CHAPTER 13 SOLVING SIMULTANEOUS EQUATIONS EXERCISE 51 Page 105 1. Solve the simultaneous equations 2 x − y = 6 x+y=6 (1) + (2) gives: 2x − y = 6 (1) x+y=6 (2) 3x = 12 From (1): 8–y=6 i.e. 8–6=y from which, x= 12 =4 3 from which, y = 2 2. Solve the simultaneous equations 2x – y = 2 x – 3y = –9 3 × (1) gives: (3) – (2) gives: 2x – y = 2 (1) x – 3y = –9 (2) 6x – 3y = 6 5x = 15 From (1): 6–y=2 i.e. 6–2=y (3) from which, x= 15 =3 5 from which, y = 4 3. Solve the simultaneous equations x – 4y = – 4 5x – 2y = 7 x – 4y = –4 2 × (2) gives: (3) – (1) gives: (1) 5x – 2y = 7 (2) 10x – 4y = 14 (3) 9x = 18 from which, 180 x= 18 =2 9 © 2014, John Bird From (1): 2 – 4y = – 4 i.e. 2 + 4 = 4y i.e. 6 = 4y from which, y = 6 = 1.5 4 4. Solve the simultaneous equations 3x – 2y = 10 5x + y = 21 2 × (2) gives: (1) + (3) gives: 3x – 2y = 10 (1) 5x + y = 21 (2) 10x + 2y = 42 (3) 13x = 52 from which, From (1): 12 – 2y = 10 i.e. 12 – 10 = 2y i.e. 2 = 2y x= from which, y = 52 =4 13 2 =1 2 5. Solve the simultaneous equations 5p + 4q = 6 2p – 3q = 7 5p + 4q = 6 (1) 2p – 3q = 7 (2) 3 × (1) gives: 15p + 12q = 18 (3) 4 × (2) gives: 8p – 12q = 28 (4) (3) + (4) gives: From (1): 23p = 46 from which, p= 46 =2 23 10 + 4q = 6 i.e. 4q = 6 – 10 i.e. 4q = –4 181 from which, q = −4 = –1 4 © 2014, John Bird 6. Solve the simultaneous equations 7x + 2y = 11 3x – 5y = –7 5 × (1) gives: 2 × (2) gives: (3) + (4) gives: From (1): 7x + 2y = 11 (1) 3x – 5y = –7 (2) 35x + 10y = 55 (3) 6x – 10y = –14 41x = 41 (4) from which, x= 41 =1 41 y= 4 =2 2 y= 58 =2 29 x= 6 =3 2 7 + 2y = 11 i.e. 2y = 11 – 7 i.e. 2y = 4 from which, 7. Solve the simultaneous equations 2x – 7y = –8 3x + 4y = 17 2x – 7y = –8 (1) 3x + 4y = 17 (2) 3 × (1) gives: 6x – 21y = –24 (3) 2 × (2) gives: 6x + 8y = 34 (4) (4) – (3) gives: From (1): 29y = 58 from which, 2x – 14 = –8 i.e. 2x = 14 – 8 i.e. 2x = 6 from which, 8. Solve the simultaneous equations a + 2b = 8 b – 3a = –3 182 © 2014, John Bird Rearranging gives: 2 × (2) gives: a + 2b = 8 (1) – 3a + b = –3 (2) – 6a + 2b = –6 (3) (1) – (3) gives: From (1): 7a = 14 from which, a= 14 =2 7 2 + 2b = 8 i.e. 2b = 8 – 2 i.e. 2b = 6 from which, b= 6 =3 2 9. Solve the simultaneous equations a + b = 7 a–b=3 a+b=7 (1) a–b=3 (2) (1) + (2) gives: 2a = 10 From (1): from which, a= 10 =5 2 5+b=7 i.e. b=7–5 from which, b=2 10. Solve the simultaneous equations 2x + 5y = 7 x + 3y = 4 2 × equation (2) gives: (3) – (1) gives: Substituting in (1) gives: 2x + 5y = 7 (1) x + 3y = 4 (2) 2x + 6y = 8 (3) y=1 2x + 5 = 7 i.e. 2x = 7 – 5 = 2 and x = 2 =1 2 Thus, x = 1 and y = 1 and may be checked by substituting into both of the original equations 183 © 2014, John Bird 11. Solve the simultaneous equations 3s + 2t = 12 4s – t = 5 2 × equation (2) gives: (1) + (3) gives: Substituting in (1) gives: 3s + 2t = 12 (1) 4s – t = 5 (2) 8s – 2t = 10 (3) 11s = 22 6 + 2t = 12 from which, i.e. s= 22 =2 11 2t = 12 – 6 = 6 and t= 6 =3 2 Thus, s = 2 and t = 3 and may be checked by substituting into both of the original equations 12. Solve the simultaneous equations 3x – 2y = 13 2x + 5y = –4 3x – 2y = 13 (1) 2x + 5y = –4 (2) 2 × equation (1) gives: 6x – 4y = 26 (3) 3 × equation (2) gives: 6x + 15y = –12 (4) (3) – (4) gives: Substituting in (1) gives: –19y = 38 3x + 4 = 13 i.e. from which, y= 3x = 13 – 4 = 9 38 = –2 −19 and x= 9 =3 3 Thus, x = 3 and y = –2 and may be checked by substituting into both of the original equations 13. Solve the simultaneous equations 5m – 3n = 11 3m + n = 8 3 × equation (2) gives: 5m – 3n = 11 (1) 3m + n = 8 (2) 9m + 3n = 24 (3) 184 © 2014, John Bird (1) + (3) gives: 14m = 35 Substituting in (1) gives: 12.5 – 3n = 11 i.e. 12.5 – 11 = 3n i.e. 1.5 = 3n from which, and m= n= 35 = 2.5 14 1.5 = 0.5 3 14. Solve the simultaneous equations 8a – 3b = 51 3a + 4b = 14 8a – 3b = 51 (1) 3a + 4b = 14 (2) 4 × (1) gives: 32a – 12b = 204 (3) 3 × (2) gives: 9a + 12b = 42 (4) (3) + (4) gives: 41a = 246 From (1): 48 – 3b = 51 i.e. 48 – 51 = 3b i.e. –3 = 3b from which, a= 246 =6 41 from which, b= −3 = –1 3 15. Solve the simultaneous equations 5x = 2y 3x + 7y = 41 5x – 2y = 0 (1) 3x + 7y = 41 (2) 3 × equation (1) gives: 15x – 6y = 0 (3) 5 × equation (2) gives: 15x + 35y = 205 (4) (4) – (3) gives: Substituting in (1) gives: 41y = 205 5x – 10 = 0 i.e. from which, 5x = 10 185 y= and 205 =5 41 x= 10 =2 5 © 2014, John Bird 16. Solve the simultaneous equations 5c = 1 – 3d 2d + c + 4 = 0 Rearranging gives: 5 × equation (2) gives: (3) – (1) gives: Substituting in (1) gives: 5c + 3d = 1 (1) c + 2d = –4 (2) 5c + 10d = –20 (3) 7d = –21 5c – 9 = 1 i.e. from which, 5c = 10 186 and d= −21 = –3 7 c= 10 =2 5 © 2014, John Bird EXERCISE 52 Page 107 1. Solve the simultaneous equations 7p + 11 + 2q = 0 –1 = 3q – 5p Rearranging gives: 7p + 2q = –11 (1) 5p – 3q = 1 (2) 3 × equation (1) gives: 21p + 6q = –33 (3) 2 × equation (2) gives: 10p – 6q = 2 (4) (3) + (4) gives: 31p = –31 Substituting in (1) gives: –7 + 2q = –11 2. Solve the simultaneous equations from which, i.e. p=–1 2q = –11 + 7 = –4 and q= x y + =4 2 3 y x – =0 9 6 x y Rearranging gives: (6) + (6) = (6)(4) 2 3 i.e. 3x + 2y = 24 (1) x y (18) − (18) = (18)(0) 6 9 i.e. 3x – 2y = 0 (2) and (1) – (2) gives: 4y = 24 Substituting in (1) gives: 3x + 12 = 24 3. Solve the simultaneous equations Rearranging gives: −4 = –2 2 i.e. from which, 3x = 24 – 12 = 12 and y=6 x= 12 =4 3 a – 7 = –2b 2 2 12 = 5a + b 3 a (2) − (2)7 = −(2)(2b) 2 i.e. 187 a + 4b = 14 (1) © 2014, John Bird 2 (3)(12) = (3)(5a) + (3) b 3 and i.e. 2 × equation (2) gives: 15a + 2b = 36 (2) 30a + 4b = 72 (3) (3) – (1) gives: 29a = 58 Substituting in (1) gives: 2 + 4b = 14 4. Solve the simultaneous equations i.e. 4b = 14 – 2 = 12 from which, and b= a=2 12 =3 4 3 s – 2t = 8 2 s + 3t = –2 4 Rearranging gives: 3 (2) s − (2)2t = (2)(8) 2 i.e. 3s – 4t = 16 (1) and s (4) + (4)3t =(4)(−2) 4 i.e. s + 12t = –8 (2) 3s + 36t = –24 (3) 3 × equation (2) gives: (1) – (3) gives: – 40t = 40 from which, t= Substituting in (1) gives: 3s + 4 = 16 5. Solve the simultaneous equations i.e. (14) 3s = 16 – 4 = 12 i.e. t = –1 and s= 12 =4 3 x 2y 49 + = 5 3 15 y 3x 5 – + =0 2 7 7 x 2y 49 Rearranging gives: (15) + (15) = (15) 5 3 15 and − 40 40 i.e. 3x + 10y = 49 3x 5 y i.e. 0 − (14) + (14) = 7 2 7 2 × equation (1) gives: 6x – 7y = –10 6x + 20y = 98 188 (1) (2) (3) © 2014, John Bird (3) – (2) gives: 