If 2x − 3y = 7 and (a + b)x − (a + b − 3)y = 4a + b represent coincident lines, then a and b satisfy the equation
(a) $a+5 b=0$
(b) $5 a+b=0$
(c) $a-5 b=0$
(d) $5 a-b=0$
The given system of equations are
$2 x-3 y=7$
$(a+b) x-(a+b-3) y=4 a+b$
For coincident lines , infinite number of solution
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b 2}=\frac{c_{1}}{c_{2}}$
$\Rightarrow \frac{2}{(a+b)}=\frac{-3}{-(a+b-3)}=\frac{7}{4 a+b}$
$\Rightarrow 2(a+b-3)=3(a+b)$
$\Rightarrow 2 a+2 b-6=3 a+3 b$
$\Rightarrow 2 a+2 b-3 a-3 b=6$
$\Rightarrow-a-b=6$
$\Rightarrow a+b=-6---(i)$
$3(4 a+b)=7(a+b-3)$
$\Rightarrow 12 a+3 b=7 a+7 b-21$
$\Rightarrow 5 a-4 b=-21---(i i)$
multiply equation $(i)$ by 5, we get $5 a+5 b=-30---(i i i)$
$\operatorname{subtract}(i i)$ from $(i i i)$
$(5 a+5 b)-(5 a-4 b)=-30+21$
$\Rightarrow 5 a+5 b-5 a+4 b=-9$
$\Rightarrow 9 b=-9$
$\Rightarrow b=-1$
substitute $b=-1$ in equation $(1)$
$a+(-1)=-6$
$\Rightarrow a=-6+1=-5$
Option A:
$a+5 b=0$
$-5+5(-1)=-5-5=-10 \neq 0$
Option B:
$5 a+b=0$
$5(-5)+(-1)=-25-1=-26 \neq 0$
Option.C:
$a-b=0$
$-5-(-1)=-4 \neq 0$
None of the option satisfies the values.
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