Basic Engineering Mathematics, Fifth Edition

Solving simultaneous equations 93

Problem 6. Solve

x

8

+

5

2

=y (1)

13 −

y

3

= 3 x (2)

Whenever fractions are involved in simultaneous equa-
tions it is often easier to firstly remove them. Thus,
multiplying equation (1) by 8 gives

8

(x

8

)

+ 8

(

5

2

)

= 8 y

i.e. x+^20 =^8 y (3)

Multiplying equation (2) by 3 gives

39 −y= 9 x (4)

Rearranging equations (3) and (4) gives

x− 8 y=− 20 (5)

9 x+y= 39 (6)

Multiplying equation (6) by 8 gives

72 x+ 8 y= 312 (7)

Adding equations (5) and (7) gives

73 x+ 0 = 292

x=

292

73

= 4

Substitutingx=4 into equation (5) gives

4 − 8 y=− 20

4 + 20 = 8 y

24 = 8 y

y=

24

8

= 3

Checking, substitutingx=4andy=3 in the original
equations, gives

(1): LHS=^4
8

+

5

2

=

1

2

+ 2

1

2

= 3 =y=RHS

(2): LHS= 13 −
3
3

= 13 − 1 = 12

RHS= 3 x= 3 ( 4 )= 12

Hence, the solution isx= 4 ,y= 3.

Problem 7. Solve

2. 5 x+ 0. 75 − 3 y= 0

1. 6 x= 1. 08 − 1. 2 y

It is often easier to remove decimal fractions. Thus,

multiplying equations (1) and (2) by 100 gives

250 x+ 75 − 300 y=0(1)

160 x= 108 − 120 y (2)

Rearranging gives

250 x− 300 y=− 75 (3)

160 x+ 120 y= 108 (4)

Multiplying equation (3) by 2 gives

500 x− 600 y=− 150 (5)

Multiplying equation (4) by 5 gives

800 x+ 600 y= 540 (6)

Adding equations (5) and (6) gives

1300 x+ 0 = 390

x=

390

1300

=

39

130

=

3

10

= 0. 3

Substitutingx= 0 .3 into equation (1) gives

250 ( 0. 3 )+ 75 − 300 y= 0

75 + 75 = 300 y

150 = 300 y

y=

150

300

= 0. 5

Checking, by substitutingx= 0 .3andy= 0 .5inequa-

tion (2), gives

LHS= 160 ( 0. 3 )= 48

RHS= 108 − 120 ( 0. 5 )= 108 − 60 = 48

Hence, the solution isx= 0. 3 ,y= 0. 5