The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 square inches. What are the dimensions of the rectangle?

If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

2x2 + 3x - 161 = 0

3x2 + 2x - 161 = 0

6x2 - 161 = 0

Question

The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 square inches. What are the dimensions of the rectangle?

 If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?


2x2 + 3x - 161 = 0

3x2 + 2x - 161 = 0

6x2 - 161 = 0

e.mccormick · Answer

Well, do you know what the area formula for a rectangle is?

anonymous · Answer

That's what you're trying to figure out. This was the question that I was asked.

e.mccormick · Answer

No, I mean in general.  Any rectangle.

anonymous · Answer

Yes.

e.mccormick · Answer

OK.  If you plug into that, what do you get?

anonymous · Answer

Ohhhh. Ok I get it now. Thank you!

e.mccormick · Answer

You will need to distribute, etc. some, but you should be able to move things around and get one of those.

If I use: LW=A and from the above take:

L=3x+2
W=x
A=161

I can put them into the LW=A and then do some algebra,