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CHAPTER 13 SOLVING SIMULTANEOUS EQUATIONS

EXERCISE 51 Page 105

1. Solve the simultaneous equations 2 6x y− = x + y = 6

2 6x y− = (1)

x + y = 6 (2)

(1) + (2) gives: 3x = 12 from which, x = 123

= 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 = 155

= 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 = 189

= 2

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From (1): 2 – 4y = – 4

i.e. 2 + 4 = 4y

i.e. 6 = 4y from which, y = 64

= 1.5

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 = 5213

= 4

From (1): 12 – 2y = 10

i.e. 12 – 10 = 2y

i.e. 2 = 2y from which, y = 22

= 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 = 4623

= 2

From (1): 10 + 4q = 6

i.e. 4q = 6 – 10

i.e. 4q = –4 from which, q = 44−

= –1

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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 = 4141

= 1

From (1): 7 + 2y = 11

i.e. 2y = 11 – 7

i.e. 2y = 4 from which, y = 42

= 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 = 5829

= 2

From (1): 2x – 14 = –8

i.e. 2x = 14 – 8

i.e. 2x = 6 from which, x = 62

= 3

8. Solve the simultaneous equations a + 2b = 8

b – 3a = –3

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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 = 147

= 2

From (1): 2 + 2b = 8

i.e. 2b = 8 – 2

i.e. 2b = 6 from which, b = 62

= 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 = 102

= 5

From (1): 5 + b = 7

i.e. b = 7 – 5 from which, b = 2

10. 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 = 22

= 1

Thus, x = 1 and y = 1 and may be checked by substituting into both of the original equations

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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 = 2211

= 2

Substituting in (1) gives: 6 + 2t = 12 i.e. 2t = 12 – 6 = 6 and t = 62

= 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 = 3819−

= –2

Substituting in (1) gives: 3x + 4 = 13 i.e. 3x = 13 – 4 = 9 and x = 93

= 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)

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(1) + (3) gives: 14m = 35 from which, m = 3514

= 2.5

Substituting in (1) gives: 12.5 – 3n = 11

i.e. 12.5 – 11 = 3n

i.e. 1.5 = 3n and n = 1.53

= 0.5

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 which, a = 24641

= 6

From (1): 48 – 3b = 51

i.e. 48 – 51 = 3b

i.e. –3 = 3b from which, b = 33−

= –1

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: 41y = 205 from which, y = 20541

= 5

Substituting in (1) gives: 5x – 10 = 0 i.e. 5x = 10 and x = 105

= 2

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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 = 217− = –3

Substituting in (1) gives: 5c – 9 = 1 i.e. 5c = 10 and c = 105

= 2

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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 from which, p = – 1

Substituting in (1) gives: –7 + 2q = –11 i.e. 2q = –11 + 7 = –4 and q = 42− = –2

2. Solve the simultaneous equations 2x +

3y = 4

6x –

9y = 0

Rearranging gives: (6) (6) (6)(4)2 3x y+ = i.e. 3x + 2y = 24 (1)

and (18) (18) (18)(0)6 9x y− = i.e. 3x – 2y = 0 (2)

(1) – (2) gives: 4y = 24 from which, y = 6

Substituting in (1) gives: 3x + 12 = 24 i.e. 3x = 24 – 12 = 12 and x = 123

= 4

3. Solve the simultaneous equations 2a – 7 = –2b

12 = 5a + 23

b

Rearranging gives: (2) (2)7 (2)(2 )2a b− = − i.e. a + 4b = 14 (1)

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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 = 124

= 3

4. Solve the simultaneous equations 32

s – 2t = 8

4s + 3t = –2

Rearranging gives: 3(2) (2)2 (2)(8)2

s t− = i.e. 3s – 4t = 16 (1)

and (4) (4)3 (4)( 2)4s t+ = −

i.e. s + 12t = –8 (2)

3 × equation (2) gives: 3s + 36t = –24 (3)

(1) – (3) gives: – 40t = 40

from which, t = 4040− i.e. t = –1

Substituting in (1) gives: 3s + 4 = 16 i.e. 3s = 16 – 4 = 12 and s = 123

= 4

5. Solve the simultaneous equations 5x + 2

3y = 49

15

37x –

2y + 5

7 = 0

Rearranging gives: 2 49(15) (15) (15)5 3 15x y+ = i.e. 3x + 10y = 49 (1)

and 3 5(14) (14) (14) 07 2 7x y− + = i.e. 6x – 7y = –10 (2)

