Can someone help me just to do the setup of this the two equations.
I can do the rest is just the step up THAT I AM HAVING DIFFICULTY.
The length of a rectangle is 2in. more than twice its width. If the perimeter of the rectangle is 52in. wHAT IS THE LENGTH AND WIDTH OF THE RECTANGLE.
l= 2w+2
P= 52= 2l+2w
Sure, I can help you with setting up the equations for this problem.
Let's start by assigning variables to the length and width of the rectangle. Let's say the width is 'w' inches.
According to the problem, the length of the rectangle is 2 inches more than twice its width. So, we can express the length as 2w + 2.
Now, let's set up the equation for the perimeter of the rectangle. The perimeter is the sum of all sides, which in this case is made up of two lengths and two widths. The perimeter is given as 52 inches.
Therefore, the equation for the perimeter can be written as:
P = 2L + 2w
Substituting the values we found earlier for the length and width:
52 = 2(2w + 2) + 2w
By distributing the 2 on the right side of the equation:
52 = 4w + 4 + 2w
Combining like terms:
52 = 6w + 4
To isolate the variable 'w,' we subtract 4 from both sides of the equation:
52 - 4 = 6w
Simplifying further gives us:
48 = 6w
Dividing both sides of the equation by 6, we find:
w = 8
Now that we know the value of the width (w = 8), we can substitute it back into the equation for the length of the rectangle to find its value.
The length (l) can be calculated as:
l = 2w + 2
l = 2(8) + 2
l = 16 + 2
l = 18
Therefore, the length of the rectangle is 18 inches, and the width is 8 inches.