Chapter : Solving Equations and Inequalities Prep Test 1.. 1 = = 1 1. (16) = 80 = 10 8 8. 90% = 90(0.01) = 0.9. 0. = 0.(100%) = % 6. x x 1 ( ) ( ) 1 = (16) + 16 1 = 8 + 1 = 6. 10x 8. 9 9. 6x (6 x) = 6x 18 + x = 9x 18 Go Figure With the donut on a table, slice the donut parallel to the table. Now cut the halved donut into quarters. Section.1 Concept Review.1 1. Sometimes true The exception is multiplying both sides of an equation by zero.. Always true. Always true. Always true Objective.1.1 Exercises 1. An expression is a combination of numbers, variables, and operation symbols. An equation consists of two expressions set equal to each other. It is important that students note that an equation contains an equals sign; an expression does not.. x = 8 () 8 8 = 8 Yes, is a solution.. b 1 = ( 1) 1 1 No, 1 is not a solution.. m = (1) No, 1 is not a solution. 9. x + = x () + () 10 + 1 1 = 1 Yes, is a solution. 11. a + = a + (0) + (0) + 0 + 0 + = Yes, 0 is a solution. 1. n = n + 10 ( ) + 10 ( ) 8 + 8 8 = 8 Yes, is a solution. 1. z +1 = + z +1 + () 9 + 1 + 9 10 1 No, is not a solution. 1. y 1 = y+ ( 1) 1 ( 1) + 1 1 + 0 1 No, 1 is not a solution. 19. x(x +1) = x + ( + 1) + () 16 + 0 1 No, is not a solution.
Chapter : Solving Equations and Inequalities 1. 8t + 1 = 1 8 1 +1 1 Yes, 1 + 1 1 1 = 1 is a solution.. m + 1 = 10m +1 10 No, + 1 1 is not a solution.. x x = x +1.89 (.1) (.1).1 + 1.89.1 8..99.99.99 No,.1 is not a solution. Objective.1. Exercises. The Addition Property of Equations is used to remove a term from one side of an equation by adding the opposite of that term to each side of the equation. 9. The value of x is less than 1. To solve the equation, 1 is subtracted from each side of the 1 equation, and 1 1 < 1. 1 1. x + = x + = x = The solution is.... 9. b =11 b + =11+ b =1 The solution is 1. + a =8 + a =8 a = 6 The solution is 6. m + 9 = m +9 9 = 9 m = 6 The solution is 6. n = n + = + n = The solution is. 1. b + = b + = b = 0 The solution is 0.. a = a + = + a = The solution is.. z + 9 = z +9 9 = 9 z = The solution is.. 10 + m = 10 10 + m = 10 m = The solution is. 9. 9 + x = 9 9 + x = 9 x = 1 The solution is 1. 1. b = b + = + b = The solution is.. = x + = x + = x The solution is.. = m 11 +11 = m 11+11 1 = m The solution is 1.. 1 = +w 1 = + w 9 = w The solution is 9. 9. = 10 +b +10 = 10 +10 + b 1 = b The solution is 1. 61. m + = 1 m + = 1 m = 1 The solution is 1. 6. x 1 = 1 x 1 + 1 = 1 + 1 The solution is 1.
Section.1 6. 8 + y = 1 8 8 8 + y = 1 8 8 y = 8 = 1 The solution is 1.. 1 = n + 1 = n + 1 9 1 = n 1 = n 6. m + 1 = 1 m + 1 1 = 1 1 m = 1 m = The solution is. 69. x + = x + = x = 9 1 8 1 x = 1 1 The solution is 1 1. 1. 6 = x 1 6 + 1 = x 1 + 1 10 1 + 1 = x 1 = x The solution is. 1. 1 1 = m + 1 1 = m + 1 1 1 1 = m 1 1 = m = m The solution is. 1 = n 1 The solution is.. w +.9 =.801 w +.9.9 =.801.9 w =1.869 The solution is 1.869. 9. 1.96 + t = 1.0 1.96 +1.96 + t = 1.0 +1.96 t = 0.88 The solution is 0.88. Objective.1. Exercises 81. If each side of an equation is multiplied by zero, the result is the identity 0 = 0. This does not help find a solution of the equation. 8. The value of x is less than 0. To solve the equation, both sides of the equation are multiplied by the reciprocal of. Therefore, we are 8 multiplying a negative number by a positive number 8 ; the product is negative, or less than 0. 8. y = 8 y = 8 y = The solution is. 8. a = 1 a = 1 a = The solution is. 89. m = 0 m = 0 m = The solution is. 91. 6n = 0 6n 6 = 0 6 n = The solution is.
