A school is building a rectangular soccer field that has an area of 6000 square yards. The soccer field must be 40 yards longer than its widht. Determine algebraically the dimensions of the soccer field, in yards

length = 100 yd, width = 60 yd

Step-by-step explanation:

To solve this problem, you must set up an equation.

We know that he length (l) is 40 yards greater than the width (w). That means that the length is equal to the width + 40 yards, which can be written as l = w + 40. If the area of the field is 6000 square yards, that means the length times the width must be equal to 6000. Here is the equation that comes from it:

w * (w + 40) = 6000

You can simplify that even further:

w ^2 + 40w = 6000

Divide both sides by 40 to get rid of the coefficient:

w ^2 + w = 150

Now subtract w from both sides, then take the square root of both sides:

w = sqrt (150-w)

You could keep simplifying this equation until you finally arrive at your answer:

w = 60.

This means that the length, which is 40 more than the width, is 100.

If you want to prove this answer, just multiply the width by the length, which is 60*100, which gives you 6000 square yards; just what you need!

Area is 6000 yd2

LW = 6000

field must be 40 yards longer than its width

L = W + 40

Replace L with W+40 in 1st equation to solve for Width

(W+40) (W) = 6000

W2 + 40W - 6000 = 0

This is a quadratic but is factorable

Factors of - 6000 that add to 40 are (100) (-60)

(W+100) (W-60) = 0

W = - 100 or W = 60

Since the width will not be negative discard - 100

The width is 60 yards

Length is W+40 = 100 yards