Number of triplets of a , b c for which the system of equations, a x-b y=2 a-b and (c+1) x+c y=10-a+3 b has infinitely many solutions and x=1, y=3 is one of the solutions, is :

Question

Number of triplets of a , b c for which the system of equations, a x-b y=2 a-b and (c+1) x+c y=10-a+3 b has infinitely many solutions and x=1, y=3 is one of the solutions, is : (A) exactly one (B) exactly two (C) exactly three (D) infinitely many

Tardigrade Admin · Accepted Answer

put x=1 y=3 in 1 text st equation ⇒ a=-2 b from 2 text nd equation c =(9+5 b /4); Now use ( a / c +1)=-( b / c )=(2 a - b /10- a +3 b ); from first two b =0 or c =1; if b=0 ⇒ a=0 c=9 / 4; if c=1 ; b=-1 ; a=2

Q. Number of triplets of $a, b & c$ for which the system of equations, $a x - b y = 2 a - b$ and $(c + 1) x + cy = 10 - a + 3 b$ has infinitely many solutions and $x = 1, y = 3$ is one of the solutions, is :

exactly one

exactly two

exactly three

infinitely many

put $x = 1$ & $y = 3$ in $1^{st}$ equation $\Rightarrow a = - 2 b &$ from $2^{nd}$ equation
$c = \frac{9 + 5 b}{4}$ ; Now use $\frac{a}{c + 1} = - \frac{b}{c} = \frac{2 a - b}{10 - a + 3 b}$ ; from first two $b = 0$
or $c = 1$ ; if $b = 0 \Rightarrow a = 0& c = 9/4$ ; if $c = 1; b = - 1; a = 2$

Q. Number of triplets of a,b&c for which the system of equations, ax−by=2a−b and (c+1)x+cy=10−a+3b has infinitely many solutions and x=1,y=3 is one of the solutions, is :