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Question

Find the value of a and b for which the following system of linear equations has infinitely many solutions:
2x+3y=7,(a+b)x+(2ab)y=3(a+b+1)

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Solution

Given, pair of equations
2x+3y=7
and (a+b)x+(2ab)y=3(a+b+1)
On comparing the given equation with the standard form
i.e. a1x+b1y+c1=0 and a2x+b2y+c2=0, we get a1=2,b1=3 and c1=7
and a2=(a+b),b2=(2ab) and c2=(a+b+1)
For infinitely many solutions,
a1a2=b1b2=c1c2
2a+b=32ab=73(a+b+1)
I II III
On taking I and II terms, we get
2a+b=32ab
2(2ab)=3(a+b)
4a2b=3a+3b
4a3a3b2b=0
a5b=0......... (1)
On taking I and III terms, we get
2a+b=73(a+b+1)
6(a+b+1)=7(a+b)
6a+6b+6=7a+7b
6a7a+6b7b=6
a+b=6......... (2)
Solving eqn (1) and (2), we get
a5b=0
a+b=6
- - -
___________
6b=6
b=1
Now, substituting the value of b in eqn (2),
we get
a+b=6
a+1=6
a=5



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