Question 130599
{{{drawing(400,400,-1,10,-1,10,
rectangle(0,0,6,8),
rectangle(1,1,5,7),
locate(1.5,4.5,6 in),
locate(6.5,4.5,6 + 2x),
line(6.5,0,6.5,4.1),
line(6.5,0,6.3,.3),
line(6.5,0,6.7,.3),
line(6.5,8,6.3,7.7),
line(6.5,8,6.7,7.7),
line(6.5,4.7,6.5,8),
locate(.5,4.5,x),
locate(2.5,1.5,4 in),
locate(2.5,-.3,4 + 2x),
line(0,-.5,2.2,-.5),
line(0,-.5,.3,-.2),
line(0,-.5,.3,-.8),
line(4,-.5,6,-.5),
line(6,-.5,5.7,-.2),
line(6,-.5,5.7,-.8),
locate(2.8,.7,x)
)}}}


The outside dimensions of the frame, as we can see from the picture are {{{6+2x}}} and {{{4 + 2x}}}, so the area of the overall thing, frame and the picture inside must be {{{(6+2x)(4+2x)}}}, so the area covered by the frame only must be {{{(6+2x)(4+2x)-(6*4)}}}, because we have to subtract the area of the picture.  It is also given that this area must be equal to the picture it surrounds, {{{6*4=24}}}, so we can write:


{{{(6+2x)(4+2x)-24=24}}}


Apply FOIL and put everything on the left:
{{{24+12x+8x+4x^2-24-24=0}}}


Collect terms:
{{{4x^2+20x-24=0}}}


Divide by 4:
{{{x^2+4x-6=0}}}


Use the quadratic formula because there are no rational factors:
{{{x = (-4 +- sqrt( 4^2-4*1*(-6) ))/(2*1) }}} 
{{{x = (-4 +- sqrt( 16+24 ))/(2) }}} 
{{{x = (-4 +- sqrt( 40 ))/(2) }}}


{{{x=-2+-sqrt(10)}}}


But {{{-2-sqrt(10)<0}}} so exclude this root -- we are looking for a positive number measure of length.


Therefore the width of the frame (x in the figure) is exactly {{{-2+sqrt(10)}}} inches or approximately 1.16 inches.