Question 144909
Sure! Let's start with the unknown ---- width = "x".
And we know the length is 3 inches longer, so length= "x+3" .
Now you attached 2-inch border around it, meaning you extended the orig. width 2inches on the left side and 2 inches on the right side right? So the new width dimension will be "x+2+2"= "x+4". Likewise the length is extended 2 inches on the top and 2 inches on the bottom, so it's new dimension will be "x+3+2+2" = "x+7".
Here we gonna use the area given with the photo and the border plus the new dimensions right? So going back in finding the area of rectangle,
A= (x+4)(x+7)  ------------ eqn 1
108= x^2 + 11x + 28
x^2 + 11x -80 = 0
so (x+16)(x-5) right?
It has 2 values, x= -16 which we can't use being negative
x= 5... perfect! This the one to use.
Therefore going back to orig. dimension we can get the size of the photo, where the width = x= 5 inches. Also the length = x+3=5+3= 8inches.
In doubt? Go back eqn 1
A= (5+4)(5+7)
108in^2 = 9*12
108in^2 = 108in^2   
thank you,
Jojo