Find Differences between Ages of A and B. Question: A says to B, I am twice as old as you were, when I was as old as you are. If the sum of ages is 63 years. Find the difference between their ages.
My Question: I understand that we need to form 2 equations in 2 variables and solve them simultaneously to get ages of A and B; and then find the difference.
Equation 1 : A + B = 63.
I am finding constructing the Equation 2 difficult. "I was as old as you are" this means we go back in time by say 'x' years when A was as old as B, 
so will it be A-x = 2*(B-x);
but it introduces a third variable 'x'.
 A: The third variable $x$ that you've introduced is just the present age difference between $A$ and $B$, so $x=A-B$. Thus, $x$ years ago, the age of person-A was
$$A'=A-x=A-(A-B)=A-A+B=B$$
(the age person-B is now!), and the age of person-B back then is found the same way to be
$$B'=B-x=B-(A-B)=B-A+B=2B-A.$$
Now use the information that $x$ years ago person-A was twice as old as person-B:
$$A'=2B'\\
\implies B=2(2B-A)\\
\implies B=4B-2A\\
\implies 2A=3B.$$
So your second equation is $2A=3B$. Combine that with the sum $A+B=63$ to solve for $A$ and $B$.
A: Let's say that $x$ years ago, A was as old as B is now.  But if $a$ is the age of A now, then $a - x$ was A's age $x$ years ago.  And that's how old B is now, so if B's age now is $b$, then $a - x = b$.
We also know that A's age now, that is, $a$, is twice what B's age was $x$ years ago, and so $a = 2(b - x)$.
And as you know, $a + b = 63$.
That's three variables, but also three linear equations, and they have (in this case) a unique solution.
A: There are $3$ (linear) equations:
$$
\left\{\begin{array}{l}
A+B=63;\\
A=2\cdot(B-x);\\
A-x=B.\\
\end{array}
\right.
$$
And hence you can find all $3$ variables $A,B,x$ (difference is $x$):
$B=63-A$;
$x=A-B=A-63+A=2A-63$;
$A=2(63-A)-2(2A-63)=-6A+252$ $\implies$ $7A=252$ $\implies$ $A=36$;
Now you can find $x$ directly.