27y = 108 Substituting in (1) gives: 3x + 40 = 49 i.e. from which, y = 3x = 49 – 40 = 9 and x= 108 =4 27 9 =3 3 u 12 v 25 u+ – =0 4 2 6. Solve the simultaneous equations v – 1 = u Rearranging gives: (12)v − (12)(1) = (12) 12 i.e. 12v – u = 12 (1) v 25 i.e. (4)u + (4) − (4) 0 = 4 2 v + 4u = 50 (2) 4 × equation (1) gives: 48v – 4u = 48 (3) (2) + (3) gives: 49v = 98 and Substituting in (1) gives: 24 – u = 12 i.e. 24 – 12 = u from which, v = and 98 =2 49 u = 12 7. Solve the simultaneous equations 1.5x – 2.2y = –18 2.4x + 0.6y = 33 Multiplying both equations by 10 gives: 15x – 22y = –180 6 × equation (1) gives: 22 × equation (2) gives: (1) 24x + 6y = 330 (2) 90x – 132y = –1080 (3) 528x + 132y = 7260 (3) + (4) gives: Substituting in (1) gives: 150 – 22y = –180 618x = 6180 (4) from which, x = 10 i.e. –22y = –180 – 150 = –330 and y = −330 = 15 −22 Thus, x = 10 and y = 15 and may be checked by substituting into both of the original equations. 189 © 2014, John Bird 8. Solve the simultaneous equations 3b – 2.5a = 0.45 1.6a + 0.8b = 0.8 Multiplying equation (1) by 100 gives: – 250a +300b = 45 (1) Multiplying equation (1) by 10 gives: (2) 4 × equation (1) gives: 150 × equation (2) gives: 16a +8b = 8 –1000a + 1200b = 180 (3) 2400a + 1200b = 1200 (4) – (3) gives: 3400a = 1020 from which, Substituting in (1) gives: –75 + 300b = 45 (4) a= 1020 = 0.3 3400 i.e. 300b = 45 + 75 = 120 and b = 190 120 = 0.4 300 © 2014, John Bird EXERCISE 53 Page 109 1. Solve the simultaneous equations 2 3 + = 14 y x 3 5 – = –2 y x Let 1 = a and x 1 =b y Thus, the equations become: 3a + 2b = 14 (1) and 5a – 3b = –2 (2) 3 × equation (1) gives: 9a + 6b = 42 (3) 2 × equation (2) gives: 10a – 6b = –4 (4) (3) + (4) gives: Substituting in (1) gives: Since 19a = 38 6 + 2b = 14 1 1 1 = a then x = = a 2 x Thus, x = i.e. and since from which, 2b = 14 – 6 = 8 a= and 38 =2 19 b= 8 =4 2 1 1 1 = b then y = = y b 4 1 1 and y = and may be checked by substituting into both of the original equations 4 2 2. Solve the simultaneous equations 4 3 – = 18 a b 2 5 + = –4 a b Let 1 1 = x and =y a b Thus, the equations become: 4x – 3y = 18 (1) and 2x + 5y = –4 (2) 4x + 10y = –8 (3) 2 × equation (2) gives: 191 © 2014, John Bird (1) – (3) gives: –13y = 26 Substituting in (1) gives: Since 4x + 6 = 18 1 1 1 = x then a = = a x 3 Thus, a = i.e. from which, 4x = 18 – 6 = 12 26 = –2 −13 y= and x= 12 =3 4 1 1 1 1 = y then b = = = − b y −2 2 and since 1 1 and b = − and may be checked by substituting into both of the original equations 2 3 3. Solve the simultaneous equations 1 3 + =5 5q 2p 5 1 35 – = p 2q 2 Let 1 1 = a and =b p q Thus, the equations become: 1 3 a+ b= 5 2 5 1 35 5a − b = 2 2 Rearranging gives: 1 3 (10) a + (10) b = (10)(5) 2 5 i.e. 5a + 6b = 50 1 35 (2)(5a ) − (2) b = (2) 2 2 i.e. 10a – b = 35 Thus, 5a + 6b = 50 (1) and 10a – b = 35 (2) 10a + 12b = 100 (3) 2 × equation (1) gives: (3) – (2) gives: Substituting in (1) gives: Since 13b = 65 5a + 30 = 50 1 1 1 = a then p = = p a 4 Thus, p = and since i.e. from which, 5a = 50 – 30 = 20 b= 65 =5 13 and a= 20 =4 5 1 1 1 = b then q = = q b 5 1 1 and q = and may be checked by substituting into both of