Question

Find the value of 'a' so that the following linear equations have no solution (3a + 1)x + 3y - 2 =0, (a2+ 1)x + (a - 2)y - 5 = 0​

Answer 1

Answer:

-1

Step-by-step explanation:

No solution =a1/a2=b1/b2 not equal to c1/c2

= 3a+1/a2+1 = 3/a-2 not equal to -2/-5

cross multiplication

=(3a+1)(a-2)= 3(a2+1)

=3a2-2+a-6a = 3a2+3

=a-6a=3+2. ( simplifying)

=-5a=5

= a=-5/5

= a=-1

Hope it helps you....