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
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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(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