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: 
$(x+6)(x+3) = 42 \\
x^2+3x+6x+12 = 42 \\
x^2+9x+12x-42 = 0 \\
\\
x^2 + 21 -42$
 A: Area $A$ of the $1$-cm-wide frame in cm$^2$ is $2\times $ height $+ 2\times $ width $+ 4$ (for the corners).
$A = 2 (x+3) + 2(x+6)+4 = 4x+22 = 42$
$\implies x=5$
The length is $(x+6) = 11$ cm 
A: The area of the frame is the difference between the area of the rectangle bordered by the outside of the frame and the area of the picture. Since the frame adds $1~\text{cm}$ on each side of the picture, the outer rectangle has length $x + 8~\text{cm}$ and width $x + 5~\text{cm}$.  Thus, the area of the frame is
\begin{align*}
(x + 8~\text{cm})(x + 5~\text{cm}) - (x + 6~\text{cm})(x + 3~\text{cm}) & = 42~\text{cm}^2\\
(x^2 + 13x~\text{cm} + 40~\text{cm}^2) - (x^2 + 9x~\text{cm} + 18~\text{cm}^2 & = 42~\text{cm}^2\\
4x~\text{cm} + 22~\text{cm}^2 & = 42~\text{cm}^2\\
4x~\text{cm} & = 20~\text{cm}^2\\
x~\text{cm} & = 5~\text{cm}
\end{align*}
Hence the length of the picture is $x + 6~\text{cm} = 5~\text{cm} + 6~\text{cm} = 11~\text{cm}$.
A: Use the equation $(x+5)(x+8) = 42$
to solve for $x$. Then substitute $x$ into $(x + 3)(x + 6)$.
A: 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
$\left( \begin{array}{ccc}
1 & -1 & 0& 0 \\
0 & 1 & 0 & -1 \\
1 & 0 & -1 & 0 \\
2 & 2 & 0& 0 \end{array} \right)^{-1} \left( \begin{array}{c}
3  \\
-2  \\
-2 \\
46 \end{array} \right) = \left( \begin{array}{c}
H  \\
W  \\
FH \\
FW \end{array} \right)$
