CHAPTER 13 SOLVING SIMULTANEOUS EQUATIONS

Transcription

1 CHAPTER 13 SOLVING SIMULTANEOUS EQUATIONS EXERCISE 51 Page Solve the simultaneous equations 2x y = 6 x + y = 6 2x y = 6 (1) x + y = 6 (2) (1) + (2) gives: 3x = 12 from which, x = 12 3 = 4 From (1): 8 y = 6 i.e. 8 6 = y from which, y = 2 2. Solve the simultaneous equations 2x y = 2 x 3y = 9 2x y = 2 (1) x 3y = 9 (2) 3 (1) gives: 6x 3y = 6 (3) (3) (2) gives: 5x = 15 from which, x = 15 5 = 3 From (1): 6 y = 2 i.e. 6 2 = y from which, y = 4 3. Solve the simultaneous equations x 4y = 4 5x 2y = 7 x 4y = 4 (1) 5x 2y = 7 (2) 2 (2) gives: 10x 4y = 14 (3) (3) (1) gives: 9x = 18 from which, x = 18 9 = 2 180

2 From (1): 2 4y = 4 i.e = 4y i.e. 6 = 4y from which, y = 6 4 = Solve the simultaneous equations 3x 2y = 10 5x + y = 21 3x 2y = 10 (1) 5x + y = 21 (2) 2 (2) gives: 10x + 2y = 42 (3) (1) + (3) gives: 13x = 52 from which, x = = 4 From (1): 12 2y = 10 i.e = 2y i.e. 2 = 2y from which, y = 2 2 = 1 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: 23p = 46 from which, p = = 2 From (1): q = 6 i.e. 4q = 6 10 i.e. 4q = 4 from which, q = 4 4 = 1 181

3 6. Solve the simultaneous equations 7x + 2y = 11 3x 5y = 7 7x + 2y = 11 (1) 3x 5y = 7 (2) 5 (1) gives: 35x + 10y = 55 (3) 2 (2) gives: 6x 10y = 14 (4) (3) + (4) gives: 41x = 41 from which, x = = 1 From (1): 7 + 2y = 11 i.e. 2y = 11 7 i.e. 2y = 4 from which, y = 4 2 = 2 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: 29y = 58 from which, y = = 2 From (1): 2x 14 = 8 i.e. 2x = 14 8 i.e. 2x = 6 from which, x = 6 2 = 3 8. Solve the simultaneous equations a + 2b = 8 b 3a = 3 182

4 Rearranging gives: a + 2b = 8 (1) 3a + b = 3 (2) 2 (2) gives: 6a + 2b = 6 (3) (1) (3) gives: 7a = 14 from which, a = 14 7 = 2 From (1): 2 + 2b = 8 i.e. 2b = 8 2 i.e. 2b = 6 from which, b = 6 2 = 3 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 which, a = 10 2 = 5 From (1): 5 + b = 7 i.e. b = 7 5 from which, b = Solve the simultaneous equations 2x + 5y = 7 x + 3y = 4 2x + 5y = 7 (1) x + 3y = 4 (2) 2 equation (2) gives: 2x + 6y = 8 (3) (3) (1) gives: y = 1 Substituting in (1) gives: 2x + 5 = 7 i.e. 2x = 7 5 = 2 and x = 2 2 = 1 Thus, x = 1 and y = 1 and may be checked by substituting into both of the original equations 183

5 11. Solve the simultaneous equations 3s + 2t = 12 4s t = 5 3s + 2t = 12 (1) 4s t = 5 (2) 2 equation (2) gives: 8s 2t = 10 (3) (1) + (3) gives: 11s = 22 from which, s = = 2 Substituting in (1) gives: 6 + 2t = 12 i.e. 2t = 12 6 = 6 and t = 6 2 = 3 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: 19y = 38 from which, y = = 2 Substituting in (1) gives: 3x + 4 = 13 i.e. 3x = 13 4 = 9 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 5m 3n = 11 (1) 3m + n = 8 (2) 3 equation (2) gives: 9m + 3n = 24 (3) 184

