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)   (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 to,




Substituting for L,




-
Now find L.




Area is wL.
Answer by MathTherapy(10552)   (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 = , or 10

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

Area:    10(35), or  

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