A 2 -digit number is seven times the sum of its digits. The number formed by reversing the digits is 18 less than the…

Let unit's digit be $x$ and ten's digit be $y$.
$\therefore$ Number $= 10y + x$
According to the first condition:
$10y + x = 7(x + y)$
$10y + x = 7x + 7y$
$3y - 6x = 0$
$y = 2x$ (i)
According to the second condition:
Number formed by reversing digits is $10x + y$.
$10x + y = (10y + x) - 18$
$10x + y - 10y - x = -18$
$9x - 9y = -18$
$x - y = -2$
$y - x = 2$ (ii)
On solving (i) and (ii):
Substitute (i) into (ii): $2x - x = 2 \Rightarrow x = 2$
Substitute $x=2$ into (i): $y = 2(2) = 4$
$\therefore$ required number is $10(4) + 2 = 42$