6 Chapter : Solving Equations and Inequalities 9. 18= t 18 = t 9 = t The solution is 9. 9. 6 = x 6 = x 8 = x The solution is 8. 9. x = 0 x = 0 x = 0 The solution is 0. 99. = x = x = x The solution is. 101. = y = y 8 = y The solution is 8. 10. 1m = 1 1m 1 = 1 1 m = 1 The solution is 1. 10. x = 1 x = () x =1 The solution is 1. 10. b = 6 109. 1 b = (6) b = 18 The solution is 18. t 6 = 6 1 6 t = 6( ) t = 18 The solution is 18. 111. c = 8 c = ( 8) c = 6 The solution is 6. 11. d = 8 11. 11. d = (8) d = 1 The solution is 1. n = n = () n = The solution is. z 8 = 9 8 8 z = 8 (9) z = The solution is. 119. 6 = y 11. ( 6) = y 9 = y The solution is 9. 9 = y 9 = y 1 = y 1 The solution is. 1. m = 6 m = 6 m = 1 The solution is 1. 1. n +n = 0 n = 0 n = 0 n = The solution is. 1. 10y y = 1 y = 1 y = 1 y = The solution is.
Section.1 19. x 1. =. 1. x 1. =1.(.) x =.8 The solution is.8. 11..a =.00.a. =.00. a =.06 The solution is.06. 1..x =..x. =.. x =.1 The solution is.1. Objective.1. Exercises 1. Employee B s salary is the highest after the raise. Explanations will vary. One possible explanation is: Because each employee receives a % raise, % of the largest original salary will be the largest raise. Adding the largest raise to the largest original salary ensures that the employee will have the largest resulting salary. 1. P(0) =1 0P 0 = 1 0 P = 0. The percent is %. 19. 0.18(0) = A.0 = A 18% of 0 is.. 11. (0.1)B = 8 0.1B 0.1 = 8 0.1 B = 00 The number is 00. 1. 1 () = A 1 % = 1 9 = A 1 % of is 9. 1. P(1) = 1P 1 = 1 P = 0. The percent is %. 1. (0.60)B = 0.60B 0.60 = 0.60 B = The number is. 19. P(6) = 1 6P 6 = 1 6 P = 1 is 00% of 6. 11. (0.0)B = 1 0.0B 0.0 = 1 0.0 B = 00 The number is 00. 1. (0.1)(0)= A.= A 1.% of 0 is.. 1. equal to since 0% 80 = 80% 0 1. Strategy To find the number of seats, solve the basic percent equation using P = 1.11% = 0.0111 and B =,00. The amount is unknown. 0.0111(,00) = A 9. = A 0 seats are reserved for wheelchair accessibility. 19. Strategy To find the total water used, solve the basic percent equation using P = 1.8% = 0.18 and A = 1.. The base is unknown. 0.18B = 1. 0.18B = 1. 0.18 0.18 B The total water used gallons a day. 161. Strategy To find the percent, solve the basic percent equation using B = 1 and A =. The percent is the unknown. P(1) = 11P 1 = 1 P 0.1 1% of the vacation bill is paid in cash. 16. There is insufficient information to solve this problem. Either the base or the amount must be given.
