Pre Calculus 10 question

[Rainbow]

I am having problems with this, can anyone help?

The perimeter of a rectangle is 64cm. Twice the width is 4cm more than the length. Find the dimensions of the rectangle.

I have to write a system of equations to solve it.

My length would be; 2(width) - 4

My first equation I think would be; x +y = 64, or would it be 2x + 2y = 64, and then where do I go from here?

 

[serds]

hey, first of all you have to define the varibles: x=lenght; y=width or vice versa. Secondly, the perimeter is 2x+2y because there are two sides of each, so it's 2x+2y=64

 

[Deleted member 4993]

I am having problems with this, can anyone help?

The perimeter of a rectangle is 64cm. Twice the width is 4cm more than the length. Find the dimensions of the rectangle.

I have to write a system of equations to solve it.

My length would be; 2(width) - 4

My first equation I think would be; x +y = 64, or would it be 2x + 2y = 64, and then where do I go from here?

If length = L and Width = W then the first given condition:

The perimeter of a rectangle is 64cm

translates to:

2*L + 2*W = 64.................................................................(1)

Now convert the next condition to an equation:

Twice the width is 4cm more than the length

L = 2* W - 4.......................................................................(2)

Now solve for L and W from (1) and (2)