SOLUTION: The length of a rectangle is two less than three times its width. If the area of the rectangle is 65 square units, find the length and the width. (The area is equal to its length t

Algebra ->  Surface-area -> SOLUTION: The length of a rectangle is two less than three times its width. If the area of the rectangle is 65 square units, find the length and the width. (The area is equal to its length t      Log On


   

Question 1096902: The length of a rectangle is two less than three times its width. If the area of the rectangle is 65 square units, find the length and the width. (The area is equal to its length times the width.)
Found 2 solutions by josgarithmetic, greenestamps:
Answer by josgarithmetic(39615) About Me  (Show Source):
You can put this solution on YOUR website!
%283w-2%29w=65, w for width;

3w%5E2-2w-65=0
%283w%2B13%29%28w-5%29=0
highlight%28w=5%29
.
.

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!

If w is the width, then the length is 2 less than 3 times w:
let w = width
then 3w-2 - length

The given area, 65, is length times width, so
w%283w-2%29+=+65
3w%5E2-2w+=+65
3w%5E2-2w-65+=+0

You can finish solving the problem by factoring; you will get one positive and one negative solution to the equation. Of course the negative answer makes no sense in the actual problem, so choose the positive answer.

And you SHOULD finish the problem that way, if this is an exercise for a class where you are learning to use algebra to solve problems like this.

But if you only want the answer, it is much easier to simply look at the given area of 65 and look for two numbers whose product is 65.

The only two "nice" numbers whose product is 65 are 5 and 13; and those two numbers satisfy the conditions of the problem.