Given, pair of equations
2x+3y=7
and (a+b)x+(2a−b)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=(2a−b) and c2=−(a+b+1)
For infinitely many solutions,
a1a2=b1b2=c1c2
∴2a+b=32a−b=−7−3(a+b+1)
I II III
On taking I and II terms, we get
2a+b=32a−b
⇒2(2a−b)=3(a+b)
⇒4a−2b=3a+3b
⇒4a−3a−3b−2b=0
⇒a−5b=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
⇒6a−7a+6b−7b=−6
⇒a+b=6......... (2)
Solving eqn (1) and (2), we get
a−5b=0
a+b=6
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−6b=−6
⇒b=1
Now, substituting the value of b in eqn (2),
we get
⇒a+b=6
⇒a+1=6
⇒a=5