2 × equation (1) gives: 6x + 20y = 98 (3)

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(3) – (2) gives: 27y = 108 from which, y = 10827

= 4

Substituting in (1) gives: 3x + 40 = 49 i.e. 3x = 49 – 40 = 9 and x = 93

= 3

6. Solve the simultaneous equations v – 1 = 12u

u + 4v – 25

2 = 0

Rearranging gives: (12) (12)(1) (12)12uv − = i.e. 12v – u = 12 (1)

and 25(4) (4) (4) 04 2vu + − = i.e. v + 4u = 50 (2)

4 × equation (1) gives: 48v – 4u = 48 (3)

(2) + (3) gives: 49v = 98 from which, v = 9849

= 2

Substituting in (1) gives: 24 – u = 12 i.e. 24 – 12 = u and 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 (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: 150 – 22y = –180 i.e. –22y = –180 – 150 = –330 and y = 33022

−−

= 15

Thus, x = 10 and y = 15 and may be checked by substituting into both of the original equations.

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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: 16a +8b = 8 (2)

4 × equation (1) gives: –1000a + 1200b = 180 (3)

150 × equation (2) gives: 2400a + 1200b = 1200 (4)

(4) – (3) gives: 3400a = 1020

from which, a = 10203400

= 0.3

Substituting in (1) gives: –75 + 300b = 45 i.e. 300b = 45 + 75 = 120 and b = 120300

= 0.4

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EXERCISE 53 Page 109

1. Solve the simultaneous equations 3x

+ 2y

= 14

5x

– 3y

= –2

Let 1 ax= 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 = 3819

= 2

Substituting in (1) gives: 6 + 2b = 14 i.e. 2b = 14 – 6 = 8 and b = 82

= 4

Since 1 ax= then x = 1 1

2a= and since 1 b

y= then y = 1 1

4b=

Thus, x = 12

and y = 14

and may be checked by substituting into both of the original equations

2. Solve the simultaneous equations 4a

– 3b

= 18

2a

+ 5b

= –4

Let 1 xa= 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)

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(1) – (3) gives: –13y = 26 from which, y = 2613−

= –2

Substituting in (1) gives: 4x + 6 = 18 i.e. 4x = 18 – 6 = 12 and x = 124

= 3

Since 1 xa= then a = 1 1

3x= and since 1 y

b= then b = 1 1 1

2 2y= = −−

Thus, a = 13

and b = 12

− and may be checked by substituting into both of the original equations

3. Solve the simultaneous equations 12 p

+ 35q

= 5

5p

– 12q

= 352

Let 1 ap= and 1 b

q=

Thus, the equations become: 1 3 52 5

a b+ =

1 3552 2

a b− =

Rearranging gives: 1 3(10) (10) (10)(5)2 5

a b+ = i.e. 5a + 6b = 50

1 35(2)(5 ) (2) (2)2 2

a b− = i.e. 10a – b = 35

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 = 6513

= 5

Substituting in (1) gives: 5a + 30 = 50 i.e. 5a = 50 – 30 = 20 and a = 205

= 4

Since 1 ap= then p = 1 1

4a= and since 1 b

q= then q = 1 1

5b=

Thus, p = 14

and q = 15

and may be checked by substituting into both of the original equations

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4. Solve the simultaneous equations 5x

+ 3y

= 1.1

3x

– 7y

= –1.1

Let 1 ax= 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 = 8.8 144 5

=

Substituting in (1) gives: 5a + 0.6 = 1.1 i.e. 5a = 1.1 – 0.6 = 0.5 and a = 0.5 15 10

=

Since 1 ax= 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 14

c + – 23

d + + 1 = 0

15

c− + 34

d− + 1320

= 0

Rearranging gives: 1 2(12) (12) (12)(1) 04 3

c d+ +− + = i.e. 3(c + 1) – 4(d + 2) + 12 = 0

and 1 3 13(20) (20) (20) 05 4 20

c d− −+ + = i.e. 4(1 – c) + 5(3 – d) + 13 = 0

Since 3(c + 1) – 4(d + 2) + 12 = 0 then 3c + 3 – 4d – 8 + 12 = 0 i.e. 3c – 4d = –7 and 4(1 – c) + 5(3 – d) + 13 = 0 then 4 – 4c + 15 – 5d + 13 = 0 i.e. 4c + 5d = 32 Thus, 3c – 4d = –7 (1)

