1. Tardigrade
  2. Question
  3. Mathematics
  4. 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 :

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

Determinants

Solution:

put $x=1$ & $y=3$ in $1^{\text {st }}$ equation $\Rightarrow a=-2 b \&$ from $2^{\text {nd }}$ 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 $c=1 ; b=-1 ; a=2$