Find perimeter and area of a rectangle with width of 2(x-y) and length of 3(x+y)?

Question

Please answer this step by step.

Answer 1

Perimieter of rectangle = 2*L + 2*W

Perimieter = 2*(2(x-y) + 2*(3(x+Y)

=4x-4y + 6x+2y

=10x-2y

you can simplify this to 5x-y

Area = L * W

Area = 2(x-y)*3(x+y)

=6(x-y)(x+y)

=6(x^2+xy-xy-y^2)

=6(x^2-y^2)

=6x^2-6y^2

or whatever form you like it in

Answer 2

perimeter= 2(length+width)

             =2(2x-2y+3x+3y)

             =2(5x+y)   answer

Area= length * width

       =2(x-y)*3(x+y)

      =6(x^2 - y^2)   answer

Answer 3

A = 2(x-y)3(x+y)

A = 6 (x-y)(x+y)

A = 6 (x^2-y^2)

A = 6x^2 - 6y^2

P = (2 x 2)(x-y) + (2 x 3)(x+y)

P = 4(x-y) + 6(x+y)

P = 4x - 4y + 6x +6y

P = 10x +2y

Not sure if you wanted it in terms of x and y.

Answer 4

Width=2x-2y

Length=3x+3y

Area=(3x+3y)(2x-2y)

Area=6x^2-6y^2

Perimeter=2(2x-2y)+2(3x+3y)

Perimeter=4x-4y+6x+6y

Perimeter=10x+2y