Question

the rectangle had a length that is 6 inches less that twice it's width. If the perimeter is 54, find the dimensions of the rectangle​

Answer 1

11" x 16"

Explanation:

This problem can be solved with an equation

If you translate the first words into an equation, you get:

l = 2w - 6

(where w is width and l is length)

The perimeter (p) of any rectangle is 2l + 2w, so

if p = 54

then 54 = 2l + 2w

Now, you have the following two equations:

l = 2w - 6

2l + 2w = 54

In order to solve this for w, you can replace the l in the equation with (2w - 6) because of the substitution property of equality (if a = b, then a can be replaced with b at any time), so you get:

54 = 2(2w - 6) + 2w

Now, you simplify:

54 = 2(2w - 6) + 2w

= 54 = 4w - 12 + 2w

= 54 = 6w - 12

= 66 = 6w

= w = 11

So the width is 11. In order to find the length you just need to substitute it into the equation for the perimeter:

2l + 2(11) = 54

= 2l + 22 = 54

= 2l = 32

= l = 16

In conclusion, the width is 11 and the length is 16.

Answer 2

OMG HI

Explanation:

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