Arectangle is 13 meters longer than it is wide. the perimeter is 98 meters. what is the width?

Hi there!

To solve this problem, let's make two equations:

Let x be the rectangle's length.
Let y be the rectangle's width.

x = y + 13
2x + 2y = 98

Now, since x equals y + 13, we can substitute the value of x into the equation 2x + 2y = 98:

2x + 2y = 98
2(y + 13) + 2y = 98
2y + 26 + 2y = 98
4y = 98 - 26
4y = 72
y = 18

Now, let's plug the value of y into the equation to find x:

x = y + 13
x = 18 + 13
x = 31

So, the length is 31 and the width is 18.

Hope this helps!

Iam not sure , but probably 750 or 7.5