<s>[INST] find the length of rectangle based on area of frame A rectangular picture is $3$ cm longer than its width, $(x+3)$. A frame $1$ cm wide is placed around the picture. The area of the frame alone is $42 \text{cm}^2$. Find the length of the picture.
I have tries: 
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0 \\
\\
x^2 + 21 -42$
 [/INST] Let $H$,$W$,$FH$,$FW$ be the height, width, frame height and frame width.
We're given the following equations:
$H = W+3$
$W + 2 = FW$
$H + 2 = FH$
$FW \cdot FH - W\cdot H = 42$
It follows:
$(W+2)\cdot (H+2) - W\cdot H = 42$
$\Leftrightarrow$ $((H-3)+2)\cdot (H+2) - ((H-3)\cdot H)= 42$
$\Leftrightarrow$ $H^2+2H-H-2-H^2+3H  =42$
$\Leftrightarrow$ $4H = 44$
$\Leftrightarrow$  $H=11$
Alternatively, we can solve
$\left( \begin{array}{ccc}
1 & -1 & 0& 0 \\
0 & 1 & 0 & -1 \\
1 & 0 & -1 & 0 \\
2 & 2 & 0& 0 \end{array} \right)^{-1} \left( \begin{array}{c}
3  \\
-2  \\
-2 \\
46 \end{array} \right) = \left( \begin{array}{c}
H  \\
W  \\
FH \\
FW \end{array} \right)$
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