6 (1) + (3) gives: 14m = 35 from which, m = = 2.5 Substituting in (1) gives: n = 11 i.e = 3n i.e. 1.5 = 3n and n = = 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 which, a = = 6 From (1): 48 3b = 51 i.e = 3b i.e. 3 = 3b from which, b = 3 = 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: 41y = 205 from which, y = = 5 Substituting in (1) gives: 5x 10 = 0 i.e. 5x = 10 and x = 10 5 = 2 185

7 16. Solve the simultaneous equations 5c = 1 3d 2d + c + 4 = 0 Rearranging gives: 5c + 3d = 1 (1) c + 2d = 4 (2) 5 equation (2) gives: 5c + 10d = 20 (3) (3) (1) gives: 7d = 21 from which, d = 21 7 = 3 Substituting in (1) gives: 5c 9 = 1 i.e. 5c = 10 and c = 10 5 = 2 186

8 EXERCISE 52 Page Solve the simultaneous equations 7p q = 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 from which, p = 1 Substituting in (1) gives: 7 + 2q = 11 i.e. 2q = = 4 and q = 4 = Solve the simultaneous equations x y + = x y = x y Rearranging gives: (6) + (6) = (6)(4) i.e. 3x + 2y = 24 (1) 2 3 x y and (18) (18) = (18)(0) i.e. 3x 2y = 0 (2) 6 9 (1) (2) gives: 4y = 24 from which, y = 6 Substituting in (1) gives: 3x + 12 = 24 i.e. 3x = = 12 and x = 12 3 = 4 3. Solve the simultaneous equations a 7 = 2b 2 12 = 5a b a Rearranging gives: (2) (2)7 = (2)(2 b) i.e. a + 4b = 14 (1) 2 187

9 and (3)(12) = (3)(5a) + 2 (3) 3 b i.e. 15a + 2b = 36 (2) 2 equation (2) gives: 30a + 4b = 72 (3) (3) (1) gives: 29a = 58 from which, a = 2 Substituting in (1) gives: 2 + 4b = 14 i.e. 4b = 14 2 = 12 and b = 12 4 = 3 4. Solve the simultaneous equations 3 2 s 2t = 8 s + 3t = 2 4 Rearranging gives: 3 (2) (2)2 (2)(8) 2 s t = i.e. 3s 4t = 16 (1) s and (4) + (4)3 t = (4)( 2) i.e. s + 12t = 8 (2) 4 3 equation (2) gives: 3s + 36t = 24 (3) (1) (3) gives: 40t = 40 from which, t = i.e. t = 1 Substituting in (1) gives: 3s + 4 = 16 i.e. 3s = 16 4 = 12 and s = 12 3 = 4 5. Solve the simultaneous equations x 2y 49 + = x y = 0 Rearranging gives: x 2y 49 (15) + (15) = (15) i.e. 3x + 10y = 49 (1) and 3x y 5 (14) (14) + (14) = 0 i.e. 6x 7y = 10 (2) equation (1) gives: 6x + 20y = 98 (3) 188

10 (3) (2) gives: 27y = 108 from which, y = = 4 Substituting in (1) gives: 3x + 40 = 49 i.e. 3x = = 9 and x = 9 3 = 3 6. Solve the simultaneous equations v 1 = 12 u u + 4 v 25 2 = 0 Rearranging gives: (12) (12)(1) (12) 12 u v = i.e. 12v u = 12 (1) and v 25 (4) u + (4) (4) = 0 i.e. v + 4u = 50 (2) equation (1) gives: 48v 4u = 48 (3) (2) + (3) gives: 49v = 98 from which, v = = 2 Substituting in (1) gives: 24 u = 12 i.e = u and u = Solve the simultaneous equations 1.5x 2.2y = x + 0.6y = 33 Multiplying both equations by 10 gives: 15x 22y = 180 (1) 24x + 6y = 330 (2) 6 equation (1) gives: 90x 132y = 1080 (3) 22 equation (2) gives: 528x + 132y = 7260 (4) (3) + (4) gives: 618x = 6180 from which, x = 10 Substituting in (1) gives: y = 180 i.e. 22y = = 330 and y = = 15 Thus, x = 10 and y = 15 and may be checked by substituting into both of the original equations. 189

