.There is a field whose width is 6 meters less than its length. If both the length and width are doubled the perimeter will be 120 meters. What are the dimensions of the field?

Answer 1

First, express the side lengths of the field as algebraic expressions. The length is an unknown variable, and the width is six meters less than that variable.

L = x

W = x - 6

Now, we know the value of the perimeter when each of these values is doubled is 120 meters. So, first we need to double the side lengths.

2L = 2x

2W = 2(x - 6) = 2x - 12

Now we write an expression that will tell us the perimeter of this hypothetical field with the doubled side lengths. You probably know that a perimeter is twice the width plus twice the length, so the width of this doubled field will be:

P = 2(2x) + 2(2x - 12) = 4x + 4x - 24 = 8x - 24

We know the perimeter of this doubled field is 120 meters, so now we know:

120 = 8x -24

Now solve the equation.

144 = 8x

x = 18

Now we know that the length of the original field is 18 meters. The width is 6 meters less than the length, so we know the width is 12 meters.

Answer 2

Perimeter = 2(L+W)

120 =2(2L + 2W)

L+W=30

W=L-6

2L-6=30

You take it from here.

P.S. I am going to take the liberty of giving you a short English lesson as well as some math help.

The word "whose" refers to a person, not to an inanimate object such as a field. The correct way to phrase your question would be: "There is a field the width of which ...