The dimensions of a garden are in whole numbers. If the length is reduced by 3 and the width reduced 1 then the area is reduced by half. What is the are of original rectangle.
Hello, alden35!
Is some information missing?
The problem does not have a unique solution.
The dimensions of a garden are in whole numbers.
If the length is reduced by 3 and the width reduced 1, the area is reduced by half.
What is the area of original rectangle?
Let $\displaystyle L$ = the length
and $\displaystyle W$ = the width.
We have: .$\displaystyle (L-3)(W-1) \:=\:\tfrac{1}{2}LW$
Solve for $\displaystyle W\!:\;W \;=\;\frac{2(L-3)}{L-6}$
Assuming that $\displaystyle L > W$, I still get two possible areas.
. . $\displaystyle \begin{array}{ccc} \text{Length} & \text{Width} & \text{Area} \\ \hline 8 & 5 & 40 \\ 9 & 4 & 36 \\ 12 & 3 & 36 \end{array} \qquad \begin{array}{ccc}L\!-\!3 & W\!-\!1 & \text{Area} \\ \hline 5 & 4 & 20 \\ 6 & 3 & 18 \\ 9 & 2 & 18 \end{array}$