Question 780636
You said, "I can not use algebra."
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Do you mean, you have too much difficulty trying to use algebra?  An analysis of the exercise can be shown and explained, and will use algebra.
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Do you mean, the use of algebra is not permitted for this exercise?  That prohibition is unnacceptable; algebra is very well suited and is expected to this exercise.


Painted picture, w for width, y for length.
{{{y=4+w.}}}
Area for just the picture is {{{ highlight(wy=w(4+w))}}}


1 inche frame extends length and width to be now
width = w+1+1={{{highlight(w+2)}}}
length = y+1+1=(4+w)+1+1={{{highlight(w+6)}}}
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Think about those distance size changes very carefully before continuing.  We add 2, not just 1.  We extend that 1 inch in TWO opposite directions, so that the 1 inche distance frame-width goes all the way around the picture.


The extended area is {{{(w+2)(w+6)}}}


Given is that the area increases by 48 square inches.  This means the difference between the two areas is 48 square inches.


{{{highlight((w+2)(w+6)-w(4+w)=48)}}}
That appears to be a quadratic equation and just needs simplification.  Some help in starting through that is,...
{{{w^2+2w+6w+12-w^2-4w=48}}}
In fact this will not remain as a quadratic equation, since w^2-w^2=0.
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{{{8w-4w=48}}}
{{{4w=48}}}
{{{highlight(w=12)}}}
and you can compute the y value yourself.