Math, asked by t2504tushar, 10 months ago

For what value of a and b, the given system of equation has infinitely many solutions : (a – 1)x + 3y = 2 and 6x + (1 –2b) = 6.​

Answers

Answered by khausalyaselvarajah
1

Answer:

(a-1)x+3y=2

(a-1)x+3y-2=0

since a1x+b1y+(-c1)=0

so here a1=(a-1),b1=3,c1=-2

now

6x+(1-2b)y=6

6x+(1-2b)y-6=0

since a2x+b2y+(-c2)=0

here a2=6,b2=(1-2b),c2=-6

Now for infinite solution we need

a1/a2=b1/b2=c1/c2

(a-1)/6=3/(1-2b)=-2/-6

first we will take b1/b2= c1/c2

i.e. 3/(1-2b)= 2/6

2(1-2b)=18

2-4b=18

-4b=18-2=16

b=-4

Now we will take

a1/a2=b1/b2

i.e.a-1/6=3/9. (b=-4)

=>3(a-1)=6. (3/9=1/3)

3a-3=6

3a=6+3

a=3

hope this will help you

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Answered by DrNykterstein
2

(a - 1)x + 3y = 2 ; 6x + (1 - 2b)y = 6

a1 = (a - 1) ; a2 = 6

b1 = 3 ; b2 = ( 1 - 2b)

c1 = 2 ; c2 = 6

For infinitely many solutions, the lines are coinciding.

==> a1 / a2 = b1/b2 = c1/c2

a1/a2 = c1/c2 ; b1/b2 = c1/c2

(a - 1) / 6 = 2/6 ; 3 / ( 1 - 2b ) = 2/6

6a - 6 = 12 ; 2 - 4b = 18

6a = 18 ; 4b = -16

a = 3 ; b = -4

Hence, for infinitely many solutions, value of a is 3 and value of b is -4

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