Problema Solution
length of a rectangle in 3yd more than twice the width, and the area of the rectangle is 77yd^2
Answer provided by our tutors
We will solve assuming that we need to find the length and the width of the rectangle.
x = the width, x>=0
2x + 3 = the length
A = 77 yd^2 the area
A = width*length
x(2x + 3) = 77
........
click here to see the equation solved for x
........
x = 5.5 yd
We need to ignore the negative solution -7 since the width can not be negative.
2x + 3 = 2*5.5 + 3 = 14 yd
The width is 5.5 yd and the length is 14 yd.