Basic Engineering Mathematics, Fifth Edition

96 Basic Engineering Mathematics

Substitutingx=2 into equation (3) gives

5 ( 2 )+ 3 y= 1

10 + 3 y= 1

3 y= 1 − 10 =− 9

y=

− 9

3

=− 3

Checking, substitutingx=2,y=−3 in equation (4)

gives

LHS=− 2 + 3 (− 3 )=− 2 − 9 =− 11 =RHS

Hence, the solution isx= 2 ,y=− 3.

Now try the following Practice Exercise

PracticeExercise 51 Solving more difficult

simultaneous equations (answers on page

345)

In problems 1 to 7, solve the simultaneous equa-

tions and verify the results

1.

3

x

+

2

y

= 14 2.

4

a

−

3

b

= 18

5

x

−

3

y

=− 2

2

a

+

5

b

=− 4

3.

1

2 p

+

3

5 q

=54.

5

x

+

3

y

= 1. 1

5

p

−

1

2 q

=

35

2

3

x

−

7

y

=− 1. 1

5.

c+ 1

4

−

d+ 2

3

+ 1 = 0

1 −c

5

+

3 −d

4

+

13

20

= 0

7.

5

x+y

=

20

27

4

2 x−y

=

16

6.^3 r+^233
5

−

2 s− 1

4

=

11

5

3 + 2 r

4

+

5 −s

3

=

15

4

8. If 5x−

3

y

=1andx+

4

y

=

5

2

, find the value

of

xy+ 1

y

13.5 Practical problemsinvolving

simultaneous equations

There are a number of situations in engineering and

science in which the solutionof simultaneous equations

is required. Some are demonstrated in the following

worked problems.

Problem 12. The law connecting frictionFand

loadLfor an experiment is of the formF=aL+b

whereaandbare constants. WhenF= 5 .6N,

L= 8 .0NandwhenF= 4. 4 N,L= 2 .0N.Findthe

values ofaandband the value ofFwhen

L= 6 .5N

Substituting F= 5 .6andL= 8 .0intoF=aL+b

gives

5. 6 = 8. 0 a+b (1)

Substituting F= 4 .4andL= 2 .0intoF=aL+b

gives

4. 4 = 2. 0 a+b (2)

Subtracting equation (2) from equation (1) gives

1. 2 = 6. 0 a

a=

1. 2

6. 0

=

1

5

or 0. 2

Substitutinga=

1

5

into equation (1) gives

5. 6 = 8. 0

(

1

5

)

+b

5. 6 = 1. 6 +b

5. 6 − 1. 6 =b

i.e. b= 4

Checking, substitutinga=

1

5

andb=4 in equation (2),

gives

RHS= 2. 0

(

1

5

)

+ 4 = 0. 4 + 4 = 4. 4 =LHS

Hence,a=

1

5

andb= 4

WhenL= 6. 5 , F=aL+b=

1

5

( 6. 5 )+ 4 = 1. 3 +4,

i.e.F= 5 .30N.