Basic Engineering Mathematics, Fifth Edition

92 Basic Engineering Mathematics

5 ×equation (1) gives 35 x− 10 y= 130 (3)

2 ×equation (2) gives 12 x+ 10 y= 58 (4)

Adding equations (3)

and (4) gives^47 x+^0 =^188

Hence, x=^188

47

= 4

Note that when the signs of common coefficients are

differentthe two equations areaddedand when the

signs of common coefficients are thesamethe two

equations aresubtracted(as in Problems 1 and 3).

Substitutingx=4 in equation (1) gives

7 ( 4 )− 2 y= 26

28 − 2 y= 26

28 − 26 = 2 y

2 = 2 y

Hence, y= 1

Checking, by substitutingx=4andy=1inequation

(2), gives

LHS= 6 ( 4 )+ 5 ( 1 )= 24 + 5 = 29 =RHS

Thus, the solution isx= 4 ,y= 1.

Now try the following Practice Exercise

PracticeExercise 49 Solving simultaneous

equations (answers on page 345)

Solve the following simultaneous equations and

verify the results.

1. 2x−y=62.2x−y= 2
   x+y= 6 x− 3 y=− 9


2. x− 4 y=−44.3x− 2 y= 10
   5 x− 2 y= 75 x+y= 21


3. 5p+ 4 q=66.7x+ 2 y= 11
   2 p− 3 q= 73 x− 5 y=− 7


4. 2x− 7 y=−88.a+ 2 b= 8
   3 x+ 4 y= 17 b− 3 a=− 3


5. a+b= 7 10. 2x+ 5 y= 7
   a−b= 3 x+ 3 y= 4
   11. 3s+ 2 t= 12 12. 3x− 2 y= 13
   4 s−t= 52 x+ 5 y=− 4
   13. 5m− 3 n= 11 14. 8a− 3 b= 51
   3 m+n= 83 a+ 4 b= 14
   15. 5x= 2 y 16. 5c= 1 − 3 d
   3 x+ 7 y= 41 2 d+c+ 4 = 0

13.3 Further solving of simultaneous equations

Here are some further worked problems on solving

simultaneous equations.

Problem 5. Solve

3 p= 2 q (1)

4 p+q+ 11 =0(2)

Rearranging gives

3 p− 2 q=0(3)

4 p+q=− 11 (4)

Multiplying equation (4) by 2 gives

8 p+ 2 q=− 22 (5)

Adding equations (3) and (5) gives

11 p+ 0 =− 22

p=

− 22

11

=− 2

Substitutingp=−2 into equation (1) gives

3 (− 2 )= 2 q

− 6 = 2 q

q=

− 6

2

=− 3

Checking, by substituting p=−2andq=−3into

equation (2), gives

LHS= 4 (− 2 )+(− 3 )+ 11 =− 8 − 3 + 11 = 0 =RHS

Hence, the solution isp=− 2 ,q=− 3.