Answer
$13\ in\times10\ in$
Work Step by Step
Step 1. Assume the width is $x$ inch, the length is then $x+3$ before extensions.
Step 2. The area of the unframed picture is $x(x+3)=70\longrightarrow x^2+3x-70=0\longrightarrow (x+10)(x-7)=0$, thus $x=7\ in$ (discard the negative solution as the dimension can not be negative) and $x+3=10\ in$
Step 3. After making the frame with an extension of $1.5$ inch, the length becomes $10+2(1.5)=13\ in$ and the width becomes $7+2(1.5)=10\ in$, that is the outside dimensions are $13\ in\times10\ in$
