The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 square inches. What are the dimensions of the rectangle?

If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

2x^2 + 3x - 161 = 0
3x^2 + 2x - 161 = 0
6x^2 - 161 = 0

Question

The length of a rectangle is 2 more than three times the width. The area of the rectangle is 161 square inches. What are the dimensions of the rectangle?

If x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

2x^2 + 3x - 161 = 0
3x^2 + 2x - 161 = 0
6x^2 - 161 = 0

anonymous · Answer

A=L*W
L=2+3w
161=(2+3W)*(W)
161=2W+3W^2
3W^2+2W-161=0

anonymous · Answer

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anonymous · Answer

:) everyone deserves a medal :D