4c + 5d = 32 (2)

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4 × equation (1) gives: 12c – 16d = –28 (3)

3 × equation (2) gives: 12c + 15d = 96 (4)

(4) – (3) gives: 31d = 124 from which, d = 12431

= 4

Substituting in (1) gives: 3c – 16 = –7 i.e. 3c = 16 – 7 = 9 and c = 93

= 3

Thus, c = 3 and d = 4 and may be checked by substituting into both of the original equations

6. Solve the simultaneous equations 3 25

r + – 2 14

s − = 115

3 24

r+ + 53

s− = 154

Rearranging gives: 3 2 2 1 11(20) (20) (20)5 4 5

r s+ −− = i.e. 4(3r + 2) – 5(2s – 1) = 44

and 3 2 5 15(12) (12) (12)4 3 4

r s+ −+ = i.e. 3(3 + 2r) + 4(5 – s) = 45

Since 4(3r + 2) – 5(2s – 1) = 44 then 12r + 8 – 10s + 5 = 44 i.e. 12r – 10s = 31 and 3(3 + 2r) + 4(5 – s) = 45 then 9 + 6r + 20 – 4s = 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 = 12

Substituting in (1) gives: 12r – 5 = 31 i.e. 12r = 31 + 5 = 36 and r = 3612

= 3

Thus, r = 3 and s = 12

and may be checked by substituting into both of the original equations

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7. Solve the simultaneous equations 5x y+

= 2027

42x y−

= 1633

Multiplying both sides of the first equation by 27(x + y) gives: i.e.

527( )x yx y

++

= 2027( )27

x y+ 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.

433(2 )2

x yx y

−−

= 1633(2 )33

x y−

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 from which, x = 1200240

= 5

Substituting in (1) gives: 100 + 20y = 135 i.e. 20y = 135 – 100 = 35 and y = 3520

= 1 34

8. If 5x – 3y

= 1 and x + 4y

= 52

find the value of 1xyy+

If 5x – 3y

= 1 then 5x = 3y

+ 1 and x = 3 15 5y

+

and if x + 4 52y

= then x = 4 52y

− +

Equating x values gives: 3 15 5y

+ = 4 52y

− + i.e. 3 4 5 15 2 5y y

+ = −

i.e. 3 20 25 25 10y+ −

= i.e. 23 235 10y

=

and (23)(10) = (5y)(23) i.e. 230 = 115y

from which, y = 230115

= 2

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Substituting into the first equation gives: 5x – 32

= 1 i.e. 5x = 2.5 and x = 2.5 15 2

=

Thus, 1xyy+ =

( )1 2 11 12

2 2

+ + = = 1

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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 if W = 90 when P = 22, then: 22 = 90a + b (2) Equation (2) – equation (1) gives: 10 = 50a

from which, a = 10 150 5

= or 0.2

Substituting in (1) gives: 12 = 1405

+ 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 = 0.2I1 + 2(I1 – I2) 12 = 3I2 + 0.4I2 – 2(I1 – I2) Determine the values of currents I1 and I2 Rearranging ( )1 1 25 0.2 2I I I= + − gives: 1 1 25 0.2 2 2I I I= + − i.e. 1 22.2 2 5I I− = Rearranging ( )2 2 1 212 3 0.4 2I I I I= + − − gives: 2 2 1 212 3 0.4 2 2I I I I= + − + i.e.

