SOLUTION: A three-digit number is 198 less than the number when its digits are reversed. twice the sum of the digits is 5 more than 7 times the tens digit. the tens digit is 5 less than twic

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Question 690416: A three-digit number is 198 less than the number when its digits are reversed. twice the sum of the digits is 5 more than 7 times the tens digit. the tens digit is 5 less than twice the hundreds digit. find the original number.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
let a = the 100's digit
let b = the 10's
let c = the units
then
100a+10b+c = the number
and
100c+10b+a = the number reversed
:
"A three-digit number is 198 less than the number when its digits are reversed."
100a + 10b + c = 100c + 10b + a - 198
Combine like terms
100a - a + 10b - 10b + c - 100c = -198
99a - 99c = -198
Simplify, divide by 99
a - c = -2
a = c - 2
:
"twice the sum of the digits is 5 more than 7 times the tens digit."
2(a+b+c) = 7b + 5
2a + 2b + 2c = 7b + 5
2a + 2b - 7b + 2c = 5
2a - 5b + 2c = 5
:
" the tens digit is 5 less than twice the hundreds digit."
b = 2a - 5
replace a with (c-2)
b = 2(c-2) - 5
b = 2c - 4 - 5
b = 2c - 9
:
Back to the 2nd statement equation
2a - 5b + 2c = 5
replace a with c-2 and replace b with 2c-9
2(c-2) - 5(2c-9) + 2c = 5
2c - 4 - 10c + 45 + 2c = 5
2c - 10c + 2c + 41 = 5
-6c = 5 - 41
-6c = -36
c = 6
then
a = 6 - 2
a = 4
and
b = 2(6) - 9
b = 3
:
436 is the original number
:
See if that checks out; subtract from the reversed number.
634
436
----Subtract
198