Question

Jamie has 300 ft of fencing to enclose a rectangular pasture. The pasture's length is to be 10 ft less than 3 times the width. Find the width of the garden. Use a system of linear equations to solve. Use substitution or elimination.

Answer 1

If he has 300 feet of fencing, the perimeter will be 300 feet. For a rectangle, the perimeter is \(2l + 2w\) where \(l,w\) are the length and width respectively.

If the length is 10 feet less than 3 times the width, what would you write for an equation relating \(l\) and \(w\)?

Answer 2

Would the first equation be 2w+2l is less than or equal to 300 and the second equation be 3w-10 is less than or equal to 300?

Answer 3

the first one will be \[2w + 2l = 300\]I'm assuming we want the maximum possible area, subject to the prescribed relationship between length and width.

the second equation doesn't have 300 as an element. It just expresses "the length is to be 10 ft less than 3 times the width"

Answer 4

So the first equation is 2w+2l=300 and the second one is 3w-10=length ?

Answer 5

Yes, that's right.

Answer 6

Will someone help me finish this one?

Answer 7

two equations:

\[2w + 2l = 300\]\[l = 3w-10\]

Why don't you substitute \((3w-10)\) in the first equation wherever you see \(l\)? That will give you an equation in terms of \(w\) only, and you should be able to solve it. When you have the value of \(w\), plug it into \(l = 3w-10\) to get the value of \(l\). Piece of cake!

Answer 8

I got that thank you. But could you help me with my newest question?

Answer 9

Perhaps. Close this question, and award a medal if someone has helped you by clicking the blue Best Response button. Then tag me on your new question when you have it posted by putting @whpalmer4 in a response...