could someone help solve this?

Question

he length of a rectangle is 2 inches more than the width. The area is 35 square inches. Find the dimensions.

The length of the rectangle is _______ inches, while the width is, _______ inches

Answer 1

This problem can be solved by setting up 2 equations:

 

1. The first equation is the area of the rectangle. L * W = Area --> In this case the equation is L * W = 35
2. The second equation deals with relating length and width based on the information in the problem, so L = W + 2

 

 

No substitute the length equation (equation 2) into the area equation (equation 1), which gives you (W + 2) * W= 35.

When multiplying by W, we get 2W + W² = 35. By subtracting 35 from both sides we now have an equation that can be factored.

 

 

W² + 2W - 35 = 0

(W - 5)(W+7) = 0

 

 

Now we set each equal to zero to solve for W.

W - 5 = 0

W = 5

 

 

W + 7 = 0

W = -7 [since the width of a rectangle cannot be negative, we know that the width is 5]

 

Knowing that the width of the rectangle is 5, we can now plug this answer into the area equation (equation 1) to find the length.

L * 5 = 35

L = 7