Posted by **cbarnett** on Tuesday, December 19, 2006 at 11:27am.

Here's the question: Jose needs to enclose a rectangular section of his yard. The area is 35 sq. ft. and the perimeter is 27 sq.ft. Find the length and width of the section.

The issue I'm having is that I think I need to find the common fators of 35 and 27 and they have no common factors so I have no idea what the answer is. Can anyone help?

L*W=35

2L + 2W= 27

From the first, L= 35/W put that in the second.

2*35/W + 2W= 27

multipy thru by W, and put in quadratic form, and use the quadratic equation or factor.

## Answer this Question

## Related Questions

- Algebra 2( it's a tough one!) - Jose needs to enclose a rectangular section of ...
- Algebra 2 - ~Solving inequalities~ Hello everybody :-) I'm pretty good at ...
- basic geometry - A rectangle has width the same asa a side of a square whose ...
- basic geometric - 10. a rectangle has width the same as a side of a square whose...
- Math - The first question is this: Helen designs a rectangle with an area of 225...
- geometric - The perimeter of a rectangular carpet is 70feet. The with is three -...
- algebra - What is the width of a rectangle with a perimeter of 70 feet if its ...
- math - A carpenter is building rectangular walls for a room addition. The width ...
- algebra II-URGENT - the perimeter of a rectangular yard is 270 feet. If its ...
- Algebra II - The perimeter of a rectangular yard is 270 feet. If its length is ...

More Related Questions