First of all, this problem has nothing to do with rational functions (which is the category this was posted in). Please use a relevant category when posting. You'll get faster responses.
Let h = the hundred's digit
Let t = the ten's digit
Let w = the "wun's" digit ("o" is a terrible variable)
With three variables we will need 3 equations. From "The sum of the digits of a three-digit number is 11" we get:
h + t + w = 11
"If the digits are reversed, the new number is 46 more than five times the old number." refers to the value of the original number and the value of the number when the digits are reversed. First we will express these values, then we will write an equation:
100h + 10t + w = the original value (think of expanded form)
100w + 10t + h = the value with the digits reversed.
From "the new number is 46 more than five times the old number" and the expressions above we can write:
100w + 10t + h = 5(100h + 10t + w) + 46
which simplifies as follows:
100w + 10t + h = 500h + 50t + 5w + 46
-499h + (-40t) + 95w = 46
From "If the hundreds digit plus twice the tens digit is equal to the units digit" we can write:
h + 2t = w
which I will rewrite as:
h + 2t + (-w) = 0
This gives us a system of:
h + t + w = 11
-499h + (-40t) + 95w = 46
h + 2t + (-w) = 0
I don't know what methods you may have learned to solve such a system. I hope you have learned a matrix method called Gaussian elimination because that is what I will use. (If you are unfamiliar with this, then solve the above system using a method you know.)
Algebra.com does not do matrices very well. So I am just going to show you the rectangular array of numbers (without the usual brackets):
1 1 1 11
-499 -40 95 46
1 2 -1 0
Added 499 times row 1 to row 2.
1 1 1 11
0 459 594 5535
1 2 -1 0
Added -1 times row 1 to row 3.
1 1 1 11
0 459 594 5535
0 1 -2 -11
Replaced row 2 with 1/459 times row 2.
1 1 1 11
0 1 22/17 205/17
0 1 -2 -11
Added -1 times row 2 to row 1.
1 0 -5/17 -18/17
0 1 22/17 205/17
0 1 -2 -11
Added -1 times row 2 to row 3.
1 0 -5/17 -18/17
0 1 22/17 205/17
0 0 -56/17 -392/17
Replaced row 3 with -17/56 times row 3.
1 0 -5/17 -18/17
0 1 22/17 205/17
0 0 1 7
Added 5/17 times row 3 to row 1.
1 0 0 1
0 1 22/17 205/17
0 0 1 7
Added -22/17 times row 3 to row 2.
1 0 0 1
0 1 0 3
0 0 1 7
which translates into
h = 1
t = 3
w = 7
making the original three-digit number: 137