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

stops on an even number.