11 8. Solve the simultaneous equations 3b 2.5a = a + 0.8b = 0.8 Multiplying equation (1) by 100 gives: 250a +300b = 45 (1) Multiplying equation (1) by 10 gives: 16a +8b = 8 (2) 4 equation (1) gives: 1000a b = 180 (3) 150 equation (2) gives: 2400a b = 1200 (4) (4) (3) gives: 3400a = 1020 from which, a = = 0.3 Substituting in (1) gives: b = 45 i.e. 300b = = 120 and b = =

12 EXERCISE 53 Page Solve the simultaneous equations 3 x + 2 y = 14 5 x 3 y = 2 Let 1 a x = and 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: 19a = 38 from which, a = = 2 Substituting in (1) gives: 6 + 2b = 14 i.e. 2b = 14 6 = 8 and b = 8 2 = 4 Since 1 a x = then x = 1 1 a = 2 and since 1 b y = then y = 1 1 b = 4 Thus, x = 1 2 and y = 1 4 and may be checked by substituting into both of the original equations 2. Solve the simultaneous equations 4 a 3 b = 18 2 a + 5 b = 4 Let 1 x a = and 1 y b = Thus, the equations become: 4x 3y = 18 (1) and 2x + 5y = 4 (2) 2 equation (2) gives: 4x + 10y = 8 (3) 191

13 (1) (3) gives: 13y = 26 from which, y = = 2 Substituting in (1) gives: 4x + 6 = 18 i.e. 4x = 18 6 = 12 and x = 12 4 = 3 Since 1 x a = then a = 1 1 x = 3 and since 1 y b = then b = y = 2 = 2 Thus, a = 1 3 and b = 1 and may be checked by substituting into both of the original equations 2 3. Solve the simultaneous equations 1 2 p + 3 5q = 5 5 p 1 2q = 35 2 Let 1 a p = and 1 b q = Thus, the equations become: Rearranging gives: 1 3 a+ b= a b= (10) a+ (10) b= (10)(5) i.e. 5a + 6b = (2)(5 a) (2) b= (2) i.e. 10a b = Thus, 5a + 6b = 50 (1) and 10a b = 35 (2) 2 equation (1) gives: 10a + 12b = 100 (3) (3) (2) gives: 13b = 65 from which, b = = 5 Substituting in (1) gives: 5a + 30 = 50 i.e. 5a = = 20 and a = 20 5 = 4 Since 1 a p = then p = 1 1 a = 4 and since 1 b q = then q = 1 1 b = 5 Thus, p = 1 4 and q = 1 and may be checked by substituting into both of the original equations 5 192

14 4. Solve the simultaneous equations 5 x + 3 y = x 7 y = 1.1 Let 1 a x = and 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 from which, b = 8.8 = Substituting in (1) gives: 5a = 1.1 i.e. 5a = = 0.5 and a = 0.5 = Since 1 a x = then x = 1 10 a = and since 1 b y = then y = 1 b = 5 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 c 3 d = = 0 Rearranging gives: and c+ 1 d + 2 (12) (12) + (12)(1) = 0 i.e. 3(c + 1) 4(d + 2) + 12 = c 3 d 13 (20) + (20) + (20) = 0 i.e. 4(1 c) + 5(3 d) + 13 = Since 3(c + 1) 4(d + 2) + 12 = 0 then 3c + 3 4d = 0 i.e. 3c 4d = 7 and 4(1 c) + 5(3 d) + 13 = 0 then 4 4c d + 13 = 0 i.e. 4c + 5d = 32 Thus, 3c 4d = 7 (1) 4c + 5d = 32 (2) 193