the original equations 4 5 192 © 2014, John Bird 4. Solve the simultaneous equations 3 5 + = 1.1 y x 3 7 – = –1.1 y x Let 1 = a and x 1 =b y Thus, the equations become: 5a + 3b = 1.1 (1) and 3a – 7b = –1.1 (2) 3 × equation (1) gives: 15a + 9b = 3.3 (3) 5 × equation (2) gives: 15a – 35b = –5.5 (4) (3) – (4) gives: 44b = 8.8 Substituting in (1) gives: Since 5a + 0.6 = 1.1 1 1 = a then x = = 10 a x and since i.e. from which, b= 5a = 1.1 – 0.6 = 0.5 8.8 1 = 44 5 and a= 0.5 1 = 5 10 1 1 = b then y = = 5 y b Thus, x = 10 and y = 5 and may be checked by substituting into both of the original equations 5. Solve the simultaneous equations c +1 d + 2 – +1=0 3 4 13 1− c 3 − d + + =0 5 4 20 Rearranging gives: and (12) (20) c +1 d +2 − (12) + (12)(1) = 0 i.e. 3(c + 1) – 4(d + 2) + 12 = 0 4 3 1− c 3− d 13 + (20) + (20) = 0 i.e. 4(1 – c) + 5(3 – d) + 13 = 0 5 4 20 Since 3(c + 1) – 4(d + 2) + 12 = 0 then 3c + 3 – 4d – 8 + 12 = 0 i.e. 3c – 4d = –7 and then 4 – 4c + 15 – 5d + 13 = 0 i.e. 4c + 5d = 32 Thus, 4(1 – c) + 5(3 – d) + 13 = 0 3c – 4d = –7 (1) 4c + 5d = 32 (2) 193 © 2014, John Bird 4 × equation (1) gives: 12c – 16d = –28 (3) 3 × equation (2) gives: 12c + 15d = 96 (4) (4) – (3) gives: 31d = 124 Substituting in (1) gives: 3c – 16 = –7 from which, i.e. d= 124 =4 31 3c = 16 – 7 = 9 and c= 9 =3 3 Thus, c = 3 and d = 4 and may be checked by substituting into both of the original equations 3r + 2 2 s − 1 11 – = 5 5 4 6. Solve the simultaneous equations 3 + 2r 5 − s 15 + = 3 4 4 Rearranging gives: (20) and 3r + 2 2s − 1 11 i.e. 4(3r + 2) – 5(2s – 1) = 44 − (20) = (20) 5 4 5 (12) 3 + 2r 5− s 15 i.e. + (12) = (12) 4 3 4 Since 4(3r + 2) – 5(2s – 1) = 44 then and 3(3 + 2r) + 4(5 – s) = 45 Thus, 2 × equation (2) gives: (3) – (1) gives: Thus, r = 3 and s = 12r + 8 – 10s + 5 = 44 then 9 + 6r + 20 – 4s = 45 i.e. 12r – 10s = 31 i.e. 12r – 10s = 31 (1) 6r – 4s = 16 (2) 12r – 8s = 32 (3) 2s = 1 Substituting in (1) gives: 3(3 + 2r) + 4(5 – s) = 45 12r – 5 = 31 from which, i.e. s= 6r – 4s = 16 1 2 12r = 31 + 5 = 36 and r= 36 =3 12 1 and may be checked by substituting into both of the original equations 2 194 © 2014, John Bird 5 20 = x+ y 27 7. Solve the simultaneous equations 4 16 = 2x − y 33 Multiplying both sides of the first equation by 27(x + y) gives: i.e. 27( x + y ) 5 20 = 27( x + y ) x+ y 27 i.e. 135 = 20x + 20y i.e. 20x + 20y = 135 (1) Multiplying both sides of the second equation by 33(2x – y) gives: i.e. 33(2 x − y ) 4 16 = 33(2 x − y ) 2x − y 33 i.e. 132 = 32x – 16y i.e. 32x – 16y = 132 (2) 4 × equation (1) gives: 80x + 80y = 540 (3) 5 × equation (2) gives: 160x – 80y = 660 (4) (3) + (4) gives: 240x = 1200 Substituting in (1) gives: 8. If 5x – If 5x – i.e. x= 20y = 135 – 100 = 35 1200 =5 240 and y= 35 3 =1 20 4 4 3 xy + 1 5 = 1 and x + = find the value of y y y 2 3 =1 y and if x + 100 + 20y = 135 from which, 4 5 = y 2 then 5x = 3 +1 y then x = − and x= 3 1 + 5y 5 4 5 + y 2 3 1 4 5 + =− + y 2 5y 5 i.e. i.e. 3 + 20 25 − 2 = 5y 10 i.e. 23 23 = 5 y 10 and (23)(10) = (5y)(23) i.e. 230 = 115y Equating x values gives: from which, y= 3 4 5 1 + = − 5y y 2 5 230 =2 115 195 © 2014, John Bird Substituting into the first