy must run. mi or more. 7. Let d represent the amount Rob can spend. + d - - d 8 Rob can spend $8 or less. - solving inequalities by multiplying or dividing 8. a a a. 6 p > ( p ) > () p > 8 6 8. -n < -8 -n - > -8 - n > 6 6 8 9. -8 < 6t -8 6 < 6t 6 - < t t > -.. -6 - - - - - x - x ) ( -) ( x - - - - - - g - > 6 - ( g - ) < -(6) g < - -6 - -8-6 7 8 9 - - - 68 Holt McDougal Algebra
. -k < -k - > - k > -7-9 -8-7 -6-6. 7 < -9h 7-9 > -9h -9 - > h h < - -6 - - - - -. - > r (-) > ( r ) -9 > r r < -9 - -9-6 - 7. -.g > - -.g -. < - -. g <... 8. Let n represent the number of notebooks..9n.9n.9.9 n 7.9 Only a whole number of notebooks can be purchased, so,,,,,, 6, or 7 notebooks can be purchased. 9. Let n represent the number of lanyards..7n.7n.7.7 n They must sell at least lanyards. - solving two-step and multi-step inequalities. x + < 9 - - x < x < x < 6 m +. > - m + > (-) m + > -8 - - m > - - - - -9-8. 7 t - + + t t 6 t t 6 6 8. (x + ) < 8 (x) + () < 8 x + < 8 - - x < - x < - x < - - - -. -( - ) > (- ) - h -(-) > (- ) - h > 9 - h -9-9 > -h -() < -(-h) - < h h > - -6 - - - - -. x + - > - x > ( x ) > ( ) x > 6..(b - ) 7. y - >.(b) +.(-).b - + + + +.b.b. y > 7. 6 b ( y ) > ( 7 6 ) y > 6 8 8. 6 -.n < 9-6 -6 -.n < -.n -. > -. n > - - - - - 9. Let m represent the number of movies per month. + m < - - m < m < m <.7 Only a whole number of movies can be watched, so if,,,,,, 6, 7, 8, 9,,,, or movies are watched, Carl s Cable Company is cheaper than Teleview. - solving inequalities with variables on both sides. + m < -m - m -m < -m - > -m - - > m m < - - - -. y 6 + y -y - y -y 6 -y - 6 - y - - - - 69 Holt McDougal Algebra
. c - 7 > 9c + 8 -c -c -7 > c + 8-8 - 8 - > c - > c - > c c < - -6 - - - - -. -( - q) 6(q + ) -() - (-q) 6(q) + 6() -6 + q 6q + 6 - q -q -6 q + 6-6 - 6 - q - q - q q - -6 - - - - -. ( - x) < x () + (-x) < x - x < x + x +x < x < x < x x > - - -..t -.8 <.6t +.9 -.6t -.6t.9t -.8 <.9 +.8 +.8.9t <.7.9t.9 <.7.9 t < 6 6. d - < d - -d -d - < - 7 no solutions 7. ( - x) > -( + x) () + (-x) > -() - (x) - x > - - x + x + x > - all real numbers 8. ( - p) < ( + p) () + (-p) < () + (p) - p < 8 + p + p + p < 8 + 8p -8-8 - < 8p - - - 8 < 8p 8 - < p p > - 9. w + > (w - ) w + > (w) + (-) w + > w - -w -w > - all real numbers 6. ( - k) < k () + (-k) < k - k < k + k +k < k < k < k k > - - - 6. (c + ) > c + (c) + () > c + c + > c + -c -c > 7 no solutions 6. Let m represent the number of months. + 6m > 7 + m - 6m - 6m > 7 + m -7-7 > m > m 8.7 > m m < 8.7 Hanna s acount will be greater than Faith s account for 8 months. -6 Solving compound inequalities 6. - < t + 6 < -6-6 -6 - < t < - - -8-6. - + r > OR r + < - + + - - r > 7 OR r < - 7 6. -8 < k - + + + -6 < k 7-6 -8-7 8 - - 6 8 7 Holt McDougal Algebra
66. > n + > - - - - > n > n < - AND n > no solutions 67. p + 7 > - 7-7 - 7 p > - - < p - - 6 8 68. < s + 9 OR > s - -9-9 + + -6 < s OR > s s > -6 OR s < all real numbers 69. Let t represent the day s temperature. 68 t 8 7. Let n represent the heart rate for a 6-year-old.. ( - 6) n.9 ( - 6). n.9 n 8.6 7. x - 7 + 7 + 7 x - x - - 7. x + > 8 x + > 8 OR x + < -8 - - - - x > OR x < - - - ] 7. 6 x 6 x 6 6 x - x - - 6 7. x + 9 + < - - x + 9 < 9 x + 9 < 9 AND x + 9 > -9-9 - 9-9 - 9 x < AND x > -8-8 - - -9-6 - 7. x 9 x 9 x x - OR x - - - 76. x < x < x < 6 x < 6 AND x > -6 x < 6 AND x 6 > - x < AND x > - - < x < - - - 77. x -. > 8. +. +. x >.9 x < -.9 OR x >.9 78.. + x < 7.. + x < 7. AND. + x > -7. -. -. -. -. x <. AND x > -. -. < x <. 79. x - 7 + - - x - 7 x - 7 OR x - 7 - + 7 + 7 + 7 + 7 x 9 OR x 8. x - + + x 6 x 6 x x - OR x 8. x - + - - x - - no solutions 8. x +. - 6..9 + 6. + 6. x +.. x +.. AND x +. -. -. -. -. -. x.8 AND x -6.8-6.8 x.8 8. Let actual depth be d, then d - 7 d - 7 AND d - 7 - + 7 + 7 + 7 + 7 d 76 AND d 68 68 d 76 7 Holt McDougal Algebra
Chapter test. all real numbers greater than or equal to -6. all real numbers greater than. all real numbers less than or equal to -. all real numbers less than or equal to 8. 6. -6 - - - - -.. 7. y - ÇÇ y - -6 - - - - - 8. - ( + 7) h - () h -8 h h -8-6 - -8-9. d <. p -.. Let m represent the number of minutes. m 9 6 9. d - > -7 + + d > -. f + < - - - f < -7 - - - -9-8 -7-6 -.. s +. -. -.. s s..... g + (-) 9 g - 9 + + g 6. Let h represent the number of hours. 8 + h 7-8 -8 h 7 Samir needs at least 7 more hours. 9 7 Holt McDougal Algebra