MHB Simultaneous equations number problem

[mathlearn]

A certain numbers tenth place digit is $x$ & unit place digit is $1$. That number can be written as $(10x+y)$ . The sum of those two digits is 15 . When those two digits are flipped a number is made, That number less the first number results in 27.

What Have I done so far

$10x+y=15$
$(10y+x)-(10x+y)=27=9y-9x=27=y-x=3$

$y-x=3$
$10x+y=15$

Now solving the two simultaneous equations by subtraction,

11x=12

Now by substituting the thing in 1 I get y=5

Now the digit is 501-105=27 which is incorrect

Where have I done wrong ?

Many Thanks :)

 

[MarkFL]

I'm guessing you mean to say the tens digint is $x$ and the unit digit is $y$ such that the number is:

$$10x+y$$

We are told:

$$x+y=15$$ (this is where you went wrong...you just want to add the digits here)

and

$$(10y+x)-(10x+y)=27$$ which simplifies to:

$$y-x=3$$

Can you proceed?