Jack J. answered 01/01/18
Tutor
5
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Experienced and Friendly Tutor from Johns Hopkins and UC Berkeley
Length = L
Width = W
The width of a rectangle is two more than three times its length:
W = 2 + (3*L)
The area of the rectangle is 5:
Area = L*W = 5
We want to solve for a single Variable, so let’s substitute the equation we got for W into the Area:
5 = L * W
5 = L * (2+3L)
5 = 2L + 3L2
Let’s subtract 5 from both sides and rearrange the equation.
0 = 3L2 + 2L - 5
0= (3L+5)*(L-1)
Solve for L:
(3L+5) = 0 and (L-1) = 0
L = -5/3 and L = 1
Careful! Length of a rectangle cannot be negative, so (L=1) is our only option.
Plug in (L=1) to get the Width:
W = 2+ 3L
W = 2+3(1)
W = 5
Dimensions of the Rectangle:
L = 1 and W = 5
Width = W
The width of a rectangle is two more than three times its length:
W = 2 + (3*L)
The area of the rectangle is 5:
Area = L*W = 5
We want to solve for a single Variable, so let’s substitute the equation we got for W into the Area:
5 = L * W
5 = L * (2+3L)
5 = 2L + 3L2
Let’s subtract 5 from both sides and rearrange the equation.
0 = 3L2 + 2L - 5
0= (3L+5)*(L-1)
Solve for L:
(3L+5) = 0 and (L-1) = 0
L = -5/3 and L = 1
Careful! Length of a rectangle cannot be negative, so (L=1) is our only option.
Plug in (L=1) to get the Width:
W = 2+ 3L
W = 2+3(1)
W = 5
Dimensions of the Rectangle:
L = 1 and W = 5