Basic Engineering Mathematics, Fifth Edition

Solving simultaneous equations 95

Subtracting equation (5) from equation (7) gives

43 x+ 0 = 86

x=

86

43

= 2

Substitutingx=2 into equation (3) gives

2

2

+

3

5

y= 4

3

5

y= 4 − 1 = 3

y=

5

3

( 3 )= 5

Since

1

a

=x, a=

1

x

=

1

2

or 0. 5

and since

1

b

=y, b=

1

y

=

1

5

or 0. 2

Hence, the solution isa= 0. 5 ,b= 0. 2 ,whichmaybe
checked in the original equations.

Problem 10. Solve

1

x+y

=

4

27

(1)

1

2 x−y

=

4

33

(2)

To eliminate fractions, both sides of equation (1) are
multiplied by 27(x+y),giving

27 (x+y)

(

1

x+y

)

= 27 (x+y)

(

4

27

)

i.e. 27 ( 1 )= 4 (x+y)

27 = 4 x+ 4 y (3)

Similarly, in equation (2) 33= 4 ( 2 x−y)

i.e. 33 = 8 x− 4 y (4)

Equation (3)+equation (4) gives

60 = 12 xandx=

60

12

= 5

Substitutingx=5 in equation (3) gives

27 = 4 ( 5 )+ 4 y

from which 4y= 27 − 20 = 7

and y=^7

4

= 1

3

4

or 1. 75

Hence,x= 5 ,y= 1. 75 is the required solution, which

may be checked in the original equations.

Problem 11. Solve

x− 1

3

+

y+ 2

5

=

2

15

(1)

1 −x

6

+

5 +y

2

=

5

6

(2)

Before equations (1) and (2) can be simultaneously

solved, the fractions need to be removed and the

equations rearranged.

Multiplying equation (1) by 15 gives

15

(

x− 1

3

)

+ 15

(

y+ 2

5

)

= 15

(

2

15

)

i.e. 5 (x− 1 )+ 3 (y+ 2 )= 2

5 x− 5 + 3 y+ 6 = 2

5 x+ 3 y= 2 + 5 − 6

Hence, 5 x+ 3 y=1(3)

Multiplying equation (2) by 6 gives

6

(

1 −x

6

)

+ 6

(

5 +y

2

)

= 6

(

5

6

)

i.e. ( 1 −x)+ 3 ( 5 +y)= 5

1 −x+ 15 + 3 y= 5

−x+ 3 y= 5 − 1 − 15

Hence, −x+ 3 y=− 11 (4)

Thus the initial problem containing fractions can be

expressed as

5 x+ 3 y=1(3)

−x+ 3 y=− 11 (4)

Subtracting equation (4) from equation (3) gives

6 x+ 0 = 12

x=

12

6

= 2