8 Chapter : Solving Equations and Inequalities 16. Strategy To find the percent, solve the basic percent equation using B = 0 + + 00 + 190 = and A = 0 + + 00 =. The percent is unknown. P() = P = P 0.1 1% of the accidental deaths were not attributed to motor vehicle accidents. 16. Strategy To find the total electricity used, solve the basic percent equation using P = % = 0. and A = 1.. The base is unknown. 0.B = 1. 0.B 0. = 1. 0. B 96.1 The total electricity used for home lighting is 96.1 billion kilowatt-hours a year. 169. Strategy To find the simple interest rate, solve the simple interest equation using I = $, P = $100, and t = 8 months = 8 years, for r. 1 I = Prt = (100) r 8 1 = 800r = 800 r 800 800 0.09 = r The annual simple interest rate is 9%. 11. Strategy To find the interest, solve the simple interest equation for each account: First, using P = $1000, r =.% = 0.0, and t = 1 year, for I. Second, using P = 000 1000 = $000, r = 8.% = 0.08, and t = 1 year, for I. Finally, find the total interest by adding the interest earned in each account. I = Prt I = (1000)(0.0)(1) I = I = Prt I = (000)(0.08)(1) I = 16 + 16 = $0 Sal earned $0 after one year. 1. The principal for each investment is the same amount. The time the interest accrued is the same for each account. If one account earns 6% and the other earns 9%, the combined interest earned is between 6% and 9%. To find simple interest rate on the combined accounts, solve the simple interest equation for each account: First, using P = $1000, r = 9% = 0.09, and t = 1 year, for I. Second, using P = $1000, r = 6% = 0.06, and t = 1 year, for I. Finally, to find the combined interest rate, add the value of P = 1000 + 1000 = $000, and total the interest earned in both accounts, using the simple interest equation to find r. I = Prt I = (1000)(0.09)(1) I = 90 I = Prt I = (1000)(0.06)(1) I = 60 90 + 60 = (000) r(1) 10 = 000r 000 000 0.0 = r The interest rate earned on the combined accounts is between 6% and 9%. 1. Strategy To find the percent, solve the basic percent equation using B = 0 and A =. The percent is the unknown. P(0) = 0P = 0 0 P = 0.0 There is a % concentration of hydrogen peroxide.
Section.1 9 1. Strategy To find which brand has the greater concentration, solve the basic percent equation for Apple Dan s using B = and A = 8. The percent is the unknown. Then solve the basic percent equation for the generic brand using B = 0 and A = 9. The percent is the unknown. Compare the percent of concentration. P() = 8 P = 8 P = 0. Apple Dan's P(0) = 9 0 P = 9 0 0 P = 0. generic % >.% Apple Dan s has the greater concentration. 19. Strategy To find the amount that is not glycerin, solve the basic percent equation, find the percent that is not glycerin using P = 100 - % = 0. and B = 0 g. The amount is unknown. 0.(0) = AA 1. = There is 1. g of cream that is not glycerin. 181. Strategy To find the percent, solve the basic percent equation using B = 00 100 = 00 and A = 0. The percent is unknown. P(00) = 0 00P = 0 00 00 P = 0.1 The percent concentration is 1.%. Applying Concepts.1 18. y 8y = 1 y = 1 y = (1) y = 1 The solution is 1. 18. 1 1 +8 = 19 x 18. 1 1 +8 = 19 x 1 x +8 = 19 1 x +8 = 19 x +8 8 = 19 8 x = The solution is. = 6 a a a a = 6 a a = 6 a a = 6 a = 6 a = (6) a = 1 The solution is 1. 189. Strategy To find the cost of the dinner, solve the basic percent equation using P = (100% + 1%)(100% + 6%) = (11%)(106%) = (1.1)(1.06) and A = 9.. The base is unknown. (1.1)(1.06) B = 9. 1.19B = 9. 1.19B = 9. 1.19 1.19 B = 80 The cost of the dinner was $80. 191. If a quantity increases by 100%, then the same quantity is added to itself. The new value will be twice the original value. 19. One equation with a solution of is x = 10.
