Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller Chapter Equations and Inequalities Section 1.1 Linear Equations and Rational Equations e Linear; linear first solution conditional identity contradiction rational empty (or null); { } orφ a Linear; −2x = {−4} 12 − = 4+ x − 8= x {8} 11 −6x − = 20 −6x = 24 x = −4 {−4} 12 −8y + = 22 −8y = 16 −2x = −2 −2 x = −4 {−2} − x= c Linear; −2 − x = −2(8) x = −16 = − 12t = −12t 0= t {0} {− 16} 14 11 = − 2( 5p − 2) 11 = − 10p + d Nonlinear e Linear; x− 2= x − + = 8+ x = 10 11 = 11− 10p = −10p 0= p {10} {0} 4x 4x x {3} b Nonlinear c Linear; {48} y = −2 13 = − 3( 4t+ 1) = − 12t− b Nonlinear 10 a Linear; 12 = 12 = 3= 12 = 4+ x x 1 4(12) = 4 x 4 48 = x 12 = d Nonlinear 80 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 15 −6(v − 2) + = − (v + 4) −6v + 12 + = − v − 22 3 −6v + 15 = 5− v 22 4( y − 3) = 3 y + 2( y − 2) 4y − 12 = 3( y + 2y − 4) −5v = −10 v=2 {2} 4y − 12 = 3( 3y − 4) 4y − 12 = 9y − 12 16 −5(u − 4) + = 11− (u − 3) −5u + 20 + = 11− u + −5y = −5u + 22 = 14 − u {2} 17 {0} −4u = −8 u= 23 2.3 = 4.5x + 30.2 −27.9 = 4.5x −6.2 = x {− 6.2} 24 2.8 = p {2.8} 19 0.05y + 0.02( 6000 − y ) = 270 0.05y + 120 − 0.02y = 270 x− =1 5 1 6 x − = 6(1) 3 6 x − 10 = x = 16 {16} 0.03y + 120 = 270 0.03y = 150 y = 5000 25 20 0.06x + 0.04(10,000 − x) = 520 0.06x + 400 − 0.04x = 520 w − = w +2 3 1 2 12 w − = 12 w + 2 4 2 3 6w − = 8w + 24 −2w = 33 0.02x = 120 w =− x = 6000 {6000} 33 − 2 21 2( 5x − 6) = x − 3( x − 10) 10x − 12 = 4( x − 3x + 30) 10x − 12 = 4( −2x + 30) 10x − 12 = −8x + 120 18x = 132 x= x− = 3 1 4 x − = 4( 2) 2 4 x−6= x = 14 {14} 18 9.4 = 3.5p − 0.4 9.8 = 3.5p {5000} y=0 132 22 = 18 81 33 Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller 26 27 n+ n− 2 n+1 20 − = 20 − 1 10 5( n + 3) − ( n − 2) = 2( n + 1) − 20 p − = p −1 10 15 3 2 7 30 p − = 30 p − 1 10 5 15 12p − = 14p − 30 −2p = −21 21 p= 21 2 5n + 15 − 4n + = 2n + − 20 n + 23 = 2n − 18 − n = −41 {41} 30 y −1 y y + + = +1 y −1 y y+3 + = 20 + 1 20 4 4( y − 1) + 5y = 10( y + 3) + 20 4y − + 5y = 10y + 30 + 20 9y − = 10y + 50 {−54} 28 t− t+ t− − = +2 10 t− t+ t− 30 − = 30 + 2 10 10(t− 2) − 6(t+ 7) = 3(t− 4) + 60 10t− 20 − 6t− 42 = 3t− 12 + 60 4t− 62 = 3t+ 48 t= 110 {110} − y = 54 y = −54 31 a T = −1.83a + 212 T = −1.83( 4) + 212 = 204.68°F x − x x +1 + = +2 x−6 x x +1 21 + = 21 + 2 7 7( x − 6) + 3x = 7( x + 1) + 42 ≈ 205°F T = −1.83a + 212 193 = −1.83a + 212 −19 = −1.83a 10.4 ≈ a 10.4 × 103 = 10,400 Approximately10, 400 ft 32 a C = 167.95x + 94 b 7x − 42 + 3x = 7x + + 42 10x − 42 = 7x + 49 3x = 91 91 x= 29 n = 41 C = 167.95( 9) + 94 = $1605.55 b C = 167.95x + 94 91 3 n+ n− n+1 − = −1 10 2445.30 = 167.95x + 94 2351.30 = 167.95x 14 = x 14 credit-hours 33 S = 14.2t+ 149 362 = 14.2t+ 149 213 = 14.2t 15 = t 2004 + 15 = 2019 In 2019 82 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 34 41 − ( − 2w ) = 4(w + 1) − 2w − 10 −6 + 2w = 4w + − 2w − 10 S = 18t+ 232 628 = 18t+ 232 396 = 18t 22 = t −6 + 2w = 2w − 0= Identity; 42 −5 + 3x = 3( x − 1) − −5 + 3x = 3x − 3− 2000+ 22 = 2022 In 2022 35 a C = 7x b C = 105 7x = 105 x = 15 The motorist will save money beginning on the 16th working day 36 a C = 2.25x C = 89 b 2.25x = 89 x ≈ 39.6 The commuter will save money on the 40th ride 37 a S1 = 