Help With Algebra

[bfastandtea]

the width of a rectangle is 16 feet less than 3 times the length. if the area is 35 square feet, find the dimensions of the rectangle.

[bfastandtea]

i've set this problem up as
35 = l * (3l-16)
when i solve, i get l = 5/3 feet.
but then when i plug in [3(5/3)-16].. i don't arrive at a proper solution.

[jeffa]

It's (fairly obviously) 7ft x 5ft.

7 x 3 = 21
21 - 16 = 5

[bfastandtea]

thank you. i reversed my signs when i was solving for my factors.

[jeffa]

x² = ((x+y) * (x-y)) + y²

[factor-fiction]

That's an identity rather than an equation I think, jeffa.
I agree with your answer though.

bfastandtea- using your equation but replacing l by y as it's easier to read:

35 =y * (3y-16)
35= 3y² -16y
3y²-16y-35 =0

Factorising the quadratic gives:
(3y+5 )(y-7) =0

y= -5 or 7

The only positive solution is y=7
So the width (w) is 35/7 =5
Or w = 3 x 7 - 16 = 5

[factor-fiction]

Sorry, mistyped the irrelevant negative solution which is -5/3