Find a two digit number $15$ more than $4$ times its reverse. Find a two digit number 15 more than 4 times its reverse. I acknowledge the answer is 91 since the reverse of 91 is 19, 4 times 19 is 76, which is 15 less than 91. I used trial and error, but I believe there is a more algebraically method to model this problem. I have developed something that looks as follows:
$x = $ first digit
$y = $ second digit  
$10x + y = 15 + 4 * (10y + x)$
$= 15 + 4(y * 10) + 4x$
$= 15 + 40y + 4x$
$10x = 15 + 39y + 4x$
$6x = 15 + 39y$ 
I'm not sure if i am heading in the correct direction. I am confused how i can change the last expression into anything else. 
 A: You are on the right track.
Note that $x,y\in\{1,2,3,4,5,6,7,8,9\}$ 
Now it has to hold $6x=15+39y$. The LHS is at most $6\cdot 9=54$. While the RHS gives way greater values, even for small $y$. So the only value which can be $y$ is $1$, since already for $y=2$ we would get $15+78=93$.
So $y=1$ and $x=9$. The only solution.
A: With your last part
$$6x = 15 + 39y \tag{1}\label{eq1}$$
You can divide by $3$ to get a slightly simpler
$$2x = 5 + 13y \tag{2}\label{eq2}$$
Now, since $x$ and $y$ are digits, this means that $0 \le x,y \le 9$. Actually, as you're assuming $x$ is the $10$'s digit of the original number, you actually have $1 \le x \le 9$.
Checking the values, note the right hand side is thus $5$ if $y = 0$, which is odd so not divisible by $2$, or next it could be $18$ if $y = 1$, giving $x = 9$. Thus, since any larger value of $y$ would make $x \gt 9$, the only solution is $x = 9, y = 1$, as you found.
You could also check if any $3$ or more digit numbers work as well, although it gets somewhat more complicated. For example, for $3$ digits, for $x \neq 0$, and $z = 0$ only if $y = 0$, you get
$$100x + 10y + z = 15 + 4 \times (100z + 10y + x) \tag{3}\label{eq3}$$
Multiplying out and moving all of the $x,y,z$ terms to the left gives
$$96x - 30y - 399z = 15 \iff 32x - 10y = 5 + 133z \tag{4}\label{eq4}$$
Now, if $z = 0$, then as stated, $y = 0$, but this then gives $32x = 5$, which doesn't work.
Next, if $z = 1$, you then get
$$32x - 10y = 138 \iff 32x - 138 = 10y \tag{5}\label{eq5}$$
This shows that $x \ge 5$. Looking at the remainders when divided by $10$ shows that $2x$ must end in a digit $8$, which only occurs when $x = 9$. However, this gives $288 - 138 = 150 = 10y \implies y = 15$, which is too large.
Next, if $z = 2$, \eqref{eq4} becomes
$$32x - 10y = 271 \iff 32x - 271 = 10y \tag{6}\label{eq6}$$
Only $x = 9$ works, giving $288 - 271 = 17 = 10y$, which doesn't work.
For $z > 2$, you'll need $x \gt 9$. Thus, overall, this shows there's no solution using $3$ digits. I haven't checked, but I suspect many, although not necessarily all, larger number of digits will also fail.
A: We don't know how many digits the number has.
If we assume it is two we do it like this.
Let the number be $N = 10a +b$ where $a,b$ are digits.
Then the reverse is $10b + a$ and $4$ times the revers is $4(10b+a)$ and $15$ more than $4$ times its reverse is $4(10b + a) + 15$.
So $10a + b = 4(10b + a) + 15$.  So we solve
$10a + b = 40b + 4a + 15$ so 
$6a = 39b + 15$ so
$2a = 13b + 5$.
Now $2a \le 18$ so $13b + 5 \le 18$ so $13b \le 13$ so $b\le 1$.
If $b =0$ then $2a = 5$ which doesn't work. If $b =1$ then $2a = 13+5 =18$ and $a =9$.
So $N=10*9 + 1 =91$.
If there are $3$ digits though we would have
$100a + 10b + c = 4(100c + 10b + a) + 15$ so
$96a = 30b + 399c + 15$ so
$32a = 10b + 133c +5$.
This can not be solved but it really should have been stated you were specifically looking for a two digit number.