45,000 + 2250x −5 + 3x = 3x − 0= Identity; 1 x + 3= x +1 43 1 1 4 x + 3 = 4 x + 1 2 4 2x + 12 = x + x = −8 Conditional equation; {−8} 44 b S2 = 48,000 + 2000x c S1 = S2 45,000 + 2250x = 48,000 + 2000x 250x = 3000 x = 12 yr 38 a S1 = 25,000 + 0.16x y − 5= y − 2 1 6 y − 5 = 6 y − 4 3 6 4y − 30 = y − 24 3y = y=2 Conditional equation; {2} b S2 = 30,000 + 0.15x c S1 = S2 25,000 + 0.16x = 30,000 + 0.15x 0.01x = 5000 x = $500,000 39 2x − = 4( x − 1) − 1− 2x 45 + = x−5 x+ x ≠ 5,x ≠ −4 46 − = x+1 x− x ≠ −1,x ≠ 47 − = 2x − 5x − x − = 5x − x 2 x − 2 2x − = 4x − − 1− 2x 2x − = 2x − −3 = −5 Contradiction 40 4( − 5n) + = −4n − − 16n 12 − 20n + = −4n − − 16n x≠ −20n + 13 = −20n − 13 = −8 Contradiction 83 ,x ≠ 0,x ≠ Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller 48 − = 2x − x 4x − − = 5 2x − x 4 x − 4 x ≠ 0,x ≠ 6,x ≠ 49 3 c = − c− c− 3 c 4(c − 3) = 4(c − 3) − c − 3 c− 4 4c = 12 − 3( c − 3) 4c = 12 − 3c + y = 17 {} 7c = 21 c= ; The value does not check t= 25 7 d − = d−7 d−7 7 d 8( d − 7) − = 8( d − 7) d − 8 d − 7 56 − 7( d − 7) = 8d 54 w +3 w −5 + 1= 4w w w + w − 5 4w + 1 = 4w 4w w w + + 4w = 4(w − 5) 5w + = 4w − 20 w = −23 {−23} = t− t − 1 = t− (t+ 1)(t− 1) 55 56 − 7d + 49 = 8d {} (t+ 1)(t− 1) (t+ 1)(t− 1) t−11 = (t+ 1)(t− 1) {2} x+ x− + 1= 6x x x+ x − 7 6x + 1 = 6x 6x x x + + 6x = 6( x − 7) 7x + = 6x − 42 x = −44 {−44} 53 − = 3t t 1 7 3t − = 3t 3t t t− = 21 {25} 51 − = 2y y 1 5 2y − = 2y 2y y y − = 10 {17} 50 52 t+ = t= 84 −15d = −105 d=7 ; The value does not check Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller = w +2 w −4 = w + (w + 2)(w − 2) 56 (w + 2)(w − 2) (w + 2)(w − 2) w 1+ = (w + 2)(w − 2) w −2= w =7 {7} 11 − = x − x + x − 25 11 − = x − x + ( x + 5)( x − 5) 57 11 ( x + 5)( x − 5) ( x + 5)( x − 5) x − − x + = ( x + 5)( x − 5) {−4} 2( x + 5) − 1( x − 5) = 11 2x + 10 − x + = 11 x + 15 = 11 x = −4 10 − = c+ c− c − 10 − = c + c − ( c + 3)( c − 3) 58 10 ( c + 3)( c − 3) (c + 3)(c − 3) c + − c − = (c + 3)(c − 3) 2( c − 3) − 1( c + 3) = 10 2c − − c − = 10 c − = 10 {19} c = 19 85 Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller {} −14 − = x − x − 12 x − x + −14 − = ( x − 4)( x + 3) ( x − 4) ( x + 3) −14 ( x − 4)( x + 3) x − x + − x − = ( x − 4)( x + 3) x + )( ) ( ) ) ( ( −14 − ( x + 3) = 2( x − 4) −14 − x − = 2x − −17 − x = 2x − −3 = x ; The value −3 does not check {} 2 − = x + 5x + x + x + 2 − = ( x + 2)( x + 3) ( x + 2) ( x + 3) 2 ( x + 2)( x + 3) x + x + − x + = ( x + 2)( x + 3) x + )( ) ( ) ) ( ( − 2( x + 3) = ( x + 2) − 2x − = x + −4 − 2x = x + −2 = x ; The value −2 does not check 59 60 − = x − x − x − x + 3x + − = ( x − 2)( x + 1) ( x − 2)( x + 2) ( x + 2)( x + 1) ( x + 2)( x − 2)( x + 1) x − x + − x − x + = ( x + 2)( x − 2)( x + 1) x + x + )( ) ( )( ) )( ) ( ( 5( x + 2) − 2( x + 1) = 4( x − 2) 5x + 10 − 2x − = 4x − 3x + = 4x − 16 = x {16} 61 86 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller − = x − 2x − x − 16 x + 6x + − = ( x − 4)( x + 2) ( x − 4)( x + 4) ( x + 4)( x + 2) ( x + 4)( x − 4)( x + 2) x − x + − x − x + = ( x + 4)( x − 4)( x + 2) x + x + )( ) ( )( ) )( ) ( ( 4( x + 4) − 1( x + 2) = 2( x − 4) 4x + 16 − x − = 2x − 3x + 14 = 2x − x = −22 {−22} 62 63 64 3m = − m − m + 2m − m + 3m = − m − ( m + 4)( m − 2) m + 3m ( m + 4)( m − 2) m − = ( m + 4)( m − 2) m + m − − m + )( ) ( 5( m + 4) = 3m − 2( m − 2) 5m + 20 = 3m − 2m + 5m + 20 = m + 4m = −16 m = −4 { } ; The value −4 does not check 10 15n = − n − n − 2n − 24 n + 10 15n = − n − ( n − 6)( n + 4) n + 15n 10 ( n − 6)( n + 4) n − = ( n − 6)( n + 4) n − n + − n + )( ) ( 10( n + 4) = 15n − 6( n − 6) 10n + 40 = 15n − 6n + 36 10n + 40 = 9n + 36 n = −4 { } ; The value –4 does not check 87 Section 1.1 Solution Manual for College Algebra 2nd Edition By Miller 65 66 67 68 69 5x − = 3x − 5x − 3x + − x 5x −3 − = ( 3x + 1)( x − 2) 3x + x − 5x −3 ( 3x + 1)( x − 2) 3x + x − − 3x + 1 = ( 3x + 1)( x − 2) x − )( ) ( 5x − 1( x − 2) = −3( 3x + 1) 5x − x + = −9x − 4x + = −9x − 13x = −5 x=− 13 5 − 13 3x − = 2x + x − 2x + 1− x 3x −4 − = ( 2x + 3)( x − 1) 2x + x − 3x −4 ( 2x + 3)( x − 1) 2x + x − − 2x + 3 = ( 2x + 3)( x − 1) x − 1 )( ) ( 3x − 2( x − 1) = −4( 2x + 3) 3x − 2x + = −8x − 12 x + = −8x − 12 9x = −14 14 x= − 14 − 9 W = K − T forK A = lw forl 70 A lw W + T = K orK = W + T = 71 Δs = s2 − s1 fors1 w w A A Δs− s2 = −s1 = l orl= w w s1 = s2 − Δs E = IR forR 72 Δt= tf − ti forti E IR = Δt− tf = −ti I I ti = tf − Δt E E = R orR = I I P = a + b + c forc P − a − b = c orc = P − a − b 88 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 73 7x + 2y = fory 2y = −7x + 2y −7x + = 2 −7x + y= ory = − x + 2 74 3x + 5y = 15 fory 5y = −3x + 15 5y −3x + 15 = 5 −3x + 15 y= ory = − x + 5 75 5x − 4y = fory −4y = −5x + −4y −5x + = −4 −4 5x − y= ory = x − 4 76 7x − 2y = fory −2y = −7x + −2y −7x + = −2 −2 7x − y= ory = x − 2 1 x + y = fory 77 1 6 x + y = 6(1) 2 3x + 2y = 2y = −3x + 2y −3x + = 2 −3x + ory = − x + y= 2 78 x − y = fory 1 12 x − y = 12( 2) 4 3x − 8y = 24 −8y = −3x + 24 −8y −3x + 24 = −8 −8 3x − 24 ory = x − y= 8 n 79 S = ( a + d ) ford n 2( S ) = ( a + d ) 2 2S = n( a + d ) 2S = na + nd 2S − na = nd 2S − na nd = n n 2S − na =d n 2S − na 2S d= −a ord = n n n 80 S = 2a + ( n − 1) d fora n 2(S ) = 2a + ( n − 1) d 2 2S = n 2a + ( n − 1) d 2S = n ( 2a + nd − d ) 2S = 2an + n2d − nd 2S − n2d + nd = 2an 2S − n2d + nd 2an = 2n 2n 2S − n d + nd 2S − n2d + nd = a ora = 2n 2n 89 Section 1.2 Solution Manual for College Algebra 2nd Edition By Miller S Wave s x 11 = x − 10 x 11 = 9( x − 10) 9( x − 10) x − 10 9 9x = 11( x − 10) 9x = 11x − 110 2x = 480 x = 240 km 31 Let x represent the price set by the merchant x − 0.25x = 180 + 0.40(180) 0.75x = 180 + 72 0.75x = 252 x = $336 32 Let x represent the price set by the bookstore x − 0.10x = 80 + 0.35( 80) 0.90x = 80 + 28 0.90x = 108 x = $120 33 a C = 110 + 60x C = 350 b 110 + 60x = 350 x 60x = 240 x = hr 34 a C = 2400 + 80x C = 5520 b 2400 + 80x = 5520 x 80x = 3120 x = 39 hr 35 Let x represent the height of the Washington Monument x = 444 4x = 2220 x = 555 ft 36 Let x represent the height of the light post x = 300 km 30 Let x represent the distance from the epicentre to the station Distanc Rat Tim P Wave s x x 4.8 x x − = 20 4.8 x x − = 24( 20) 24 4.8 5x − 3x = 480 110 = 2x 55 = x x − 10 = 55 − 10 = 45 There were 55 Democrat and 45 Republican senators 27 Let x represent the number of deer in the population 30 = x 80 5x = 2400 x = 480 deer 28 Let x represent the number of bass in the lake 24 = x 40 4x = 960 x = 240 bass 29 Let x represent the distance