15 4 equation (1) gives: 12c 16d = 28 (3) 3 equation (2) gives: 12c + 15d = 96 (4) (4) (3) gives: 31d = 124 from which, d = = 4 Substituting in (1) gives: 3c 16 = 7 i.e. 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 6. Solve the simultaneous equations 3r r 4 2 s 1 11 = s 3 = 15 4 Rearranging gives: and 3r+ 2 2s 1 11 (20) (20) = (20) i.e. 4(3r + 2) 5(2s 1) = r 5 s 15 (12) + (12) = (12) i.e. 3(3 + 2r) + 4(5 s) = Since 4(3r + 2) 5(2s 1) = 44 then 12r s + 5 = 44 i.e. 12r 10s = 31 and 3(3 + 2r) + 4(5 s) = 45 then 9 + 6r s = 45 i.e. 6r 4s = 16 Thus, 12r 10s = 31 (1) 6r 4s = 16 (2) 2 equation (2) gives: 12r 8s = 32 (3) (3) (1) gives: 2s = 1 from which, s = 1 2 Substituting in (1) gives: 12r 5 = 31 i.e. 12r = = 36 and r = = 3 Thus, r = 3 and s = 1 2 and may be checked by substituting into both of the original equations 194

16 7. Solve the simultaneous equations 5 x+ y 4 2x y = = Multiplying both sides of the first equation by 27(x + y) gives: i.e. 5 27( x + y ) = x + y 20 27( x+ y) i.e. 135 = 20x + 20y i.e. 20x + 20y = 135 (1) 27 Multiplying both sides of the second equation by 33(2x y) gives: i.e. 4 33(2 x y) 2 x y = 16 33(2 x y) i.e. 132 = 32x 16y i.e. 32x 16y = 132 (2) 33 4 equation (1) gives: 80x + 80y = 540 (3) 5 equation (2) gives: 160x 80y = 660 (4) (3) + (4) gives: 240x = 1200 from which, x = = 5 Substituting in (1) gives: y = 135 i.e. 20y = = 35 and y = = If 5x 3 y = 1 and x + 4 y = 5 2 find the value of xy + 1 y If 5x 3 y = 1 then 5x = 3 y + 1 and x = 3 1 5y + 5 and if x y = 2 then x = 4 5 y + 2 Equating x values gives: i.e y + 5 = 4 5 y + 2 i.e = i.e. 5y = 5y y y = 10 and (23)(10) = (5y)(23) i.e. 230 = 115y from which, y = = 2 195

17 Substituting into the first equation gives: 5x = 1 i.e. 5x = 2.5 and x = = 5 2 Thus, xy y + ( ) = = =

18 EXERCISE 54 Page 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 if W = 90 when P = 22, then: 22 = 90a + b (2) Equation (2) equation (1) gives: 10 = 50a from which, a = 10 = 1 or Substituting in (1) gives: 12 = b i.e. 12 = 8 + b from which, 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 = 0.2I 1 + 2(I 1 I 2 ) 12 = 3I I 2 2(I 1 I 2 ) Determine the values of currents I 1 and I 2 Rearranging 5 0.2I1 2( I1 I2) = + gives: 5 = 0.2I1+ 2I1 2I2 i.e. 2.2 I1 2 I2 = 5 Rearranging 12 3I2 0.4I2 2( I1 I2) = + gives: 12 = 3I I2 2I1+ 2I2 i.e. 2 I I = Thus, 2.2 I1 2 I2 = 5 (1) and 2 I I2 = 12 (2) 2 equation (1) gives: 4.4 I1 4 I2 = 10 (3) 2.2 equation (2) gives: 4.4 I I2 = 26.4 (4) 36.4 (3) + (4) gives: 7.88 I 2 = 36.4 from which, I 2 = =

