Question 1082717
.
7 by 4 units.


7 and 4 are the only positive (integer) factors of 28 that differ by 3.


So, this problem is for mental solution.


<pre>
But if you want to see how to solve it using equation, look into the lesson 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Surface-area/Problems-on-the-area-and-the-perimeter-of-a-rectangle.lesson>Problems on the area and the dimensions of a rectangle</A>

in this site.


With equations, the solution is THIS:

Area A = LW (the product of the length L and the width W).

From the condition, W = L-3. Substitute it into the equation for area.
You will get the quadratic equation

L*(L-3) = 28,

L^2 - 3L - 28 = 0.

Factor left side:

(L-7)*(L+4) = 0.

The only positive root is L=3, which gives you the answer: L = 7 units, W = 7-3 = 4 units.
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Solved.



Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this online textbook under the topic 
"<U>Dimensions and the area of rectangles and circles and their elements</U>".