from the epicenter to the station Distanc Rat Tim P Wave x s S Wave x s x x − = 40 x x 15 − = 15( 40) 5 5x − 3x = 600 2x = 600 4.8 x x light post x ft man ft 100 40 ft 20 ft Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller x = 1.5 + x + x x = 20 40 + 20 x = 20 60 20x = 360 x = 18 ft 37 Let x represent the height of the pole Then, x is the length of the pole that is in the ground, and x is the length of the pole that is in the snow x − x − x = 1.5 24 x − x − x = 24(1.5) 24x − 16x − 3x = 36 5x = 36 x = 7.2 The pole is 7.2 ft long, and the snow is 4.8 ft deep C = ( F − 32) 38 F = ( F − 32) 5 9( F ) = ( F − 32) 9 9F = 5F − 160 4F = −160 F = −40 −40°C = −40°F 39 Let x represent the amount of 20% fertilizer solution (in litres) to be drained (and therefore the amount of water to be added) 40 L is the amount of the resulting 15% fertilizer solution Therefore, ( 40 − x ) is the amount of 20% fertilizer solution that is not drained Amount of Solution Pure fertilizer 0% Solution x 20% Solution 0( x) 0.20( 40 − x) 40 − x 15% Solution 40 0.15( 40) 0( x) + 0.20( 40 − x) = 0.15( 40) − 0.20x = −0.20x = −2 x = 10 10 L should be drained and replaced by water 40 Let x represent the amount of water (in litres) to be evaporated Therefore, ( 200 − x) is the amount of the final 25% salt solution 0% Solution 5% Solution 25% Solution 200 − x 200 Amount of Solution x 0( x) 0.05( 200) 0.25( 200 − x) Pure salt 101 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.2 Solution Manual for College Algebra 2nd Edition By Miller 0( x) + 0.05( 200) = 0.25( 200 − x) 10 = 50 − 0.25x −40 = −0.25x 160 = x 160 mL should be evaporated 41 The length of the lot is l= 128 + 2x The width of the lot is w = 60 + 2x P = 2l+ 2w 440 = 2(128 + 2x) + 2( 60 + 2x) 440 = 256 + 4x + 120 + 4x 440 = 8x + 376 64 = 8x 8= x The width of the easement is ft 42 The length of the play area is l= 78 + 2x The width of the play area is w = 36 + 2x P = 2l+ 2w 396 = 2( 78 + 2x) + ( 36 + 2x) 396 = 156 + 4x + 72 + 4x 396 = 8x + 228 168 = 8x 21 = x The width of the border is 21 ft 43 a The width of the kitchen is w The length of the kitchen is l= w + P = 2l+ 2w 48 = 2(w + 4) + 2w 48 = 2w + + 2w 48 = 4w + 40 = 4w 10 = w The kitchen is 14 ft by 10 ft b A = lw + 0.1lw = 1.1lw A = 1.1(14)(10) = 154 ft2 c C = 1.06(12)(154) = $1958.88 44 a The width of the porch is w The length of the porch is l= 2w + P = 2l+ 2w 64 = 2( 2w + 2) + 2w 64 = 4w + + 2w 64 = 6w + 60 = 6w 10 = w l= 2w + = 2(10) + = 22 The porch is 22 ft by 10 ft b A = lw + 0.1lw = 1.1lw A = 1.1( 22)(10) = 242 ft2 c C = 1.075( 5.85)( 242) ≈ $1521.88 45 Aliyah had 8000 − 0.28( 8000) = 8000 − 2240 = 5760 to invest Let x represent the amount she invested at 11% Then, ( 5760 − x ) is the amount she invested at 5% 11% Investment 5% Investment Total 5760 − x Principal x Interest (I = Prt) x ( 0.11)(1) ( 5760 − x)( 0.05)(1) 453.60 102 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller x ( 0.11) + ( 5760 − x)( 0.05) = 453.60 0.11x + 288 − 0.05x = 453.60 0.06x + 288 = 453.60 0.06x = 165.60 x = 2760 5760 − x = 5760 − 2760 = 3000 Aliyah invested $2760 in the stock returning 11% and $3000 in the stock returning 5% 46 Let x represent the amount Caitlin invested in the balanced fund Then, ( 