19 14.24 Substituting in (1) gives: 2.2 I = 5 i.e. 2.2 I 1 = and I 1 = = Thus, I 1 = 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 if v = 40 when t = 7, then: 40 = u + 7a (2) Equation (2) equation (1) gives: 20 = 5a from which, a = 20 5 = 4 Substituting in (1) gives: 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) = Three new cars and four new vans supplied to a dealer together cost and five new cars and two new vans of the same models cost Find the cost of a car and a van. Let a car = C and a van = V then 3C + 4V = (1) 5C + 2V = (2) 2 equation (2) gives: 10C + 4V = (3) (3) (1) gives: 7C = from which, C = Substituting in (1) gives: V = i.e. 4V = from which, V = Hence, a car costs and a van costs = = =

20 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 2 = 2m + c (1) When x = 5 and y = 0.5, then 0.5 = 5m + c (2) (1) (2) gives: 1.5 = 3m i.e. m = = 0.5 In equation (1): 2 = 1 + c i.e. c = = 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 R 0 is the resistance at 0 C and α is the temperature coefficient of resistance. If R = Ω at 30 C and R = Ω at 100 C, find α and R 0 R = R0 ( 1+ αt) thus when R = Ω at t = 30 C, then: = R0 ( 1 30α ) and when R = Ω at t = 100 C, then: = R0 ( 1 100α ) + (1) + (2) Dividing equation (2) by equation (1) gives: ( α ) ( + α ) R0 = R i.e. ( α ) ( + α ) = from which, (32.17)( α) = (25.44)( α) and α = α i.e = 2544α 965.1α i.e = α from which, α = = Substituting in (1) gives: = R0[ 1 + (30)( )] = R0( ) 199

21 from which, R 0 = = Thus, α = and R 0 = 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: 52 = a + 100b (1) and if c = 172 when T = 400, then: 172 = a + 400b (2) Equation (2) equation (1) gives: 120 = 300b from which, b = = 0.40 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 (1) If q = 22 when p = 5 then 22 = 5a + b/5 or 110 = 25a + b (2) (2) (1) gives: 84 = 21a from which, a = = 4 Substituting in (1) gives: 26 = 16 + b i.e. b = = In a system of forces, the relationship between two forces F 1 and F 2 is given by: 5F 1 + 3F = 0 3F 1 + 5F = 0 Solve for F 1 and F 2 5F 1 + 3F = 0 (1) 200

22 3F 1 + 5F = 0 (2) 5 (1) gives: 25F F = 0 (3) 3 (1) gives: 9F F = 0 (4) (3) (4) gives: 16F 1 24 = 0 i.e. 16F 1 = 24 from which, F 1 = = 1.5 Substituting in (1) gives: F = 0 i.e. 3F 2 = = 13.5 from which, F 2 = = For a balanced beam, the equilibrium of forces is given by: R1+ R2 = 12.0kN As a result of taking moments: 0.2R = 0.8R2 Determine the values of the reaction forces R 1 and R 2 Rearranging gives: R1+ R2 = 12.0 (1) 0.2R 0.8R = 3.9 (2) (2) gives: R1 4.0R2 = 19.5 (3) (1) (3) gives: 5.0 R 2 = 31.5 from which, R 2 = = 6.3 kn Substituting in (1) gives: R = 12.0 Hence, R 1 = = 5.7 kn 201

23 EXERCISE 55 Page Solve the simultaneous equations x+ 2y+ 4z = 16 2x y+ 5z = 18 3x+ 2y+ 2z = 14 x+ 2y+ 4z = 16 (1) 2x y+ 5z = 18 (2) 3x+ 2y+ 2z = 14 (3) (1) (3) gives: 2x + 2z = 2 (4) 2 (2) gives: 4x 2y + 10z = 36 (5) (1) + (5) gives: 5x + 14z = 52 (6) 7 (4) gives: 14x + 14z = 14 (7) (6) (7) gives: 19x = 38 from which, x = = 2 Substituting in (6) gives: z = 52 i.e. 14z = = 42 from which, z = = 3 Substituting in (1) gives: 2 + 2y + 12 =16 i.e. 2y = = 2 and y = 2 2 = 1 2. Solve the simultaneous equations 2x+ y z = 0 3x+ 2y+ z = 4 5x+ 3y+ 2z = 8 2x+ y z = 0 (1) 3x+ 2y+ z = 4 (2) 5x+ 3y+ 2z = 8 (3) (1) + (2) gives: 5x + 3y = 4 (4) 2 (1) gives: 4x + 2y 2z = 0 (5) 202

