Question 91488
Let W = the width of the rectangle and L = the length.
The problem description tells you that the length, L, is (=) 1 yd less than three times its width (L = 3W-1).
You also know that the area, A = 24 sq.yds.
Starting with the formula for the area of a rectangle: A = L*W
Making the appropriate substitutions into the formula, you get:
{{{(3W-1)(w) = 24}}} Simplifying this:
{{{3W^2-W = 24}}} Subtract 24 from both sides.
{{{3W^2-W-24 = 0}}} You can solve this quadratic equation by factoring.
{{{(3w+8)(W-3) = 0}}} Apply the zero products principle:
{{{3W+8 = 0}}} or {{{W-3 = 0}}}, so...
{{{3W+8 = 0}}} Subtract 8 from both sides.
{{{3W = -8}}} Divide both sides by 3.
{{{W = -8/3}}} Discard this solution a negative width is not meaningful.

{{{W-3 = 0}}} Add 3 to both sides.
{{{W = 3}}} The width is 3 yards.
{{{L = 3W-1}}}
{{{L = 3(3)-1}}}
{{{L = 9-1}}}
{{{L = 8}}} The length is 8 yards.

Check:
{{{A = L*W}}}
{{{A = 8(3)}}}
{{{A = 24}}}sq.yds.