SOLUTION: Find the area of a rectangle whose length is five less than four times its width and whose perimeter is thirty more than six times its width.

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Question 893526: Find the area of a rectangle whose length is five less than four times its width and whose perimeter is thirty more than six times its width.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
w and L for width and length, in that order.

L=-5+4w and 2w+2L=30+6w. These are just the symbolism translations of the description.

The perimeter description equation simplifies tow%2BL=15%2B3w,
L=15%2B2w
2w-L=-15
L-2w=15

Substituting for L,
%284w-5%29-2w=15
2w-5=15
2w=20
highlight%28w=10%29
-
Now find L.
L=4w-5
L=4%2A20-5
highlight%28L=75%29

Area is wL.

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Find the area of a rectangle whose length is five less than four times its width and whose perimeter is thirty more than six times its width.


Let width be W

Then length = 4W - 5

Perimeter = 2W + 2L, or 2W + 2(4W - 5), or 2W + 8W - 10, or 10W - 10

With given perimeter, we have: 10W - 10 = 6W + 30

10W - 6W = 30 + 10

4W = 40

W, or with = 40%2F4, or 10

Length:  4(10) - 5, or 40 - 5, or 35

Area:    10(35), or highlight_green%28350%29