Question 640694
The three digits; x, y, z
then
100x+10y+z = "the original number"
:
Write an equation for each statement:
:
"the sum of the digit of a three-place number is 16."
x + y + z = 16
:
"if the digit are reversed and the resulting number added to the original
 number the sum is 1049."
(100x + 10y + z) + (100z + 10y + x) = 1049
Combine like terms
100x + x + 10y + 10y + 100z + z = 1049
101x + 20y + 101z = 1049
 
"if the resulting number is subtracted from the original number the difference
 is 297."
(100x + 10y + z) - (100z + 10y + x) = 297
100x + 10y + z - 100z - 10y - x = 297 
100x - x + 10y - 10y + z - 100z = 297
99x - 99z = 297
Simplify divide by 99
x - z = 3
:
Multiply the 1st equation by 101 and subtract the 2nd equation
101x + 101y + 101z = 1616
101x + 20y + 101z = 1049
---------------------------Subtraction eliminates x and z, find y
 0 + 81y + 0 = 567
y = {{{567/81}}}
y = 7
:
Back to the 1st equation, replace y with 7
x + 7 + z = 16
x + z = 9; subtracted 7 from both sides
then
x + z = 9
x - z = 3
------------addition eliminates z, find x
2x = 12
x = 6
then we know x + z = 9
6 + z = 9
z = 3
:
original number: 673
:
:
Confirm this in the statement:
"the resulting number is subtracted from the original number the difference
 is 297."\
673 - 376 = 297