Question 1169317
let a = the 100's digit
let b = the 10's digit
let c = the unite
then
100a + 10b + c = the original number
:
write an equation for each statement, simplify
:
A three-digit number is such that the sum of the digits is 12.
a + b + c = 12
:
 If the digits are reversed, the resulting number is 198 less than the original number.
100a + 10b + c = 100c + 10b + a + 198
Combine like terms on the left
100a - a + 10b - 10b + c - 100c = 198
99a - 99c = 198
simplify, divide by 99
a - c = 2
:
Also, the hundreds digit is equal to the sum of the ones digit and the tens digit. Find the original number.
a = b + c
rewrite to
a - b - c = 0
:
Use elimination on the 1st and last equations
a + b + c = 12
a - b - c = 0
---------------addition eliminates b and c, find a
2a = 12
a = 6
Replace a with 6 in the 2nd simplified equation
6 - c = 2
-c = 2 - 6
-c = -4
c = 4
find b
6 + b + 4 = 12
b = 12 - 10
b = 2
:
the original number: 624
:
:
Check in the statement
"If the digits are reversed, the resulting number is 198 less than the original number."
Subtract
624
426
----
198