CBSE Class 10 Answered

Dear Student,

(a+2b)x + (2a-b)y = 3...............(1)

&

2 (a-2b)x + (2a+b)y = 3............(2)

Multiplying eqn (1) by 2.(a-2b); and eqn (2) by (a+2b), we get

   2.(a-2b).(a+2b)x +2.(a-2b).(2a-b)y = 3.2.(a-2b)...................(3)

& 2.(a-2b).(a+2b)x + (2a+b).(a+2b)y = 3.(a+2b)....................(4)

Subtract eqn(4) from eqn(3), and hence eliminating 'x'; we get:

(2a²-15ab+2b²)y = 18b-3a

=> y = 3.(6b-a)/(2a²-15ab+2b²)

Similarly, multiplying eqn(1) by (2a+b) and eqn(2) by (2a-b);

(a+2b).(2a+b)x + (2a-b).(2a+b)y = 3.(2a+b)..................(5)

2 (a-2b).(2a-b)x + (2a+b).(2a-b)y = 3.(2a-b)..................(6)

Subtracting eqn(6) from eqn(5), and hence eliminating 'y, we get:

(-2a²+11ab-2b²)x= 6ab

So, x = 6ab/(-2a²+11ab-2b²)

& y = 3.(6b-a)/(2a²-15ab+2b²)

Regards Topperlearning.