0 Chapter : Solving Equations and Inequalities 19. Students should refer to the two sections of the table, one for single taxpayers and one for married taxpayers filing jointly. The first two columns of a table are used to locate the row containing the taxable income; the last two columns are used to calculate the income tax due. A single taxpayer s taxable income of $,000 falls between $6,0 and $16,0. 1,6 +.0(,000 6,0) = 1,6 + 0.0(90) = 1,6 + 88. = 1,. The income tax is $1,.. Section. Concept Review. 1. Always true. Never true First use the Addition Property of Equations to remove the constant term from the left side of the equation. Then use the Multiplication Property of Equations to multiply both sides of the equation by 1 a.. Never true Division by zero is undefined.. Sometimes true If a = c, the equation will have no variables when cx is subtracted from both sides. Objective..1 Exercises 1. x +1 =10 x +1 1 =10 1 x = 9 x = 9 x = The solution is.. a = a + = + a = 1 a = 1 a = 6 The solution is 6.. = x + 9 9 = x + 9 9 = x = x 1 = x The solution is 1.. 1 = 9 + z 1 9 = 9 9 +z = z = z 1 = z The solution is 1. 9. x =11 x =11 x = 9 1( x) = 1 9 x = 9 The solution is 9. 11. 6x = 1 6x = 1 6x = 18 6x = 18 6 6 x = The solution is. 1. d + = 1 d + = 1 d = 1 d = 1 d = The solution is. 1. n = n + = + n = 1 n = 1 n = The solution is. 1. 1 = 11y+ 9 1 9 = 11y+ 9 9 = 11y 11 = 11y 11 = y The solution is. 19. = 11 n 11= 11 11 n 8 = n 8 = n = n The solution is. 1. 8x + = 9 8x + = 9 8x = 8x 8 = 8 x = The solution is.
Section. 1. x = x + = + x = 6 x = 6 x = 6 The solution is 6.. 9a = 9a = 9a = 9a 9 = 9 a = 9 = 1 The solution is 1.. 6a + = 9 6a + = 9 6a = 6a 6 = 6 a = 6 = The solution is.. 10 = 18x + 10 = 18x + = 18x 18 = 18x 18 18 = x 1 6 = x. 11= 1+ n 11 1 = 1 1+ n = n = n 1 = n The solution is 1. 9. 9 x = 6 9 9 x = 6 9 x = x = x = The solution is. 1. 1 x = 0 1 1 x = 0 1 x = 1 x = 1 x = 1 The solution is 1.. 8b = 9 8b + = 9 + 8b = 6 8b 8 = 6 8 The solution is 1 6. 9. x 6 = 1 6 x 6 + 6 = 1 6 + 6 x = 18 6 x = x = x = 1 The solution is 1. 1. 9x + = 9x + = 9x = 0 1 9 (9x) = 1 9 (0) x = 0 The solution is 0.. = 9 a 9 = 9 9 a = a = a = a The solution is. b = 6 8 = The solution is.
Chapter : Solving Equations and Inequalities. x + = 9 x + = 9 x = 6 x = 6. 9. x = 6 = The solution is. 1 m 1 = 1 m 1 +1 = +1 1 m = 6 1 m = 6 m = 18 The solution is 18. n + =1 n + =1 n = 6 n = (6) n = 8 The solution is 8. 1. b + =10 8. b + =10 8 8 b = 6 8 8 b = 8 (6) b = 16 The solution is 16. y = y + = + y = 1 y = y = The solution is.. c 1 = 8 c 1+1 = 8 +1 c = 9 c = (9) c = 1 The solution is 1.. w = 9 w = 9 w = 1 w = ( 1) w =1 The solution is 1. 9. 1+ 8 x = 61. 6. 1 1+ x = 1 8 x = 10 8 8 8 x = 8 ( 10) x = 16 The solution is 16. = y 1 + 1 = y 1 + 1 6 + 6 = y 6 = y The solution is. 6 8 = 1 1 b 8 1 = 1 1 1 b 9 10 = 1 b 1 = 1 b 1 = 1 b 1 8 = b 1 The solution is. 8
