Question

The length of a rectangle is 3 yd more than twice the width, and the area of the rectangle is 35 yd^2. Find the Dimensions of the rectangle.

Answer 1

Let w = width, now express length in terms of width (w).

Answer 2

The problem tells you that the length is 3 yd more than twice the width.
twice width means 2w............do you follow this?

and 3yd more than twice would be 2w + 3........does than sound reasonable?

Answer 3

Earth to @help78347845879 "??

Answer 4

I kind of still don't get it

Answer 5

I have made three points.
I have labeled width as w, w=width, I could of called it anything, and so can you, what do you want to call width, use a single letter for convenience.

Answer 6

2nd point, I have taken the instructions provided by the problem and expressed length in terms of width.

Answer 7

The third point I was asking if you understood what was going on, or are you capable of understanding.?

Answer 8

This will not be a monolog.

Answer 9

@help78347845879 can you show us where you are having problems or tell us what you don't understand? And don't say you don't understand anything because that isn't true.

Answer 10

thanks @myininaya

Answer 11

When assured that you have followed these points, we will write the equation to solve for the width.

Answer 12

and the length (which are the dimensions of the rectangle)

Answer 13

@help78347845879 no one is trying to belittle you. We just want you to talk to us and tell us what you don't understand.

Answer 14

draw a rect let w be width then length =2w +3

multiply together(w)( 2w+3) = area =35 given
solve w =3.5