Question 1074358
A key to this question is that they are assuming the 
width of the frame is uniform all around
Let the width of the frame  = {{{ x }}}
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I will subtract the width of the frame from BOTH sides
of the overall width AND the overall height to get the 
dimensions of the painting itself
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{{{ ( 32 - 2x )*( 29 - 2x ) = 550 }}}
{{{ 928  - 58x - 64x + 4x^2 = 550 }}}
{{{ 4x^2 - 122x + 378 = 0 }}}
{{{ 2x^2 - 61x + 189 = 0 }}}
Use the quadratic formula
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
{{{ a = 2 }}}
{{{ b = -61 }}}
{{{ c = 189 }}}
{{{x = (-(-61) +- sqrt( (-61)^2-4*2*189 ))/(2*2) }}}
{{{x = ( 61 +- sqrt( 3721 - 1512 ))/4 }}}
{{{x = ( 61 +- sqrt( 2209 ))/4 }}}
{{{ x = ( 61 - 47 )/4 }}} ( the positive square root is too big )
{{{ x = 14/4 }}}
{{{ x = 3.5 }}}
The width of the frame is 3.5 inches
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check:
{{{ ( 32 - 2x )*( 29 - 2x ) = 550 }}}
{{{ ( 32 - 2*3.5 )*( 29 - 2*3.5 ) = 550 }}}
{{{ ( 32 - 7 )*( 29 - 7 ) = 550 }}}
{{{ 25*22 = 550 }}}
{{{ 550 = 550 }}}
OK