Section. 6. = x + = x + = x () = x 1 = x 1 The solution is. 6. 9 y = 9 y = 9 9 y = 9 9 9 y = 9 () y = 18 The solution is 18. 69. y +9 + y = y + 9 = y+ 9 9 = 9 y = 1 y = 1 y = The solution is. 1. 11z z = 9 z = 9 z + = 9 + z =1 z = 1 z = The solution is.. b 8b +1 = 6 b +1 = 6 b +1 1 = 6 1 b = b = b = 1 The solution is 1.. 8 = n 6 +n 8 = n 6 8 + 6 = n 6 + 6 1 = n 1 = n = n The solution is.. 1.x. = 1. 1.x. +. = 1.+. 1.x =. 1.x 1. =. 1. x =.9 The solution is.9. 9. 6.=.y.06 6.+.06 =.y.06+.06. =.y.. =.y. 0.8= y The solution is 0.8. 81. x + = x + = x = 9 x = 9 x = The answer is 11. 8. x =11 x =11 x = 9 x = 9 x = The answer is 0. Objective.. Exercises 8. 6y + = y +1 6y y + = y y +1 y + =1 y + =1 y =1 y = 1 y = The solution is. 8. 11n + =10n +11 11n 10n + =10n 10n +11 n + =11 n + =11 n = 8 The solution is 8. 89. 9a 10 = a + 9a a 10 = a a + 6a 10 = 6a 10 +10 = +10 6a = 1 6a 6 = 1 6 a = The solution is. x = ( ) = 6 = 11 x + x = ( ) + ( ) = 9 6 = = 0
Chapter : Solving Equations and Inequalities 91. 1b 1 = b 19 1b b 1 = b b 19 9b 1 = 19 9b 1 +1 = 19+1 9b = 18 9b 9 = 18 9 b = The solution is. 9. a = a 0 a a = a a 0 a = 0 a + = 0 + a = 1 a = 1 a = The solution is. 9. n = 6 n n + n = 6 n + n n = 6 n + = 6 + n =8 n = 8 n = The solution is. 9. y = 16 y y + y = 16 y +y y = 16 y + = 16 + y = 1 y = 1 y = The solution is. 99. m + =m +8 m m + =m m +8 m + =8 m + =8 m = m = m = The solution is. 101. 6d = d + 6d d = d d + d = d + = + d = ( 1)( d) = ( 1)() d = The solution is. 10. a + = a + a a + = a a + a + = a + = a = 0 a = 0 a = 0 The solution is 0. 10. 10 n =16 n 10 n + n =16 n +n 10 n =16 10 10 n =16 10 n = 6 n = 6 n = The solution is. 10. b 10 = b b b 10 = b b 10 = b 10 = b = b The solution is. 109. 9y = y +16 9y y = y y +16 y = 16 y = 16 y = The solution is. 111. 8 x = 18 x 8 x +x = 18 x + x 8 + x = 18 8 8 + x = 18 8 x = 10 The solution is 10. 11. m = 6m m + 6m = 6m +6m m = m = m = 1( ) = 1( ) m = The solution is. 11. 6y 1 = y + 6y y 1 = y y + y 1 = y 1+1 = +1 y = y = y = The solution is.
Section. 11. 10x = x 1 10x x = x x 1 x = 1 x + = 1+ x = x = x = The solution is. 1. b = 6 b b + 6 b = 6 b + 6 b 6 b + 6 b = 9 6 b = 6 9 9 6 b = 6 9 () 119. 8a = a 8a a = a a a = a + = + a = a = 11. 1. a = The solution is. x = 1 1 x + x 1 1 x = 1 1 x 1 1 x + 9 1 x 1 1 x = 8 1 x = 1 8 8 1 x = 1 8 () x = The solution is. c = 1 10 c c c = 1 10 c c = 1 10 c 8 10 c = 10 c 10 ( ) = 10 10 c 10 = c The solution is 10. b = The solution is. 1..x. =.x.x.x. =.x.x. = 1.8x. 1.8 = 1.8x 1.8 = x The solution is. 19. x = x 8 x x = x x 8 x = 8 x = 8 x = The answer is 1. 11. 6a = a 6a +a = a +a a = a = a = a = a = 1 The answer is. Objective.. Exercises 1.x + (x +1) = x + x + = x + = x + = x = 1 x = 1 x = The solution is. 1. 9n (n 1) =1 9n 6n + =1 n + =1 n + =1 n =1 n = 1 n = The solution is. x + = ( )+ = 16+ = 1 a a +1 = ( 1) ( 1) +1 = (1) ( 1) +1 = + +1 = 6 +1 =