equation gives: 5x – Thus, 3 =1 2 i.e. 5x = 2.5 and x = 2.5 1 = 5 2 1 ( 2) + 1 1 + 1 xy + 1 2 = =1 = y 2 2 196 © 2014, John Bird EXERCISE 54 Page 112 1. In a system of pulleys, the effort P required to raise a load W is given by P = aW + b, where a and b are constants. If W = 40 when P = 12 and W = 90 when P = 22, find the values of a and b. P = aW + b, hence if W = 40 when P = 12, then: 12 = 40a + b (1) and 22 = 90a + b (2) if W = 90 when P = 22, then: Equation (2) – equation (1) gives: 10 = 50a from which, a= 10 1 or 0.2 = 50 5 1 12 = 40 + b 5 Substituting in (1) gives: from which, i.e. 12 = 8 + b b=4 Thus, a = 0.2 and b = 4 and may be checked by substituting into both of the original equations 2. Applying Kirchhoff's laws to an electrical circuit produces the following equations: 5 = 0.2I1 + 2(I1 – I 2 ) 12 = 3I 2 + 0.4I 2 – 2(I 1 – I 2 ) Determine the values of currents I 1 and I 2 Rearranging 5 = 0.2 I1 + 2 ( I1 − I 2 ) gives: 5= 0.2 I1 + 2 I1 − 2 I 2 i.e. 2.2 I1 − 2 I 2 = 5 Rearranging 12 =3I 2 + 0.4 I 2 − 2 ( I1 − I 2 ) gives: 12 = 3I 2 + 0.4 I 2 − 2 I1 + 2 I 2 i.e. −2 I1 + 5.4 I 2 = 12 Thus, and 2 × equation (1) gives: 2.2 × equation (2) gives: (3) + (4) gives: 2.2 I1 − 2 I 2 = 5 (1) −2 I1 + 5.4 I 2 = 12 (2) 4.4 I1 − 4 I 2 = 10 (3) −4.4 I1 + 11.88 I 2 = 26.4 7.88 I 2 = 36.4 197 (4) from which, I 2 = 36.4 = 4.62 7.88 © 2014, John Bird Substituting in (1) gives: 2.2 I1 − 9.24 = 5 i.e. 2.2 I1 = 14.24 and I1 = 14.24 = 6.47 2.2 Thus, I1 = 6.47 and I 2 = 4.62 and may be checked by substituting into both of the original equations 3. Velocity v is given by the formula v = u + at. If v = 20 when t = 2 and v = 40 when t = 7, find the values of u and a. Hence find the velocity when t = 3.5 v = u + at, hence if v = 20 when t = 2, then: 20 = u + 2a (1) and 40 = u + 7a (2) if v = 40 when t = 7, then: Equation (2) – equation (1) gives: 20 = 5a from which, a= Substituting in (1) gives: 20 =4 5 20 = u + 8 from which, u = 12 Thus, a = 4 and u = 12 and may be checked by substituting into both of the original equations When t = 3.5, velocity, v = u + at = 12 + (4)(3.5) = 26 4. Three new cars and four new vans supplied to a dealer together cost £97 700 and five new cars and two new vans of the same models cost £103 100. Find the cost of a car and a van. Let a car = C and a van = V then 2 × equation (2) gives: (3) – (1) gives: Substituting in (1) gives: from which, 3C + 4V = 97 700 (1) 5C + 2V = 103 100 (2) 10C + 4V = 206 200 (3) 7C = 108 500 from which, C = 46 500 + 4V = 97 700 i.e. 4V = V= 108 500 = 15 500 7 108 500 97 700 – 46 500 = 51 200 7 51 200 = 12 800 4 Hence, a car costs £15 500 and a van costs £12 800 198 © 2014, John Bird 5. y = mx + c is the equation of a straight line of slope m and y-axis intercept c. If the line passes through the point where x = 2 and y = 2, and also through the point where x = 5 and y = 0.5, find the slope and y-axis intercept of the straight line. When x = 2 and y = 2, then When x = 5 and y = 0.5, then 2 = 2m + c (1) 0.5 = 5m + c (2) i.e. m = 1.5 = –0.5 −3 (1) – (2) gives: 1.5 = –3m In equation (1): 2 = –1 + c i.e. c = 2 + 1 = 3 Hence, slope m = –0.5 and the y-axis