Determine the value of 'a' for which the following system of linear equation has infinitely many solutions: ax+3y=a-3 ; 12x+ay=a

ax + 3y = a-3 12x + ay = a

For infinite solutions-----

a/12 = 3/a = (a-3)/a

Case i

a/12 = 3/a

= a^2 = 36

a= either 6 or -6

Case ii

3/a = (a-3)/a

=3a= a^2 - 3a

6a = a^2

i.e. a can either be 0 or 6

As 6 satisfies both the conditions, hence the value of a is 6

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since linear equation has infinitly many solutions : a1/a2 = b1/b2 =/= c1/c2

using equations:

a/12 = 3/a

a2 = 36

a = +6 or -6

now substitute the value of a in equations:

when a = 6

a1/a2= 6/12 = 1/2....i

b1/b2= 3/6 = 1/2.....ii

c1/c2 = 3/6 = 1/2 ...iii

so a =/= +6

hence a= -6,

a1/a2= -6/12 = -1/2....iv

b1/b2= 3/-6 = -1/2.....v

c1/c2= -9/6 = -3/2 ...vi

SO, VALUE FOR a = - 6...!

hope it helps u ....!! best of luck...!!

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