The length of a rectangle is 3 inches more than the width. If the area of the rectangle is 54 in. squared, find the length and width.
2007-10-19 02:47:29 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Width = x
Length = 3 inches more than width = 3 + x
Area = length * width
54 = (3 + x)(x)
54 = 3x + x^2
0 = x^2 + 3x - 54
0 = (x - 6)(x + 9)
x = 6 and -9
can eliminate the -9 (can't have a negative measurement).
so...
width = x = 6
length = 3 + x = 9
2007-10-19 02:56:49 · answer #1 · answered by Mathematica 7 · 0⤊ 0⤋
L=W+3
L.W=54
(W+3).W=54
W^2+3W-54=0
W1=(-3+15)/2=6
W2=(-3-15)/2=-9 (negative, does not satisfy)
Hence
Width=6
Length=W+3=6+3=9
2007-10-19 03:01:49 · answer #2 · answered by iyiogrenci 6 · 0⤊ 0⤋
w+3 = length
w = width
w(w+3) = 54
w^2 + 3w - 54 = 0
(w-6)(w+9) = 0
w = 6 or w = -9 width can't be negative so w = 6
L = 6 +3 = 9
9 * 6 -= 54
2007-10-19 02:55:45 · answer #3 · answered by Marcus C 3 · 1⤊ 0⤋
L = W + 3
A = LW = 54 sq.in.
(W + 3)(W) = 54
Solve for W.
W^2 + 3W - 54 = 0
(W + 9)(W - 6) = 0
W = 6 in. is the only positive solution.
L = W + 3 = 6 + 3 = 9in.
2007-10-19 02:57:42 · answer #4 · answered by Darlene 4 · 0⤊ 0⤋
Two equations, two unknowns. Solve one for the other, then solve the other.
Area=Lw=54
L-w=3, so L=3+w
So insert 3+w for L and you get 54=w(3+w)=3w+w^2
Solve for w, you get 6.
So L=3+w=9
2007-10-19 02:58:15 · answer #5 · answered by bwbear_ 2 · 0⤊ 0⤋
L=3+W
A=54
A=LW
now subsitiute the variable L
2007-10-19 02:55:53 · answer #6 · answered by MoOnliGhtSeductress 4 · 0⤊ 0⤋