6 Chapter : Solving Equations and Inequalities 1. a (a ) =1 a a + =1 a + =1 a + =1 a = 8 a = 8 a = The solution is. 19. ( y)+ y = 1 10y + y = 1 6y = 1 1 6y = 1 6y = 1 6y 6 = 1 6 y = The solution is. 11. 10x +1 = (x + ) 1 10x +1 = 6x +10 1 10x +1 = 6x + 9 10x 6x +1 = 6x 6x +9 x +1 = 9 x +1 1 = 9 1 x =8 x = 8 x = The solution is. 1. a = (a + ) a = a 10 a = a a + a = a + a + a = + a = a = The solution is. 1. y = (y ) + y =10y 1+ y =10y 11 y 10y =10y 10y 11 y = 11 y + = 11+ y = y = y = The solution is. 19. [ ( y 1)] = (y +8) [ y +]= 6y + [6 y]= 6y + 18 1y = 6y + 18 1y 6y = 6y 6y+ 18 18y = 18 18 18y = 18 18y = 6 18y 18 = 6 18 y = 1 The solution is 1. 11. a +[ + (a 1)] = (a + ) a +[ + a ] = 6a + 8 a + [ 1 +a]= 6a + 8 a +6a = 6a + 8 9a = 6a + 8 9a 6a = 6a 6a + a = 8 a + = 8 + a = 10 a = 10 a = 10 The solution is 10. 1. [ (b + )]= (b + 6) [ b ]= 6b 1 [ b]= 6b + 6b = 6b + 6b + 6b = 6b + 6b +1b = + +1b = + 1b = 1b 1 = 1 b = 1 The solution is 1. 1. 0.x (1.6x) 8 = (1.9x.1) 0.x. x 8 =.x 1..9x 8 =.x 1..9x.x 8 =.x.x 1. 8.6x 8 = 1. 8.6x 8+ 8 = 1.+ 8 8.6x =. 8.6x 8.6 =. 8.6 x = 0. The solution is 0.. 8
Section. 1. a = (a + ) a = a 10 a = a a + a = a + a + a = + a = a = The answer is 0. 19. z = (z +) z = 1z +1 z 1z = 1z 1z +1 10z = 1 10z + = 1+ 10z = 0 10z 10 = 0 10 z = The answer is 1. Objective.. Exercises a +a = ( ) + ()( ) = 9+ ()( ) = 9 9 = 0 z z = ( ) = = 1 161. Strategy To find the distance, replace C by and solve for D. C = 1 D = 1 D + = 1 D + = 1 D () = 1 D 168 = D The car will slide 168 ft. 16. Strategy To find the depth, replace P by and solve for D. P = 1 D +1 = 1 D +1 1 = 1 D +1 1 0 = 1 D (0) = 1 D 0 = D The depth of the diver is 0 ft. 16. Strategy To find the initial velocity, replace s by 80 and t by and solve for v. s = 16t + vt 80 = 16() +v() 80 = 16() +v 80 = 6 + v 80 6 = 6 6 + v 16 = v 16 = v 8 = v The initial velocity is 8 ft/s. 16. Strategy To find the length, replace H by 66 and solve for L. H =1.L +.8 66 =1.L +.8 66.8 =1.L +.8.8 8. =1.L 8. 1. = 1.L 1. 1.8 L The approximate length of the humerus is 1.8 in. 169. Strategy To find the number miles, replace F by 6. and solve for m. F =1.0 + 0.9(m 1) 6. =1.0 + 0.9m 0.9 6. = 0. +0.9m 6. 0. = 0. 0.+ 0.9m.0 = 0.9m.0 0.9 = 0.9m 0.9 6 = m The passenger was driven 6 mi. 11. Strategy To find the number of errors, replace S by and W by 90 and solve for e. S = W e 10 90 e = 10 90 e 10()= 10 10 0 = 90 e 0 90 = 90 90 e 0 = e 0 = e 8 = e The candidate made 8 errors.
