Arectangle field has a perimeter of 220 meters and an area of 2856 square meters. what are the dimensions?

The dimensions are 68 meters by 42 meters.

Step-by-step explanation:

Let the length be x

Let the width be y

Perimeter = 220 meter

So,

or or    ...........(1)

Area = 2856 square meter

So,     ............(2)

Substituting the value of x from (1) in (2)

=>

=>

Solving this equation, we get roots as y = 68 and y = 42

So, if we put y = 68, we get x = 42

If we put y = 42, we get x = 68

As length is longer than width, we will take length = 68 meters and width = 42 meters.

Hence, the dimensions are 68 meters by 42 meters.

a

step-by-step explanation:

given the 2 equations

- x + y = - 2 → (1)

2x - 3y = 4 → (2)

multiplying (1) by 2 and adding to (2) will eliminate the x- term

- 2x + 2y = - 4 → (3)

add (2) and (3) term by term

(2x - 2x) + (- 3y + 2y) = (4 - 4), simplifying gives

- y = 0 ⇒ y = 0

substitute y = 0 in either of the 2 equations for corresponding value of x

substituting in (2) gives

2x = 4 ( divide both sides by 2 )

x = 2

solution is (2, 0 ) → a