2x) is the amount she invested in the stock fund Balanced Fund (3.5%) Principal x Interest (I = Prt) x ( 0.035)(1) Stock Fund (17%) 2x ( 2x)( 0.17)(1) Total 1125 x ( 0.035) + ( 2x)( 0.17) = 1125 0.035x + 0.34x = 1125 0.375x = 1125 x = 3000 2x = 2( 3000) = 6000 Caitlin invested $3000 in the balanced fund and $6000 in the stock fund 103 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller larger number x + 25 = 4+ x x 1 x + 25 x = x + x x x + 25 = 4x + 24 = 3x 8= x x+ 25 = 8+ 25 x = 12.8 8x = 89.6 x = 11.2 12.8 = y 12 12.8y = 96 y = 7.5 x = 11.2 ft and y = 7.5 cm x 1.2 = 48 0.96 1.04 0.96x = 1.248 x = 1.3 0.5 1.2 = y 0.96 1.2y = 0.48 y = 0.4 x = 1.3 m and y = 0.4 in 49 No If x represents the measure of the smallest angle, then the equation x + ( x + 2) + ( x + 4) = 180 does not result in an odd integer value for x Instead the measures of the angles would be even integers 50 No If x represents the number of each type of bill, then the solution to the equation 20x + 10x + 5x = 100 is not a whole number 51 Let x represent the smaller number Then, ( x + 16) is the 47 = 33 The numbers are and 33 53 Let x represent the tens digit of the number Then, (14 − x) is the ones digit 10(14 − x) + 1( x) = 10( x) + 1(14 − x) + 18 140 − 10x + x = 10x + 14 − x + 18 140 − 9x = 9x + 32 108 = 18x 6= x 14 − x = 14 − =8 The original number is 68 54 Let x represent the tens digit of the number Then, ( − x) is the ones digit 10( − x) + 1( x) = 10( x) + 1( − x) − 45 90 − 10x + x = 10x + − x − 45 90 − 9x = 9x − 36 126 = 18x 7= x 9− x = 9− =2 The original number is 72 m 1x1 + m 2x2 = 55 larger number x + 16 = 3+ x x 2 x + 16 x = x 3+ x x x + 16 = 3x + 14 = 2x 7= x x+ 16 = + 16 ( 30)( −1.2) + ( 20) x2 = 56 20x2 = 36 x2 = 1.8 m m 1x1 + m 2x2 = ( 64) x1 + (80)( 2) = = 23 The numbers are and 23 52 Let x represent the smaller number Then, ( x + 25) is the 64x1 = −160 x1 = −2.5 m 104 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 57 m 1x1 + m 2x2 = 17 (10)( −3.2) + m (8) = −6 −14 = i ⋅ i 14 = i2 84 8m = −32 m = kg 58 = ( −1) 22 ⋅ 21 = −2 21 m 1x1 + m 2x2 = m ( −10) + ( 6)( 7) = −10m = −42 m = 4.2 kg 18 = i2 150 = ( −1) 52 ⋅ = −5 Section 1.3 Complex Numbers 19 −1 i b real; imaginary conjugate −121 = i 121 = 11i −100 = i 100 = 10i −98 = i 98 = 7i −63 − i 63 = 3i −19 = i 19 10 −10 −15 = i 10 ⋅ i 15 20 −23 = i 23 21 11 − −16 = − i 16 = −4i −98 −2 −45 −5 −63 12 − −25 = − i 25 = −5i 13 − −9 = i ⋅ i = 2i⋅ 3i= 6i2 = 6( −1) = −6 14 22 −1 −36 = i ⋅ i 36 = 1i⋅ 6i= 6i2 = 6( −1) = −6 15 −10 −5 = i 10 ⋅ i i 98 = i 45 = i 63 = i 80 i 98 = = 49 = i 45 = = 9=3 63 =i = i = 3i 80 =i = i 16 = 4i 23 Real part: 3; Imaginary part: −7 24 Real part: 2; Imaginary part: −4 25 Real part: 0; Imaginary part: 19 26 Real part: 0; Imaginary part: 40 27 Real part: − ; Imaginary part: 4 28 Real part: − ; Imaginary part: = i2 50 = ( −1) 52 ⋅ = −5 16 −80 = −6 −15 = i ⋅ i 15 = i2 90 = ( −1) 32 ⋅10 29 −4 = 4⋅ 2i = 8i= + 8i = −3 10 105 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller b i = i ⋅ i 30 −144 = 2⋅12i = 24i= + 24i 31 + −12 = + 3i or + 2i ( 29 28 = (1) ⋅ i1 = i c i = i ⋅ i 50 ) 32 − −24 = + −2 ior − 2i 48 = (1) ⋅ i2 = − + 3i = + i 14 14 14 = + i 14 + 5i 34 = + i 6 = + i −4 − 6i −4 −6 i 35 = + −2 −2 −2 = + 3i − 15i 15 