Question

$\begin{array}{c}\text{The difference of two numbers is}3\text{and the}\text{difference of their squares is}69.\text{Find the}\text{numbers.}\end{array}$

Answer 1

Let $a\ne b$ are two numbers $a>b$

and $a-b=3\Rightarrow a=b+3$

and $a^2-b^2=69$

Now $(b+3{)}^2-b^2=60=b^2+6b+9-b^2=69$

$6b=60\Rightarrow b=10$

$a=10+3$

$a=13$

So, Number are $3\nless 10$