intercept c = 3 6. The resistance R ohms of copper wire at t°C is given by R = R 0 (1 + αt), where R0 is the resistance at 0°C and α is the temperature coefficient of resistance. If R = 25.44 Ω at 30°C and R = 32.17 Ω at 100°C, find α and R 0 R = R0 (1 + α t ) thus when R = 25.44 Ω at t = 30°C, then: and when R = 32.17 Ω at t = 100°C, then: Dividing equation (2) by equation (1) gives: i.e. 32.17 (1 + 100α ) = 25.44 (1 + 30α ) and i.e. i.e. from which, Substituting in (1) gives: from which, 25.44 = R0 (1 + 30α ) (1) 32.17 = R0 (1 + 100α ) (2) 32.17 R0 (1 + 100α ) = 25.44 R0 (1 + 30α ) (32.17)(1 + 30α= ) (25.44)(1 + 100α ) 32.17 + 965.1α = 25.44 + 2544α 32.17 – 25.44 = 2544α – 965.1α 6.73 = 1578.9α α= 6.73 = 0.00426 1578.9 25.44 = R0 [1 + (30)(0.00426) ] = R0 (1.1278 ) 199 © 2014, John Bird from which, R0 = 25.44 = 22.56 1.1278 Thus, α = 0.00426 and R0 = 22.56 Ω and may be checked by substituting into both of the original equations. 7. The molar heat capacity of a solid compound is given by the equation c = a + bT. When c = 52, T = 100 and when c = 172, T = 400. Find the values of a and b. c = a + bT, hence if c = 52 when T = 100, then: and if c = 172 when T = 400, then: Equation (2) – equation (1) gives: 52 = a + 100b (1) 172 = a + 400b (2) 120 = 300b 120 = 0.40 300 from which, b= Substituting in (1) gives: 52 = a + 40 from which, a = 12 Thus, a = 12 and b = 0.40 and may be checked by substituting into both of the original equations 8. In an engineering process two variables p and q are related by: q = ap + b/p, where a and b are constants. Evaluate a and b if q = 13 when p = 2 and q = 22 when p = 5 If q = 13 when p = 2 then 13 = 2a + b/2 or 26 = 4a + b If q = 22 when p = 5 then 22 = 5a + b/5 or 110 = 25a + b (2) – (1) gives: 84 = 21a from which, a = Substituting in (1) gives: 26 = 16 + b (1) (2) 84 =4 21 i.e. b = 26 – 16 = 10 9. In a system of forces, the relationship between two forces F1 and F2 is given by: 5F1 + 3F2 + 6 = 0 3F1 + 5F2 + 18 = 0 Solve for F1 and F2 5F1 + 3F2 + 6 = 0 (1) 200 © 2014, John Bird 3F1 + 5F2 + 18 = 0 (2) 5 × (1) gives: 25F1 + 15F2 + 30 = 0 (3) 3 × (1) gives: 9F1 + 15F2 + 54 = 0 (4) (3) – (4) gives: 16F1 – 24 = 0 i.e. 16F1 = 24 Substituting in (1) gives: 7.5 + 3F2 + 6 = 0 from which, F2 = from which, F1 = 24 = 1.5 16 i.e. 3F2 = –7.5 – 6 = –13.5 −13.5 = –4.5 3 10. For a balanced beam, the equilibrium of forces is given by: R1 + R2 = 12.0 kN As a result of taking moments: 0.2 R1 + 7 × 0.3 + 3 × 0.6 = 0.8 R2 Determine the values of the reaction forces R1 and R2 Rearranging gives: 5 × (2) gives: (1) – (3) gives: from which, Substituting in (1) gives: Hence, R1 + R2 = 12.0 (1) 0.2 R1 − 0.8 R2 = −3.9 (2) R1 − 4.0 R2 = −19.5 (3) 5.0 R2 = 31.5 R2 = 31.5 = 6.3 kN 5 R1 + 6.3 = 12.0 R1 = 12.0 – 6.3 = 5.7 kN 201 © 2014, John Bird EXERCISE 55 Page 113 1. Solve the simultaneous equations x + 2 y + 4z = 16 2 x − y + 5z = 18 3x + 2 y + 2 z = 14 (1) – (3) gives: 2 × (2) gives: (1) + (5) gives: 7 × (4) gives: (6) – (7) gives: Substituting in (6) gives: i.e. (1) 2 x − y + 5z = 18 (2) 3x + 2 y + 2 z = 14 (3) –2x + 2z = 2 i.e. (4) 4x – 2y + 10z = 36 (5) 5x + 14z = 52 (6) –14x + 14z = 14 (7) 19x = 38 from which, x = 38 =2 19 from which, z = 42 =3 14 10 + 14z = 52 14z = 52 – 10 = 42 Substituting in (1) gives: 2. x + 2 y + 4z = 16 2 + 2y + 12 =16 2y = 16 – 2 – 12 = 2 and y = Solve the simultaneous equations 2 =1 2 2x + y − z = 0 3x + 2 y + z = 4 5x + 3 y + 2 z = 8 (1) + (2) gives: 2 × (1) gives: 2x + y − z = 0 (1) 3x + 2 y + z = 4 (2) 5x + 3 y + 2 z = 8 (3) 5x + 3y = 4 (4) 4x + 2y – 2z = 0 (5) 202 © 2014, John Bird (3) + (5) gives: 9x + 5y = 8 (6) 5 × (4) gives: 25x + 15y = 20 (7) 3 × (6) gives: 27x + 15y = 24 (8) (8) – (7) gives: Substituting in (4) gives: i.e. i.e. from which, x = 4 =2 2 from which, y = −6 = –2 3 10 + 3y = 4 3y = 4 – 10 = –6 Substituting in (1) gives: 3. 2x = 4 4 + –2 – z = 0 z=2 Solve the simultaneous equations 3x + 5 y + 2 z = 6 x − y + 3z = 0 2 x + 7 y + 3z = −3 3x + 5 y + 2 z = 6 (1) x − y + 3z = 0 (2) 2 x + 7 y + 3z = −3 (3) x + 8y = –3 (4) 3 × (1) gives: 9x + 15y + 6z = 18 (5) 2 × (2) gives: 2x – 2y + 6z = 0 (6) (3) – (2) gives: (5) – (6) gives: 7x + 17y = 18 (7) 7 × (4) gives: 7x + 56y = – 21 (8) (7) – (8) gives: Substituting in (7) gives: i.e. Substituting in (1) gives: –39y = 39 from which, y = 39 = –1 − 39 7x – 17 = 18 7x = 18 + 17 = 35 from which, x = 35 =5 7 15 – 5 + 2z = 6 203 © 2014, John Bird i.e. 4. 2z = 6 – 15 + 5 = –4 Solve the simultaneous equations from which, z = −4 = –2 2 2 x + 4 y + 5z = 23 3x − y − 2 z = 6 4 x + 2 y + 5z = 31 2 × (2) gives: (3) + (4) gives: 2 x + 4 y + 5z = 23 (1) 3x − y − 2 z = 6 (2) 4 x + 2 y + 5z = 31 (3) 6x – 2y – 4z = 12 (4) 10x + z = 43 (5) 8x + 4y + 10z = 62 (6) (6) – (1) gives: 6x + 5z = 39 (7) 5 × (5) gives: 50x + 5z = 215 (8) 2 × (3) gives: (8) – (7) gives: Substituting in (7) gives: i.e. 44x = 176 5z = 39 – 24 = 15 8 + 4y + 15 = 23 i.e. 4y = 23 – 8 – 15 Solve the simultaneous equations 176 =4 44 24 + 5z = 39 Substituting in (1) gives: 5. from which, x = from which, z = 15 =3 5 from which, y = 0 2x + 3y + 4z = 36 3x + 2y + 3z = 29 x + y + z = 11 2 × (2) gives: 2x + 3y + 4z = 36 (1) 3x + 2y + 3z = 29 (2) x + y + z = 11 (3) 2x + 2y + 2z = 22 (4) 204 © 2014, John Bird (1) – (4) gives: y + 2z = 14 3 × (3) gives: 3x + 3y + 3z = 33 (6) – (2) gives: y=4 Substituting in (5) gives: 4 + 2z = 14 (5) (6) i.e. 2z = 14 – 4 = 10 Substituting in (1) gives: 2x + 12 + 20 = 36 i.e. 2x = 36 – 12 – 20 = 4 6. Solve the simultaneous equations from which, z = 10 =5 2 from which, x = 4 =2 2 4x + y + 3z = 31 2x – y + 2z = 10 3x + 3y – 2z = 7 (1) + (2) gives: 4x + y + 3z = 31 (1) 2x – y + 2z = 10 (2) 3x + 3y – 2z = 7 (3) 6x + 5z = 41 (4) 6x – 3y + 6z = 30 (5) 9x + 4z = 37 (6) 3 × (4) gives: 18x + 15z = 123 (7) 2 × (6) gives: 18x + 8z = 74 (8) 3 × (2) gives: (3) + (5) gives: (7) – (8) gives: Substituting in (4) gives: 7z = 49 from which, z = 49 =7 7 6x + 35 = 41 i.e. 6x = 41 – 35 = 6 Substituting in (1) gives: 4 + y + 21 = 31 i.e. y = 31 – 4 – 21 = 6 205 from which, x = 6 =1 6 © 2014, John Bird 7. Solve the simultaneous equations 5x + 5y – 4z = 37 2x – 2y + 9z = 20 –4x + y + z = – 14 2 × (3) gives: (2) + (4) gives: 5 × (3) gives: (1) – (6) gives: 5x + 5y – 4z = 37 (1) 2x – 2y + 9z = 20 (2) –4x + y + z = –14 (3) –8x + 2y + 2z = –28 (4) –6x + 11z = –8 (5) –20x + 5y + 5z = –70 (6) 25x – 9z = 107 (7) 9 × (5) gives: –54x + 99z = –72 (8) 11 × (7) gives: 275x – 99z = 1177 (9) (8) + (9) gives: 221x = 1105 Substituting in (7) gives: 125 – 107 = 9z Substituting in (1) gives: 25 + 5y – 8 = 37 8. 