SOLUTION: The perimeter of a rectangle is 46 ft. The width is 3ft less than twice the length. Find the length and width of the rectangle.

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Question 941139: The perimeter of a rectangle is 46 ft. The width is 3ft less than twice the length. Find the length and width of the rectangle.
Answer by laoman(51) About Me  (Show Source):
You can put this solution on YOUR website!
Let the length be l and width be w
From first statement, perimeter = 2(l+w)
2l+2w = 46
w = 2*l - 3
solving the simultaneous eqns
2l+2w = 46
2l-w = 3
Solved by pluggable solver: SOLVE linear system by SUBSTITUTION
Solve:
+system%28+%0D%0A++++2%5Cl+%2B+2%5Cw+=+46%2C%0D%0A++++2%5Cl+%2B+-1%5Cw+=+3+%29%0D%0A++We'll use substitution. After moving 2*w to the right, we get:
2%2Al+=+46+-+2%2Aw, or l+=+46%2F2+-+2%2Aw%2F2. Substitute that
into another equation:
2%2A%2846%2F2+-+2%2Aw%2F2%29+%2B+-1%5Cw+=+3 and simplify: So, we know that w=14.3333333333333. Since l+=+46%2F2+-+2%2Aw%2F2, l=8.6666666666667.

Answer: system%28+l=8.6666666666667%2C+w=14.3333333333333+%29.


length = 8.67 ft
width = 14.33 ft