The length of a rectangle is 4ft less than twice its width. the perimeter of the rectangle is 256 ft. find the length and width of the rectangle.

For problems like this, change the wording into simple algebraic formulas. We know that the area of a rectangle is its width times its length, or A=LW. The problem tells you that the area is 48, so LW=48. Also, if the length = twice the width minus 4, then L=2W-4.

From here, substitute your new L value for the L in the first equation:
48=LW --> 48=(2W-4)W

Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0. Solving this equation gives us W values of 6 and -4, but since the width of a rectangle cannot be negative, the width must be 6.

Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.

To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.

Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.

Hopefully this helps. If you need any more help or if I went over something too quickly, just let me know.

48ft^2

Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0. Solving this equation gives us W values of 6 and -4, but since the width of a rectangle cannot be negative, the width must be 6.

Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.

To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.

Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.

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