Question 185501
Let L = length of photo and W = width of photo.
From the problem...
{{{L = W+4}}} "A photo is 4 inches longer than it is wide"
A 3-inch border is placed around the photo making the total area 165 sq.ins.
The total area of the photo plus 3-inch border can be expressed as:
{{{A = (L+6)(W+6)}}} because the addition of the 3-inch border adds a total of 6 inches to the length and 6 inches to the width.
Now Substitute A = 165 and L = W+4.
{{{((W+4)+6)(W+6) = 165}}} Simplify the left side.
{{{W^2+16W+60 = 165}}} Subtract 165 from both sides.
{{{W^2+16W-105 = 0}}} Factor this quadratic equation.
{{{(W-5)(W+21) = 0}}} so that...
{{{W = 5}}} or {{{W = -21}}} Discard the negative solution as lengths are positive quantities.
The width is 5 inches and the length is 9 inches.