Word problem for math

Question

A new youth activity center is being built in Del rio. The perimeter of the rectangular playing field is 336 yards. The length of the field 6 yards less than double the width. What are the dimensions of the playing field

Answer 1

Hello Brittany,

First, in order to solve the problem you present, we need to develop algebraic expressions to represent both the length and the width since no values are given. Therefore:

Let x equal the width and,

Let 2x - 6 equal the length. <----"The length of the field is 6 yards less than double the width."

Second, the formula for determining the perimeter of a rectangle is as follows:

P = 2(l + w) or P = 2(l) + 2(w). You will notice that the second formula is the first formula with the exception that we applied the distributive property of math, okay? I prefer the second formula, so let's use that one and now all we need do is to substitute the expressions we developed above for "l," "w," and "P."

P = 2(l) + 2(w)

336 = 2(2x - 6) + 2(x)

336 = 4x - 12 + 2x <--------Combine like terms.

336 = 6x - 12 <--------Add 12 to both sides of the equation.

348 = 6x <---------Divide both sides by 6 leaving x to itself.

x = 58 <--------This is the width of the rectangular field.

Now, we want to substitute 58 for the value of "x" in the other equation.

2x - 6 =

2(58) - 6 =

116 - 6 =

110 <-------This is the length of the rectangular field.

Finally, we want to check our work by substituting our values we arrived it with the formula for the perimeter setting one side equal to 336 yards..

P = 2(l) + 2(w)

336 = 2(110) + 2(58)

336 = 220 + 116

336 = 336

The values check out and the solution is valid. I hope I helped you with this solution and wish you a great week. Please feel free to leave feedback should you wish, seek clarity on this solution and need some answers provided to questions, Do so by clicking the "Add Comment" prompt and add your question or feedback. It will appear directly beneath this solution. Best!