Question

The length of a rectangle is 4 inches more than twice the width. If the perimeter of

the rectangle is 38 inches, what is the length?

Answer 1

To find the length of the rectangle with a perimeter of 38 inches and a width that is 4 inches less than twice its width, we set up an equation 2(2w + 4) + 2w = 38. Solving for w gives us 5 inches for the width, and subsequently, the length is found to be 14 inches.

Step-by-step explanation:

The student has asked about finding the length of a rectangle given the relationship between length and width and the perimeter of the rectangle. To solve for the length of the rectangle, let's denote the width as w inches. According to the problem, the length is 4 inches more than twice the width, so we can express the length as 2w + 4 inches. The perimeter of a rectangle is given by 2l + 2w, where l is the length and w is the width.

Since the perimeter is 38 inches, we can set up the equation: 2(2w + 4) + 2w = 38. Simplifying this equation gives us 4w + 8 + 2w = 38, which further simplifies to 6w + 8 = 38. Subtracting 8 from both sides gives us 6w = 30, and dividing both sides by 6 gives us w = 5. Now, we can find the length by substituting w into the length expression: 2(5) + 4 = 14 inches.

Therefore, the length of the rectangle is 14 inches.

Answer 2

Step-by-step explanation:

2X+4)+(2X+4)+X+X=38

Lets add our X's and then add our whole numbers. 6X+8=38

Lets work our sentence by subtracting 8 from each side of the equal sign.

6X+8–8=38–8

6X=30

Lets divide each side by 6.

6X/6=30/6

X=5 now lets prove 5 is X.

(2×5+4)+(2×5+4)+5+5=38

(14)+(14)+5+5=38

28+10=38

38=38

X is equal to 5.