Im trying to solve the following Simultaneous congruence.
$2x ≡ 3(mod\ 5) $
$3x ≡ 2(mod\ 4)$
$4x ≡ 3(mod\ 9) $
by Chinese remainder theorem
$x$ = $B_1c_1x_1 \ + \ B_2c_2x_2 \ + B_3c_3x_3 \ $
Where
$c_1 = 3$
$c_2 = 2 $
$c_3 = 3$
$B_1 = 2$x$3 $
$B_2 = 3$x$3 $
$B_3 = 3$x$2 $
$x_n=b_n(mod \ b_1)$
$x_1 = 1$
$x_2 = 1 $
$x_3 = no \ solution $
This is all i know and this is when $ x = c (mod \ n)$
but since there is a coefficient infront of x what should i change?
Thanks in advance.