Question 99799
For a number of two digits, 
Let x = 10's digit
Let y = units digit
The number would = (10x+y)
:
the sum of its digits is equal to 1/7th of the number
x + y  = {{{1/7}}}(10x + y)
:
Multiply equation by 7 to get rid of the denominator, results
7x + 7y = 10x + y
Simplify
7y - y  =  10x - 7x
6y = 3x
or
3x = 6y
x = 6y/3
x = 2y
:
 and the product of its digits is equal to 10 less than half the number. 
x * y  =  {{{1/2}}}(10x + y) - 10
:
Multiply equation by 2 to get rid of the denominator, results:
2xy = 10x + y  - 20
:
Substitute 2y for x
2*(2y)*y  =  10(2y) + y - 20
4y^2 = 20y + y - 20
4y^2 = 21y - 20
:
Arrange as a quadratic equation, solve for y
4y^2 - 21y + 20 = 0
:
This factors to:
(4y - 5)(y - 4) = 0
two solutions
4y = 5
y = 5/4, not our solution (it has to be an integer)
and
y = 4; our solution
:
x = 2y
x = 8
:
The number is:  84

:
Check our solution using the statement:
"the sum of its digits is equal to 1/7th of the number"
8  +  4   = {{{1/7}}}*84
12 = 12; confirms our solution
;
You can do the same by checking it in the statement:
"the product of its digits is equal to 10 less than half the number."
:
HOw about this? Could you follow my procedure OK?