1 22 5.4 12I I− + = Thus, 1 22.2 2 5I I− = (1) and 1 22 5.4 12I I− + = (2) 2 × equation (1) gives: 1 24.4 4 10I I− = (3)

2.2 × equation (2) gives: 1 24.4 11.88 26.4I I− + = (4)

(3) + (4) gives: 27.88 36.4I = from which, 236.47.88

I = = 4.62

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Substituting in (1) gives: 12.2 9.24 5I − = i.e. 12.2 14.24I = and 114.24

2.2I = = 6.47

Thus, 1I = 6.47 and 2I = 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 = 205

= 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) = 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 3C + 4V = 97 700 (1) 5C + 2V = 103 100 (2) 2 × equation (2) gives: 10C + 4V = 206 200 (3)

(3) – (1) gives: 7C = 108 500 from which, C = 108 5007

= 15 500

Substituting in (1) gives: 46 500 + 4V = 97 700 i.e. 4V = 108 5007

97 700 – 46 500 = 51 200

from which, V = 51 2004

= 12 800

Hence, a car costs £15 500 and a van costs £12 800

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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 = 1.53−

= –0.5

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 = R0(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 R0

R = ( )0 1R tα+ thus when R = 25.44 Ω at t = 30°C, then: 25.44 = ( )0 1 30R α+ (1) and when R = 32.17 Ω at t = 100°C, then: 32.17 = ( )0 1 100R α+ (2)

Dividing equation (2) by equation (1) gives: ( )( )

0

0

1 10032.1725.44 1 30

RR

αα

+=

+

i.e. ( )( )1 10032.17

25.44 1 30αα

+=

+ from which, (32.17)(1 30 ) (25.44)(1 100 )α α+ = +

and 32.17 + 965.1α = 25.44 + 2544α i.e. 32.17 – 25.44 = 2544α – 965.1α i.e. 6.73 = 1578.9α

from which, α = 6.731578.9

= 0.00426

Substituting in (1) gives: 25.44 = [ ] ( )0 01 (30)(0.00426) 1.1278R R+ =

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200

from which, 0R = 25.441.1278

= 22.56

Thus, α = 0.00426 and 0R = 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: 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 = 120300

= 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 = 8421

= 4

Substituting in (1) gives: 26 = 16 + b 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)

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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 from which, F1 = 2416

= 1.5

Substituting in (1) gives: 7.5 + 3F2 + 6 = 0 i.e. 3F2 = –7.5 – 6 = –13.5

from which, F2 = 13.53

−

= –4.5

10. For a balanced beam, the equilibrium of forces is given by: 1 2 12.0kNR R+ =

As a result of taking moments: 1 20.2 7 0.3 3 0.6 0.8R R+ × + × =

Determine the values of the reaction forces 1R and 2R

Rearranging gives: 1 2 12.0R R+ = (1)

1 20.2 0.8 3.9R R− = − (2)

5 × (2) gives: 1 24.0 19.5R R− = − (3)

(1) – (3) gives: 5.0 2R = 31.5

from which, 2R = 31.55

= 6.3 kN

Substituting in (1) gives: 1 6.3 12.0R + =

Hence, 1R = 12.0 – 6.3 = 5.7 kN

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EXERCISE 55 Page 113

1. Solve the simultaneous equations 2 4 16x y z+ + =

2 5 18x y z− + =

3 2 2 14x y z+ + = 2 4 16x y z+ + = (1)

2 5 18x y z− + = (2)

3 2 2 14x y z+ + = (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 = 3819

= 2

Substituting in (6) gives: 10 + 14z = 52

i.e. 14z = 52 – 10 = 42 from which, z = 4214

= 3

Substituting in (1) gives: 2 + 2y + 12 =16

i.e. 2y = 16 – 2 – 12 = 2 and y = 22

= 1

2. Solve the simultaneous equations 2 0x y z+ − =

3 2 4x y z+ + =

5 3 2 8x y z+ + = 2 0x y z+ − = (1)

3 2 4x y z+ + = (2)

5 3 2 8x y z+ + = (3)

(1) + (2) gives: 5x + 3y = 4 (4) 2 × (1) gives: 4x + 2y – 2z = 0 (5)

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(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 = 42

= 2

Substituting in (4) gives: 10 + 3y = 4

i.e. 3y = 4 – 10 = –6 from which, y = 63− = –2

Substituting in (1) gives: 4 + –2 – z = 0 i.e. z = 2 3. Solve the simultaneous equations 3 5 2 6x y z+ + =

3 0x y z− + =

2 7 3 3x y z+ + = − 3 5 2 6x y z+ + = (1)

3 0x y z− + = (2)

2 7 3 3x y z+ + = − (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 = 3939−

