PLEASE HELP ME WITH THIS PROBLEM
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Given:
Length of a rectangle = (3x-3)
Width of the rectangle = (2x+1)
To find:
The area of the rectangle.
Solution:
We know that, area of a rectangle is
[tex]Area=Length \times width[/tex]
Putting Length = (3x-3) and width = (2x+1), we get
[tex]Area=(3x-3)\times (2x+1)[/tex]
[tex]Area=(3x-3)(2x+1)[/tex]
Using distributive property, we get
[tex]Area=3x(2x+1)-3(2x+1)[/tex]
[tex]Area=3x(2x)+3x(1)-3(2x)-3(1)[/tex]
[tex]Area=6x^2+3x-6x-3[/tex]
[tex]Area=6x^2-3x-3[/tex]
Therefore, the area of rectangle is [tex](3x-3)(2x+1)\equiv 6x^2-3x-3[/tex] sq. units.