The area of a rectangular room is 750 square feet. The width of the room is 5 feet less than the length of the room. The area of a rectangular room is 750 square feet. The width of the room is 5 feet less than the length of the room. Which equation can be used to solve for y, the length of the room? Check all that apply. a) y (y + 5) = 750b) y^2 - 5y = 750c) 750 - y (y - 5) = 0d) y (y - 5) + 750 = 0e) (y + 25) (y - 30) = 0

The correct options are

(b) y²-5y=750

(c) 750-y (y-5) = 0

(e) (y+25) (y-30) = 0

Step-by-step explanation:

Given that,

The area of rectangular room is 750 square feet.

Let the length of the rectangular room be y feet.

Since the width of the rectangular 5 less than the length of the room.

Then the width of the given rectangular room is (y-5) feet.

We know the area of a rectangular plot is = Length*width.

The area of the rectangular room is = y (y-5) square feet

According to problem,

y (y-5) = 750 ... (1)

⇒y²-5y=750 ... (2)

⇒y²-5y-750=0

⇒y²-30y+25y-750=0

⇒y (y-30) + 25 (y-30) = 0

⇒ (y-30) (y+25) = 0 ... (3)

⇒y-30=0 or, y+25=0

⇒y = 30, - 25

The length of a rectangle can't negative.

So, x=30.

We can rewrite the equation (1) in form of

(i)

y (y-5) = 750

⇒750 = y (y-5)

⇒750-y (y-5) = 0 ... (4)

(ii)

y (y-5) = 750

⇒y (y-5) - 750=0 ... (5)

The correct options are

(b) y²-5y=750

(c) 750-y (y-5) = 0

(e) (y+25) (y-30) = 0