= –1

Substituting in (7) gives: 7x – 17 = 18

i.e. 7x = 18 + 17 = 35 from which, x = 357

= 5

Substituting in (1) gives: 15 – 5 + 2z = 6

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i.e. 2z = 6 – 15 + 5 = –4 from which, z = 42− = –2

4. Solve the simultaneous equations 2 4 5 23x y z+ + =

3 2 6x y z− − =

4 2 5 31x y z+ + = 2 4 5 23x y z+ + = (1)

3 2 6x y z− − = (2)

4 2 5 31x y z+ + = (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 = 17644

= 4

Substituting in (7) gives: 24 + 5z = 39

i.e. 5z = 39 – 24 = 15 from which, z = 155

= 3

Substituting in (1) gives: 8 + 4y + 15 = 23 i.e. 4y = 23 – 8 – 15 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)

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(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 = 102

= 5

Substituting in (1) gives: 2x + 12 + 20 = 36

i.e. 2x = 36 – 12 – 20 = 4 from which, x = 42

= 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 = 497

= 7

Substituting in (4) gives: 6x + 35 = 41

i.e. 6x = 41 – 35 = 6 from which, x = 66

= 1

Substituting in (1) gives: 4 + y + 21 = 31 i.e. y = 31 – 4 – 21 = 6

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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 = 1105221

= 5

Substituting in (7) gives: 125 – 9z = 107

i.e. 125 – 107 = 9z from which, z = 189

= 2

Substituting in (1) gives: 25 + 5y – 8 = 37

i.e. 5y = 37 – 25 + 8 = 20 from which, y = 204

= 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)

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(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 = 24060

− = –4

Substituting in (5) gives: –32 – 4z = –40

i.e. 40 – 32 = 4z from which, z = 84

= 2

Substituting in (1) gives: –24 + 7y + 16 = 13

i.e. 7y = 13 + 24 – 16 = 21 from which, y = 217

= 3

9. Solve the simultaneous equations 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) 2 × (4) gives: 22x – 2y = 28 (6)

(6) – (5) gives: 15x = 22.5 from which, x = 22.515

= 1.5

Substituting in (4) gives: 16.5 – y = 14 i.e. 16.5 – 14 = y i.e. y = 2.5 Substituting in (1) gives: 4.5 + 5 + z = 14 i.e. z = 14 – 4.5 – 5 = 4.5

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10. Kirchhoff’s laws are used to determine the current equations in an electrical network and result

in the following:

1 2 3

1 2 3

1 2 3

8 3 313 2 52 3 2 6

i i ii i ii i i

+ + = −− + = −− + =

Determine the values of 1i , 2i and 3i

1i + 8 2i + 3 3i = –31 (1)

3 1i – 2 2i + 3i = –5 (2)

2 1i – 3 2i + 2 3i = 6 (3)

(2) × (2) gives: 6 1i – 4 2i + 2 3i = –10 (4) (4) – (3) gives: 4 1i – 2i = –16 (5) 3 × (2) gives: 9 1i – 6 2i + 3 3i = –15 (6) (6) – (1) gives: 8 1i – 14 2i = 16 (7) 2 × (5) gives: 8 1i – 2 2i = –32 (8)

(7) – (8) gives: –12 2i = 48 from which, 2i = 4812

− = –4

Substituting in (5) gives: 4 1i + 4 = –16

i.e. 4 1i = –16 – 4 = –20 i.e. 1i = 204

− = –5

Substituting in (1) gives: –5 – 32 + 3 3i = –31

i.e. 3 3i = –31 + 5 + 32 = 6 i.e. 3i = 63

= 2

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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. 1.4F1 + 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)

(1) – (2) gives: –2.8F1 + 4.2F3 = 11.2 (4) 2 × (2) gives: 8.4F1 – 2.8F2 + 11.2F3 = 70.0 (5) (1) + (5) gives: 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: 80.36F3 = 321.44 from which, 3F = 321.4480.36

= 4

Substituting in (4) gives: –2.8F1 + 16.8 = 11.2

i.e. 16.8 – 11.2 = 2.8F1 i.e. 1F = 5.62.8

= 2

Substituting in (1) gives: 2.8 + 2.8F2 + 11.2 = 5.6

i.e. 2.8F2 = 5.6 – 2.8 – 11.2 = –8.4 i.e. 2F = 8.42.8

= –3