A photograph measures $30cm$ by $20cm$. It is put into a frame such that there is a border of uniform width around the photo. Together the photo and the border have an area of $816cm^2$. What is the width of the border?
What I have interpreted this as is two rectangles, we know the dimensions of one of them and the area of the other. What I'm confused about is how they are even related to each other. I would understand if it said the border is a third bigger than the photo, then I could make a second equation and solve by elimination or substitution. Any help is greatly appreciated.
We can write the length and width of the frame as the length and width of the photograph, plus some constant width of the photo frame, which we'll call $d$. The frame wraps around the entire photograph, so we'll add two frame-widths to each dimension. The area of the entire thing is $816\ \text{cm}^2$. The photo is $30\ \text{cm}\times20\ \text{cm}$. The area of the framed photo can be written as those lengths, plus our constants: $816 = (30+2d)(20+2d)$. Using FOIL, we can expand this: $$816=4d^2+100d+600$$
Re-arrange the equation; divide by 4$$0=d^2+25d-54$$ Now we can solve for $d$. This can be done without the quadratic formula, if you'll notice that $27\cdot2=54$ and $27-2=25$. $$0=(d+27)(d-2)$$ $$d=-27,\ d=2$$ Obviously, width cannot be negative, so we are left with the answer: $d=2$.