24 (3) + (5) gives: 9x + 5y = 8 (6) 5 (4) gives: 25x + 15y = 20 (7) 3 (6) gives: 27x + 15y = 24 (8) (8) (7) gives: 2x = 4 from which, x = 4 2 = 2 Substituting in (4) gives: y = 4 i.e. 3y = 4 10 = 6 from which, y = 6 3 = 2 Substituting in (1) gives: z = 0 i.e. z = 2 3. Solve the simultaneous equations 3x+ 5y+ 2z = 6 x y+ 3z = 0 2x+ 7y+ 3z = 3 3x+ 5y+ 2z = 6 (1) x y+ 3z = 0 (2) 2x+ 7y+ 3z = 3 (3) (3) (2) gives: x + 8y = 3 (4) 3 (1) gives: 9x + 15y + 6z = 18 (5) 2 (2) gives: 2x 2y + 6z = 0 (6) (5) (6) gives: 7x + 17y = 18 (7) 7 (4) gives: 7x + 56y = 21 (8) (7) (8) gives: 39y = 39 from which, y = = 1 Substituting in (7) gives: 7x 17 = 18 i.e. 7x = = 35 from which, x = 35 7 = 5 Substituting in (1) gives: z = 6 203

25 i.e. 2z = = 4 from which, z = 4 = Solve the simultaneous equations 2x+ 4y+ 5z = 23 3x y 2z = 6 4x+ 2y+ 5z = 31 2x+ 4y+ 5z = 23 (1) 3x y 2z = 6 (2) 4x+ 2y+ 5z = 31 (3) 2 (2) gives: 6x 2y 4z = 12 (4) (3) + (4) gives: 10x + z = 43 (5) 2 (3) gives: 8x + 4y + 10z = 62 (6) (6) (1) gives: 6x + 5z = 39 (7) 5 (5) gives: 50x + 5z = 215 (8) (8) (7) gives: 44x = 176 from which, x = = 4 Substituting in (7) gives: z = 39 i.e. 5z = = 15 from which, z = 15 5 = 3 Substituting in (1) gives: 8 + 4y + 15 = 23 i.e. 4y = from which, y = 0 5. Solve the simultaneous equations 2x + 3y + 4z = 36 3x + 2y + 3z = 29 x + y + z = 11 2x + 3y + 4z = 36 (1) 3x + 2y + 3z = 29 (2) x + y + z = 11 (3) 2 (2) gives: 2x + 2y + 2z = 22 (4) 204

26 (1) (4) gives: y + 2z = 14 (5) 3 (3) gives: 3x + 3y + 3z = 33 (6) (6) (2) gives: y = 4 Substituting in (5) gives: 4 + 2z = 14 i.e. 2z = 14 4 = 10 from which, z = 10 2 = 5 Substituting in (1) gives: 2x = 36 i.e. 2x = = 4 from which, x = 4 2 = 2 6. Solve the simultaneous equations 4x + y + 3z = 31 2x y + 2z = 10 3x + 3y 2z = 7 4x + y + 3z = 31 (1) 2x y + 2z = 10 (2) 3x + 3y 2z = 7 (3) (1) + (2) gives: 6x + 5z = 41 (4) 3 (2) gives: 6x 3y + 6z = 30 (5) (3) + (5) gives: 9x + 4z = 37 (6) 3 (4) gives: 18x + 15z = 123 (7) 2 (6) gives: 18x + 8z = 74 (8) (7) (8) gives: 7z = 49 from which, z = 49 7 = 7 Substituting in (4) gives: 6x + 35 = 41 i.e. 6x = = 6 from which, x = 6 6 = 1 Substituting in (1) gives: 4 + y + 21 = 31 i.e. y = = 6 205