8 Chapter : Solving Equations and Inequalities 1. Strategy To find the population, replace P 1 by 8,000, N by 1,100,000, and d by, and solve for P. N =.1P 1P d 1,100,000 =.1(8,000)P 1,100,000 = 10,80P 6 6 6 (1,100,000) = 10,80 10,80 10,80P 6 1, P The population is approximately 1,000 people. 1. Strategy To find the break-even point, replace the variables P, C, and F in the cost equation by the given values and solve for x. Px = Cx + F 0x = 16x + 9,0 0x 16x = 16x 16x + 9,0 8x = 9,0 8x 8 = 9,0 8 x = 0 The break-even point is 0 televisions. 1. In the lever-system equation F 1 x = F (d x), x represents the distance from force F 1 to the fulcrum, d represents the length of the lever, and d x represents the distance from the force F to the fulcrum. 19. Strategy To determine whether or not the see-saw is balanced, replace the variables F 1, F, d, and x in the lever system equation by the given values. Evaluate each side of the equation. If it is a true equation, the see-saw is balanced. If it is not a true equation, the see-saw is not balanced. F 1 x = F (d x) 60(.) 0(8.) 10 0(.) 10 The see-saw is not balanced. 181. Strategy To find the force when the system balances, replace the variables F 1, x, and d in the lever system equation by the given values and solve for F. F 1 x = F (d x) 100 = F (10 ) 100 = F 8 00 = 8F 00 8 = 8F 8 = F A -lb force must be applied to the other end. 18. Strategy To find the location of the fulcrum when the system balanced, replaces the variables F 1, F, and d in the lever system equation by the given values and solve for x. F 1 x = F (d x) 18x =160(18 x) 18x = 880 160x 18x +160x = 880 160x +160x 88x = 880 88x 88 = 880 88 x =10 The fulcrum is 10 ft from the 18 lb acrobat. Applying Concepts. 18. (x 1) (6x ) = 9 6x 6x + = 9 6x 6x +1 = 9 1 9 No solution
Section. 9 18. [(w + ) (w +1)] = (+w) [w+ 8 w 1] = 10+ w [w + ] = 10+ w 9w+ 1= 10+ w 9w w+ 1= 10+ w w w+ 1= 10 w +1 1= 10 1 w = 11 w = 11 w = 11 The solution is 11. 189. To solve, remember Dividend = Quotient Divisor + Remainder,166 = 18 x + 0,166 0 = 18x + 0 0,116 = 18x,116 = 18x 18 18 6 = x 191. One possible answer is x 6 = x x = 19. The sentence Solve x (x 1) does not make sense because x (x 1) is not an equation; it is an expression. Equations can be solved. Expressions cannot be solved. 19. Look for the following explanations in your students work. x = x + = + By the addition property of Equations, the same number can be added to each side of the equation without changing the solution of the equation. Add to each side of the equation the opposite of the constant term. x + 0= 9 By the Inverse Property of Addition, the sum of a number and its opposite is 0. x = 9 By the Addition Property of Zero, the sum of a number and zero is the number. x = 9 By the Multiplication Property of Equations, each side of an equation can be multiplied by the same nonzero number without changing the solution of the equation. Multiply each side of the equation by the reciprocal of the coefficient of x. 1x = 1 By the Inverse Property of Multiplication, the product of a number and its reciprocal is 1. x = 1 By the Multiplication Property of One, the product of a number and 1 is the number. Section. Concept Review. 1. Always true. Always true. Always true. Never true The number of dimes is represented by 0 n.
0 Chapter : Solving Equations and Inequalities 9. Always true Objective..1 Exercises 1. the unknown number: n The difference between a number and 1 The number is.. the unknown number: n is n 1= n 1 +1= +1 n = seven The product of is negative twenty-one seven and a number n = 1 n = 1 n = The number is.. the unknown number: n Four less than three is times a number n = n + = + n = 9 n = 9 n = The number is.. the unknown number: n five Four times the sum of is twelve twice a number and three ( n + ) = 1 8n +1 = 1 8n +1 1 = 1 1 8n = 0 8n 8 = 0 8 n = 0 The number is 0. 9. the unknown number: n 11. the unknown number: n Twice the difference between a number and twenty-five (n ) = n n 0 = n n n 0 = n n 0 = n The number is 0. 1. the smaller number: n the larger number: 0 n Three times the smaller = 0 n =8 0 n = 0 8 = 1 The smaller number is 8. The larger number is 1. 1. the smaller number: n the larger number: 18 n is three times the number is equal to two times the larger n = (0 n) n = 0 n n +n = 0 n + n n = 0 n The total of three times the smaller and twice the is forty-four larger n +(18 n) = n +6 n = n + 6 = n + 6 6 = 6 n = 8 18 n = 18 8 = 10 The smaller number is 8. The larger number is 10. six times the difference Twelve is between a number and three 1 = 6( n ) 1 = 6n 18 1+18 = 6n 18+18 0 = 6n 0 6 = 6n 6 = n The number is.