Question 1181295
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The length of the rectangle is one more than the width. 
If the dimensions are both decreased by 2 units, the area of the new rectangle 
is 30 sq. units less than the area of the original rectangle. 
Find the area of the original rectangle.
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<pre>
Let w be the width of the rectangle; then its length is (w+1) unit, according to the condition.


The hypothetical rectangle has the width of (w-2) units and the length of ((w+1)-2) = (w-1) units;

so its area is this product  (w-2)*(w-1).


And now, we have THIS equation for the areas

    (w-2)*(w-1) = w*(w+1) - 30.


Simplify; then solve for w

    w^2 - 2w - w + 2 = w^2 + w - 30

            -3w + 2  = w - 30

             2 + 30  = w + 3w

              32     = 4w

               w     = 32/4 = 8


Thus we found that the dimensions of the original rectangle are  w= 8 (the width)  and  8+1 = 9 (the length).


Hence, the area of the original triangle is  8*9 = 72 square units.
</pre>

Solved.


<pre>
    To <U>CHECK</U> : the dimensions of the smaller rectangle are 6 and 7 units;  its area is 6*7 = 42 square units;

    it is 30 units less than 72 square units, the area of the original rectangle.   ! Checked !
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