36 = − i −3 −3 −3 = −3+ 5i 37 33 −41 = i−44 ⋅ i3 d i 42 a = (1) ⋅ i3 = − i i3 = b i = i ⋅ i 47 44 = (1) ⋅ i3 = − i 66 64 c i = i ⋅ i = (1) ⋅ i2 = − −27 = i−28 ⋅ i1 d i 43 a −18 + −48 −18 + 3i = 4 18 3i =− + 4 9 = − + 3ior − + i 2 −20 + −50 −20 + 2i 38 = −10 −10 −20 2i = + −10 −10 2 ior − i = 2− 2 14 − −98 14 − 2i 39 = −7 −7 14 2i =− + 7 = −2 + 2ior − + i 40 −10 + −125 −10 + 5i = 5 10 5i =− + 5 = −2 + 5ior − + i = (1) ⋅ i1 = i i37 = i36 ⋅ i1 −37 b i =i = i−40 ⋅ i3 = (1) ⋅ i3 = −i c i = i ⋅ i 82 80 = (1) ⋅ i2 = −1 −82 −84 d i = i ⋅ i = (1) ⋅ i2 = −1 44 a i = i ⋅ i 103 100 = (1) ⋅ i3 = − i −103 b i = i−104 ⋅ i1 = (1) ⋅ i1 = i c i = 52 − 52 d i = 45 ( − 7i) + ( 8− 3i) = ( + 8) + ( −7 − 3) i = 10 − 10i 46 ( − 10i) + ( 8+ 4i) 41 a i = 20 106 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller = ( + 8) + ( −10 + 4) i 55 2i( 5+ i) = 10i+ 2i = 10i+ 2( −1) = 14 − 6i 47 (15 + 21i) − (18 − 40i) = (15− 18) + 21− ( −40) i = −2 + 10i 56 4i( + 5i) = 24i+ 20i2 = ( 250 − 80) + (100 − 25) i 57 = −3+ 61i 48 ( 250 + 100i) − ( 80 + 25i) −3 = 170 + 75i 1 5 49 + i − + i 12 ( = 24i+ 20( −1) = −20 + 24i ) 11 − −7 = i ( 11 − i ) = i 33 − i 21 = i 33 − ( −1) 21 5 = − + − i 12 5 = − + − i 6 12 12 =− + i 12 =− + i 12 3 50 − i − + i 10 −2 58 ( ) = 21 + i 33 13 + −5 = i ( ) 13 + i = i 26 + i2 10 = i 26 + ( −1) 10 = − 10 + i 26 59 ( 3− 6i)(10 + i) = 3(10) + 3( i) + ( −6i)(10) + ( −6i)( i) = 30 + 3i− 60i− 6i2 = 30 − 57i− 6( −1) = 36 − 57i 60 ( − 5i)( 8+ 2i) 1 = − +− − i 10 7 4 = − +− − i 10 10 24 24 =− − i 10 24 51 ( 2.3 + 4i) − ( 8.1− 2.7i) + ( 4.6 − 6.7i) = 2( 8) + 2( 2i) + ( −5i)( 8) + ( −5i)( 2i) = 16 + 4i− 40i− 10i2 = 16 − 36i− 10( −1) = 26 − 36i = ( 2.3 − 8.1+ 4.6) + ( + 2.7 − 6.7) i 61 ( − 7i) = ( 3) − 2( 3)( 7i) + ( 7i) = −1.2 + 0i 0.05 −0.12 0.07 52 + − −0.03i +0.08i +0.05i = ( 0.05 − 0.12 − 0.07) 2 = − 42i+ 49i2 = − 42i+ 49( −1) = − 42i− 49 = −40 − 42i + ( −0.03+ 0.08 − 0.05) i = −0.14 + 0i 53 − (16 + 24i) = −2 − 3i 54 − ( 60 − 30i) = −10 + 5i 62 (10 − 3i) = (10) − 2(10)( 3i) + ( 3i) 2 = 100 − 60i+ 9i2 ( = 100 − 60i+ 9( −1) = 100 − 60i− = 91− 60i )( 63 3− −5 + −5 107 ) Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller ( )( ) = 3( 4) + 3( i 5) + ( −i 5) ( 4) + ( −i 5)( i 5) 69 a + 6i = 3− i + i b ( 3− 6i)( 3+ 6i) = ( 3) + ( 6) = + 36 b ( − 5i)( + 5i) = ( 4) + ( 5) = 16 + 25 = 12 − i − 5( −1) = 17 − i ( )( ) = ( + i )(10 + i 7) = 2(10) + 2( i 7) + i (10) 64 + −7 10 + −7 b ( − 8i)( + 8i) = ( 0) + ( 8) = + 64 ( ) b ( − 9i)( + i) = ( 0) + ( 9) = + 81 = 20 + 12i + 7( −1) 73 (10 − 4i)(10 + 4i) = (10) + ( 4) = 100 + 16 = 24 + 8i− 15i+ 35i2 74 ( 3− 9i)( 3+ 9i) = ( 3) + ( 9) = + 81 2 = −24 − 3i− 6( −1) = −18 − 3i 76 ( −5i)( 5i) = ( 5) = 25 2 77 = ( 2) − 2( 2)( i) + i2 + ( 2) ( )( + 3i = + 2( 2)( i) + i2 = − 4i+ i2 + + 4i+ i2 78 = + 2i ( ) + 3i ( ) + ( 3) 2 = 2+ 3= )( + 7i = = + 2( −1) = ) − 7i ( 5) + ( ) 2 = + = 12 = ( 3) − 2( 3)( 2i) + ( 2i) + ( 3) 2 = 90 75 ( 7i)( −7i) = = 49 = −24 + 9i− 12i− 6i2 2 = 116 = 24 − 7i+ 35( −1) = −11− 7i 66 −3( 8− 3i) − 6i( + i) 68 ( 3− 2i) + ( 3+ 2i) = 81 = 13+ 12i 65 4( + 2i) − 5i( 3− 7i) 2 = 64 72 a − 9i = 20 + 2i + 10i + 7i2 = 41 71 a − 8i +i i 67 ( − i) + ( + i) = 45 70 a + 5i = 12 + 3i − 4i − 5i2 2 + 2( 3)( 2i) + ( 2i) = − 12i+ 4i2 + + 12i+ 4i2 = 18 + 8i2 = 18 + 8( −1) = 10 108 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller 79 + 2i ( + 2i)( + i) = − i ( − i)( + i) = 18 + 6i+ 6i+ 2i2 = ( 3) + (1) 18 + 12i+ 2( −1) = ( ) = 16 + 19 + 9i = 17 19 = + i 17 17 − 5i (8 − 5i)(13 − 2i) 81 = 13 + 2i (13 + 2i)(13 − 2i) ) )( ) − 5i ( 6) + ( ) − 5i 36 + − 5i = 41 i = − 41 41 −1 84 − 3i = − 3i + 3i = − 3i + 3i = ( 104 − 16i− 65i+ 10i2 (13) + ( 2) 104 − 81i+ 10( −1) = ( ( ( 4) + (1) 20 + 9i+ 1( −1) = = 121+ 16 98 − 73i = 137 98 73 = − i 137 137 −1 83 + 5i = + 5i − 5i = + 5i − 5i 20 + 5i+ 4i+ i2 110 − 40i− 33i+ 12i2 (11) + ( 4) 110 − 73i+ 12( −1) = 9+1 16 + 12i = 10 16 12 = + i 10 10 = + i 5 5+ i ( + i)( + i) = 80 − i ( − i)( + i) = 10 − 3i (10 − 3i)(11− 4i) = 11+ 4i (11+ 4i)(11− 4i) 82 ) ( ( 169 + 94 − 81i = 173 94 81 = − i 173 173 = ) )( ) + 3i ( 4) + ( 3) + 3i 16 + + 3i = 19 = + i 19 19 = 109 Full file at https://TestbankDirect.eu/Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.3 Solution Manual for College Algebra 2nd Edition By Miller 5⋅ i = 13i 13i⋅ i 5i 5i = = 13i 13( −1) 5i = =− i 13 −13 = 0− i 13 6( − i) = 86 7i 7i( − i) −6i −6i = = −7i −7( −1) 6 = − i= − i 7 −1 −1 = 87 3i −3 −1⋅ 3i − 3i = = 3i⋅ 3i 3i − 3i − 3i = = 3( −1) −3 85 = 88 −2 −11 = = 91 89 92 b2 − 4ac = 93 a = 4i x + 25 = −25 + 25 = 0= b x2 + 25 = ( −5i) + 25 = 25( −1) + 25 = −25 + 25 = 0= 94 a x2 + 49 = ( 7i) + 49 = 49( −1) + 49 = −49 + 49 = 0= b x2 + 49 = ( −7i) + 49 = 49( −1) + 49 = − ( 2)( 6) −49 + 49 = 0= x2 − 4x + = 95 a = 4i ( −5) − 4( 2)( 4) = −32 = i 32 b2 − 4ac = ( 5i) + 25 = 25( −1) + 25 = = 16 − 48 90 ( 4) = −16 = i 16 11 11 i= + i 11 11 − 4( 2)( 5) = 16 − 32 11i −2 ⋅ 11i ( 4) = −4 = i = 2i 3i 3i = 0+ 3 −2 b2 − 4ac = ( −6) = 36 − 40 11i⋅ 11i −2 11i −2 11i = = 11( −1) 11i2 = b2 − 4ac = ( + i 3) − 4( + i 3) + = 2 − 4( 5)(10) + 4i + 3i2 − − 4i + = = 25 − 200 + 3( −1) − + = = −175 = i 175 − 3− = = 5i 0= 110 Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller b logic for simplification would be −9 ⋅ −4 = i ⋅ i x2 − 4x + = ( ) ( ) 2− i − 2− i + = = i2 36 − 4i + 3i − + 4i + = = −1⋅ = −6 100 The product ( a + b)( a − b) + 3( −1) − + = − 3− = simplifies to a2 − b2 The product (a + bi)( a − bi) simplifies to 0= 96 a x2 − 6x + 11 = a2 − (bi) , which simplifies to (3+ i 2) − 6(3+ i 2) + 11= a2 + b2 101 Any real number For example: 102 Any complex number and its conjugate For example: + 5i and − 5i In general, for real numbers, a and b, ( a + bi)( a − bi) = a2 + b2 , which is a + 6i + 2i2 − 18 − 6i + 11 = + 2( −1) − 18 + 11 = 9− 2− = 0= b real number 103 z ⋅ z = ( a + bi)( a − bi) = a2 + b2 x2 − 6x + 11 = ( ) ( ) 3− i − − i + 11 = 104 z2 − z2 − 6i + 2i − 18 + 6i + 11 = = ( a + bi) − ( a − bi) + 2( −1) − 18 + 11 = 2 = a2 + 2abi+ (bi) − a2 − 2abi+ (bi) = a2 + 2abi+ b2i2 − a2 + 2abi− b2i2 9− − = 0= = ( 4ab) i 97 ( a + bi)( c + di) 105 a x2 − = ( x + 3)( x − 3) = ac + adi+ bci+ bdi2 b x2 + = ( x + 3i)( x − 3i) = ac + ( ad + bc) i+ bd ( −1) 106 a x2 − 100 = ( x + 10)( x − 10) = ( ac − bd) + ( ad + bc) i 98 ( a + bi) = ( a) + 2( a)(bi) + (bi) = a2 + ( 2ab ) i+ b2i2 2 b x2 + 100 = ( x + 10i)( x − 10i) 107 a x2 − 64 = ( x + 8)( x − 8) b x2 + 64 = ( x + 8i)( x − 8i) = a + ( 2ab ) i+ b ( −1) ( 108 a x2 − 25 = ( x + 5)( x − 5) ) = a − b + ( 2ab ) i 2 b x2 + 25 = ( x + 5i)( x − 5i) 99 The second step does not follow because the multiplication property of radicals can be applied only if the individual radicals are real numbers Because −9 and −4 are imaginary numbers, the correct ( )( ) b x + = ( x + i 3)( x − i 3) 110 a x − 11 = ( x + 11)( x − 11) b x + 11 = ( x + i 11)( x − i 11) 109 a x2 − = x + x − 2 111 Section 1.4 Solution Manual for College Algebra 2nd Edition By Miller 111 113 112 114 10 Section 1.4 Quadratic Equation y2 + 7y − 18 = ( y + 9)( y − 2) = quadratic linear ± k 100 y + = or y − = y = −9 y=2 {−9,2} −b ± b2 − 4ac x = 2a b − 4ac ( x − 3)( x + 7) = x − = or x + = x=3 x = −7 { 11 8t(t+ 3) = 2t− 8t2 + 24t= 2t− 8t2 + 22t+ = ( 2t+ 5)( 4t+ 1) = 2t+ = or 4t+ = 2t= −5 4t= −1 t= − t= − 1 − ,− 4 } 3,− (t+ 4)(t− 1) = t+ = or t− = t = −4 t= {−4,1} 12 n + 5n = 24 y2 = 18 − 7y 6m ( m + 4) = m − 15 6m + 24m = m − 15 n + 5n − 24 = 6m + 23m + 15 = ( n + 8)( n − 3) = ( 6m + 5)( m + 3) = n + = or n − = n = −8 n= 6m + = or m + = m = −3 6m = −5 m =− − ,− 3 {−8,3} 13 112 40p2 − 90 = Chapter Equations and Inequalities Solution Manual for College Algebra 2nd Edition By Miller ( ) 10 4p2 − = 10( 2p − 3)( 2p + 3) = n − 4n + 2n − = 27 n2 − 2n − 35 = 2p − = or 2p + = 2p = 2p = −3 3 p= p=− 2 3 ,− 2 ( n + 5)( n − 7) = n + = or n − = 19 x2 = 81 32n − 162 = ( x = ± 81 = ±9 9,− ) 16n2 − 81 = { 2( 4n − 9)( 4n + 9) = 15 n= − w = ± 121 = ±11 {11,−11} 9 ,− 4 3x2 = 12x 21 5y2 − 35 = 5y2 = 35 y2 = 3x2 − 12x = 3x ( x − 4) = 3x = or x − = {0,4} x=0 { v2 = z = 25z z − 25z = { or z − 25 = } 5,− 4u2 = −64 u2 = −16 ( m + 4)( m − 5) = −8 u = ± −16 = ±4i {4i,−4i} m + 4m − 5m − 20 = −8 m − m − 12 = 24 8p2 + 72 = ( m + 3)( m − 4) = {−3,4} v=± 23 4u2 + 64 = z = 25 {0,25} 17 } 7,− 6v2 = 30 z( z − 25) = z= y=± 22 6v2 − 30 = x=4 16 } 20 w = 121 4n − = or 4n + = 4n = 4n = −9 n= n = −5 {−5,7} 14 ( n + 2)( n − 4) = 27 18 8p2 = −72 m + = or m − = m = −3 m =4 p2 = −9 p = ± −9 = ±3i {3i,−3i} 113 n= Section 1.4 Solution Manual for College Algebra 2nd Edition By Miller 25 ( k + 2) = 28 17 =± − 17 t= ± − i 17 = ± 2 17 i = ± 2 17 i ± t− k + = ± 28 k = −2 ± 28 = −2 ± {−2 ± 7} 26 3( z + 11) − 10 = 110 3( z + 11) = 120 ( z + 11) = 40 z + 11 = ± 40 47 1 30 a − = − 3 47 a− = ± − z = −11± 40 {−11± 10} = −11± 10 47 ± − i 47 = ± 3 47 = ± i 3 47 i ± a= 27 2(w − 5) + = 23 2(w − 5) = 18 w − 5= ± w = 5± w = 5± w = + or w = − w =8 {8,2} 28 ( c − 3) = 49 w =2 1 31 x + 14x + n = x + 14x + (14) 2 2 = x + 14x + ( 7) 2 c − = ± 49 = x2 + 14x + 49 c = 3± 49 = ( x + 7) c = 3± n = 49;( x + 7) c = 3+ or c = 3− c = 10 {10,−4} 2 c = −4 2 1 32 y + 22y + n = y + 22y + ( 22) 2 2 = y + 22y + (11) 2 17 1 29 t− = − 2 = y2 + 22y + 121 = ( y + 11) n = 121;( y + 11) 114 2 ... resulting 75% sand mixture 100% Sand 70% Sand 75% Sand 96 Full file at https://TestbankDirect.eu /Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution... 400 mph and the plane to New York City travels 460 mph 97 Full file at https://TestbankDirect.eu /Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Section 1.2 Solution Manual for College... 16(1.75) 16 2x + 5x = 28 7x = 28 x= The loop is mi 98 Full file at https://TestbankDirect.eu /Solution-Manual-for-College-Algebra-2nd-Edition-By-Miller Chapter Equations and Inequalities Solution