5y = 37 – 25 + 8 = 20 Solve the simultaneous equations 1105 =5 221 125 – 9z = 107 i.e. i.e. from which, x = from which, z = 18 =2 9 from which, y = 20 =4 4 6x + 7y + 8z = 13 3x + y – z = –11 2x – 2y – 2z = –18 6x + 7y + 8z = 13 2 × (2) gives: (1) 3x + y – z = –11 (2) 2x – 2y – 2z = –18 (3) 6x + 2y – 2z = –22 (4) 206 © 2014, John Bird (3) + (4) gives: 7 × (2) gives: 8x – 4z = –40 (5) 21x + 7y – 7z = –77 (6) (1) – (6) gives: –15x + 15z = 90 (7) 15 × (5) gives: 120x – 60z = –600 (8) 4 × (7) gives: –60x + 60z = 360 (9) (8) + (9) gives: Substituting in (5) gives: i.e. i.e. 9. from which, z = 8 =2 4 –24 + 7y + 16 = 13 7y = 13 + 24 – 16 = 21 Solve the simultaneous equations 240 = –4 60 –32 – 4z = –40 40 – 32 = 4z Substituting in (1) gives: from which, x = − 60x = –240 from which, y = 21 =3 7 3x + 2y + z = 14 7x + 3y + z = 22.5 4x – 4y – z = – 8.5 3x + 2y + z = 14 (1) 7x + 3y + z = 22.5 (2) 4x – 4y – z = – 8.5 (3) (2) + (3) gives: 11x – y = 14 (4) (3) + (1) gives: 7x – 2y = 5.5 (5) 22x – 2y = 28 (6) 2 × (4) gives: (6) – (5) gives: Substituting in (4) gives: 15x = 22.5 from which, x = 22.5 = 1.5 15 16.5 – y = 14 i.e. 16.5 – 14 = y Substituting in (1) gives: 4.5 + 5 + z = 14 i.e. z = 14 – 4.5 – 5 = 4.5 207 i.e. y = 2.5 © 2014, John Bird 10. Kirchhoff’s laws are used to determine the current equations in an electrical network and result in the following: −31 i1 + 8 i2 + 3 i3 = −5 3 i1 − 2 i2 + i3 = 2 i1 − 3 i2 + 2 i3 = 6 Determine the values of i1 , i2 and i3 i1 + 8 i2 + 3 i3 = –31 (1) 3 i1 – 2 i2 + i3 = –5 (2) 2 i1 – 3 i2 + 2 i3 = 6 (3) (2) × (2) gives: 6 i1 – 4 i2 + 2 i3 = –10 (4) (4) – (3) gives: 4 i1 – i2 = –16 (5) 9 i1 – 6 i2 + 3 i3 = –15 (6) 3 × (2) gives: (6) – (1) gives: 2 × (5) gives: (7) – (8) gives: Substituting in (5) gives: 8 i1 – 14 i2 = 16 (7) 8 i1 – 2 i2 = –32 –12 i2 = 48 (8) from which, i2 = − 48 = –4 12 4 i1 + 4 = –16 i.e. 4 i1 = –16 – 4 = –20 Substituting in (1) gives: –5 – 32 + 3 i3 = –31 i.e. 3 i3 = –31 + 5 + 32 = 6 208 i.e. i1 = − i.e. i3 = 20 = –5 4 6 =2 3 © 2014, John Bird 11. The forces in three members of a framework are F1 , F2 and F3 . They are related by the simultaneous equations shown below. 1.4F1 + 2.8F2 + 2.8F3 = 5.6 4.2F1 – 1.4F2 + 5.6F3 = 35.0 4.2F1 + 2.8F2 – 1.4F3 = –5.6 Find the values of F1, F2 and F3 . (1) – (2) gives: 2 × (2) gives: (1) + (5) gives: 1.4F1 + 2.8F2 + 2.8F3 = 5.6 (1) 4.2F1 – 1.4F2 + 5.6F3 = 35.0 (2) 4.2F1 + 2.8F2 – 1.4F3 = –5.6 (3) –2.8F1 + 4.2F3 = 11.2 (4) 8.4F1 – 2.8F2 + 11.2F3 = 70.0 (5) 9.8F1 + 14F3 = 75.6 (6) 9.8 × (4) gives: –27.44F1 + 41.16F3 = 109.76 (7) 2.8 × (6) gives: 27.44F1 + 39.20F3 = 211.68 (8) (7) + (8) gives: Substituting in (4) gives: i.e. Substituting in (1) gives: i.e. 80.36F3 = 321.44 from which, F3 = 321.44 =4 80.36 –2.8F1 + 16.8 = 11.2 16.8 – 11.2 = 2.8F1 i.e. F1 = 5.6 =2 2.8 2.8 + 2.8F2 + 11.2 = 5.6 2.8F2 = 5.6 – 2.8 – 11.2 = –8.4 209 i.e. F2 = 8.4 = –3 2.8 © 2014, John Bird