Question

A two digit number is such that the product of its digits is 18.when 63 is substracted from the number the digit interchange their places

Answer 1

let the two digit number be 10x+ y
ATQ,. x × y= 18..........(1)
10x + y - 63 = 10 y + x
9x - 9y - 63= 0
9x - 9y = 63
x-y = 7..........(.2)
x= 7+y
put value of x in eq 1b, we get
( 7+y ). y = 18
y² + 7y = 18
y² +7y-18=0 
y² +9y -2y -18= 0
y(y+9) -2 ( y +9) =0
( y+9) (y-2). = 0
y=-9 , y = 2
y= 2........
so, x= 7+2=9
therefore the two digit number is 10(9)+2
= 92

Answer 2

${\green {\boxed {\mathtt {☆Solution}}}}$

$\rm \: let \: the \: tens \: and \: unit \: digit \: of \: the \: required \: number \: be \: x \: and \: y \: respectively \: then \\ \rm \: xy = 18 \implies \: y = \frac{18}{x} \\ \rm \: \purple {\: and \: (10x + y) - 63 = 10y + x }\\ \rm\implies \: 9x - 9y = 63 \implies \: x - y = 7 \: \: \: \: .....(1) \\ \rm \orange{ \: putting \: y = \frac{18}{x } \: into \: (1) }\\ \rm \: x - \frac{18}{x} = 7 \\ \rm \: x {}^{2} - 18 - 7x \implies \: x {}^{2} - 7x - 18 \\ \rm \implies \: x {}^{2} - 9x + 2x - 18 = 0 \implies \: x(x - 9) + 2(x - 9) = 0 \\ \rm \implies(x - 9)(x + 2) = 0 \\ \rm \: x = 9 \: or \: x = - 2 \: \: \: \: ( but \: a \: digit \: cannot \: be \: negative) \\ \rm \red {\: \boxed{\therefore \: x = 9}} \\ \rm \: putting \: x = 9 \: in \: (1) we \: get \: y = 2 \\ \rm \: thus \: the \: tens \: digit \: is \: 9 \: and \: the \: unit \: digit \: is \: 2 \\ \rm hence \: the \: required \: number \: is \: 92$