27 7. Solve the simultaneous equations 5x + 5y 4z = 37 2x 2y + 9z = 20 4x + y + z = 14 5x + 5y 4z = 37 (1) 2x 2y + 9z = 20 (2) 4x + y + z = 14 (3) 2 (3) gives: 8x + 2y + 2z = 28 (4) (2) + (4) gives: 6x + 11z = 8 (5) 5 (3) gives: 20x + 5y + 5z = 70 (6) (1) (6) gives: 25x 9z = 107 (7) 9 (5) gives: 54x + 99z = 72 (8) 11 (7) gives: 275x 99z = 1177 (9) (8) + (9) gives: 221x = 1105 from which, x = = 5 Substituting in (7) gives: 125 9z = 107 i.e = 9z from which, z = 18 9 = 2 Substituting in (1) gives: y 8 = 37 i.e. 5y = = 20 from which, y = 20 4 = 4 8. Solve the simultaneous equations 6x + 7y + 8z = 13 3x + y z = 11 2x 2y 2z = 18 6x + 7y + 8z = 13 (1) 3x + y z = 11 (2) 2x 2y 2z = 18 (3) 2 (2) gives: 6x + 2y 2z = 22 (4) 206

28 (3) + (4) gives: 8x 4z = 40 (5) 7 (2) gives: 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: 60x = 240 from which, x = Substituting in (5) gives: 32 4z = = 4 60 i.e = 4z from which, z = 8 4 = 2 Substituting in (1) gives: y + 16 = 13 i.e. 7y = = 21 from which, y = 21 7 = 3 9. Solve the simultaneous equations 3x + 2y + z = 14 7x + 3y + z = x 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) 2 (4) gives: 22x 2y = 28 (6) (6) (5) gives: 15x = 22.5 from which, x = = 1.5 Substituting in (4) gives: 16.5 y = 14 i.e = y i.e. y = 2.5 Substituting in (1) gives: z = 14 i.e. z = =

29 10. Kirchhoff s laws are used to determine the current equations in an electrical network and result in the following: i + 8i + 3i = i 2i + i = i 3i + 2i = Determine the values of i 1, i 2 and i 3 i i 2 + 3i 3 = 31 (1) 3i 1 2i 2 + i 3 = 5 (2) 2i 1 3i i 3 = 6 (3) (2) (2) gives: 6i 1 4i 2 + 2i 3 = 10 (4) (4) (3) gives: 4i 1 i 2 = 16 (5) 3 (2) gives: 9i 1 6i 2 + 3i 3 = 15 (6) (6) (1) gives: 8i 1 14 i 2 = 16 (7) 2 (5) gives: 8i 1 2i 2 = 32 (8) (7) (8) gives: 12 i 2 = 48 from which, i 2 = Substituting in (5) gives: 4i = 16 i.e. 4i 1 = 16 4 = 20 i.e. i 1 = Substituting in (1) gives: i 3 = = = 5 4 i.e. 3i 3 = = 6 i.e. i 3 = 6 3 = 2 208

30 11. The forces in three members of a framework are F 1, F 2 and F 3. They are related by the simultaneous equations shown below. 1.4F F F 3 = F 1 1.4F F 3 = F F 2 1.4F 3 = 5.6 Find the values of F 1, F 2 and F F F F 3 = 5.6 (1) 4.2F 1 1.4F F 3 = 35.0 (2) 4.2F F 2 1.4F 3 = 5.6 (3) (1) (2) gives: 2.8F F 3 = 11.2 (4) 2 (2) gives: 8.4F 1 2.8F F 3 = 70.0 (5) (1) + (5) gives: 9.8F F 3 = 75.6 (6) 9.8 (4) gives: 27.44F F 3 = (7) 2.8 (6) gives: 27.44F F 3 = (8) (7) + (8) gives: 80.36F 3 = from which, F 3 = = 4 Substituting in (4) gives: 2.8F = 11.2 i.e = 2.8F 1 i.e. F 1 = = 2 Substituting in (1) gives: F = 5.6 i.e. 2.8F 2 = = 8.4 i.e. 2 F = = 3 209