Algebra 2 Homework Question?
"Write three equations in three variables for each problem. Then solve using any method of your choice."
The sum of the digits of a three digit number is 11. The hundreds digit exceeds the sum of the tens digit and the units digit by 1. When the digits are reversed, the new number is 396 less than the original number. Find the new number."
Okay, well I already have my solution but I used the trial-and-error/guess-and-check method. The directions instruct me to write three equations; I only came up with 2.
These are my equations:
x + y + z = 11
x = y + z + 1 (or) x - y - z = 1
My work (using elimination):
x + y + z = 11
x - y - z = 1
-----------------------
2x = 12
x = 6
- x + y + z = -1 (multiplied entire equation by -1)
x + y + z = 11
-----------------------
2y + 2z = 10
2y = -2z + 10
y = -z + 5
This is where I did my guess and check; luckily my first guess was right.
y = -(2) +5
y = 3
z = 2
x+y+z = 11
6+3+2 = 11
reversed is supposed to be 396 less than 632
632 - 396 = 236
I am not sure how to get my third equation. I am thinking it has something to do with the 396 statement. Someone help? Thank you.