Find Differences between Ages of A and B.

Question

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'.

Answer 1

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.

Answer 2

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 $B$.

Answer 3

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.