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 equations can be used to solve for y, the length of the room? Select three options. y(y + 5) = 750 y2 – 5y = 750 750 – y(y – 5) = 0 y(y – 5) + 750 = 0 (y + 25)(y – 30) = 0

y^{2}-5y=750y

2

−5y=750

750-y(y-5)=0750−y(y−5)=0

(y + 25)(y -30) = 0(y+25)(y−30)=0

Step-by-step explanation:

Givens

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.

Let's call ww the width and ll the length. According to the problem they are related as follows

w=l-5w=l−5 , because the width is 5 feet less than the lenght.

We know that the area of the room is defined as

A=w\times lA=w×l

Where A=750 ft^{2}A=750ft

2

Replacing the given area and the expression, we have

\begin{gathered}750=(l-5)l\\750=l^{2}-5l \\l^{2}-5l-750=0\end{gathered}

750=(l−5)l

750=l

2

−5l

l

2

−5l−750=0

We need to find two number which product is 750 and which difference is 5, those numbers are 30 and 25.

l^{2}-5l-750=(l-30)(l+25)l

2

−5l−750=(l−30)(l+25)

Using the zero property, we have

\begin{gathered}l=30\\l=-25\end{gathered}

l=30

l=−25

Where only the positive number makes sense to the problem because a negative length doesn't make any sense.

Therefore, the length of the room is 30 feet.

Also, the right answers are the second choice where y=ly=l , the third choice and the last choice

y² - 5y = 750  

or

750 - y(y - 5) = 0

or

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

Step-by-step explanation:

length = y  ft

width = y - 5

Area of the rectangular room = 750 square feet

length * width = 750

y*(y-5) = 750

⇒ y² - 5y = 750

y(y-5) = 750

⇒ 0 = 750 - y(y - 5)

⇒ 750 - y(y - 5) = 0

y(y - 5) = 750

y² - 5y -750 = 0

Sum = -5

Product = -750

Factors = 25 , (-30)

y²  + 25y - 30y - 750 = 0

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

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

cant open file sorry!

step-by-step explanation: