Question 484352
The dimensions of a rectangle are such that its length is 3 inches more than its width. If length were doubled and width decreased by 1, area would be increased by 176 inches squared. What are the length and width?
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Given: l = w + 3 [1]
The original area = A = lw
For the new area we can write
2l*(w-1) = A + 176
2(w+3)*(w-1) = A + 176
Substitute the value for l in [1] into the expressions for the two areas:
A = w(w+3) [2]
A + 176 = 2(w+3)(w-1) -> A = 2(w+3)(w-1) - 176
Since the LHS's are equal, we can equate the RHS's:
w(w+3) = 2(w+3)(w-1) - 176
Simplify and solve for w:
w^2 + 3w = 2w^2 + 4w - 6 - 176
w^2 + w - 182 = 0
w = (-1 +/- sqrt(1 + 4*182))/2
w = (-1 +/- 27)/2 
Taking the positive solution, we have w = 13
Therefore l = 13 + 3 = 16
Ans: w = 13, l = 16