How to solve a word problem when given width and height of the following? The width of a room is $4$ feet shorter than its length, and its height is $3$ feet less than its length. The area of $4$ walls is larger than the sum of the areas of the floor and ceiling by $134$ square feet. What is the length of the room?
This is what i got from the question 

I don't get about the area thing? how would I solve this problem?
 A: So you know that the Area of the Ceiling and Floor together is $$A_{f,c} = 2L(L-4)$$
you also know that the sum of the other four walls is $$A_w = 2L(L-3) + 2(L-4)(L-3)$$ and by the constraint you laid out we have that $$A_w-A_{f,c}=134$$ so then we have to solve the quadratic $$2L(L-3)+2(L-4)(L-3)-2L(L-4)-134=0$$
which can be simplified down to $$2L^2-6L +2(L^2-7L+12)-2L^2+8L-134=2L^2-12L-110=0$$
so you just need to solve $$L^2-6L-55=0$$ and that factors to $$(L-11)(L+5)=0$$ and you get that $$L=11$$ because length can't be negative.
A: The first sentence of the problem translates mathematically to
\begin{align*}
w &= l-4 & h&=l-3\tag{1}
\end{align*}
The second sentence translates to
$$
2\,lh+2\,wh=134+2\,lw\tag{2}
$$
Dividing equation $(2)$ through by $2$ gives
$$
\frac{2\,lh+2\,wh}{2}=\frac{134+2\,lw}{2} \implies lh+wh=67+lw\tag{3}
$$
Substituting the relations in $(1)$ into equation $(3)$ gives
$$
l(l-3)+(l-4)(l-3)=67+l(l-4)\tag{4}
$$
Simplifying $(4)$ gives
$$
l^2-6\,l-55=0\tag{5}
$$
Can you solve equation $(5)$ for $l$?
