I am thinking of a two digit number... (System of Equations Question) I am thinking of a two digit number. If the digits of my number are reversed, the new number is 36 greater than my original number. If the tens digit of my original number is doubled and the units digit is halved, the new number is 17 greater than my original number. What is my original number?
What I'm thinking of doing:
Let x = $tens$
Let y = $ones$  (two digit number, so two places)
And go on from there, but I can't see how to apply "digits of a number" being "reversed" to algebra. The only thing I could think of is raising to to a negative 1 power, but that's not so much reversing the digits. They mean 123 = 321, etc.
Could anyone help me with this? Would be really good. This one is particularly tricky. 
Thanks 
 A: From the question we have 
$$10y +x - 36 = 20x + \dfrac{y}{2} -17$$
multiply both sides by 2
$$20y + 2x - 72 = 40x +y - 34 \Rightarrow 19 y = 38x +38 \Rightarrow y = 2 x + 2$$
We know $x$ and $y$ are integers in the range 0 to 9 so possible solutions are
$$
\begin{array}{c|c}
x & y \\
\hline
0 & 2 \\
1 & 4 \\
2 & 6 \\
3 & 8 \\ 
\end{array}
$$
We have so far reduced the number of possible solutions to 4 check them all 
$$20 - 2 = 18 \neq 36 $$
$$41 -14 = 27 \neq 36 $$
$$62 -26 = 36, 43 - 26 =17$$
$$ 83 - 38 = 45 \neq 36$$
Only one of these works our original number was 26.
A: You are on the right track by setting the x and y to digits. 
A hint to get you started: the value of the number "xy" is $10x + y$, while the value of the number "yx" is $10y + x$
A: It's a nice question, there is something satisfying about using variables to represent actual digits.
The first step I took was to make two equations from your question:
10x + y + 36 = 10y + x
10x - y/2 = 17

Then by rearranging and solving as simultaneous equations (either through methods of elimination or substitution) I got my solution.
x - y = -4
20x - y = 34

20x - x - y + y = 38
19x = 38
x = 2

2 - y = -4
y = 6

hence x = 2, y = 6

Therefore the original number is 26.
