Question

Which equations can be used to solve for y, the length of the room? Check all that apply. y(y + 5) = 750 y2 – 5y = 750 750 – y(y – 5) = 0 y(y – 5) + 750 = 0 (y + 25)(y – 30) = 0

Answer 1

Let's assume length of the room as y
so the width of the room = y - 5

Area
y * (y - 5) = 750

Answer 2

Answer with explanation:

We have to find ,those equation which is used to solve for, y ,the length of room.

Measurement of length is not given,but surely it will be a positive Integer , not equal to 0.

We will check from option 1,which of these equation gives positive Integral value.

1. y×(y+5)=750

→y² + 5 y - 750=0

→y² + 30 y - 25 y - 750=0

→y× (y +30) -25 × (y +30)=0

→ (y - 25)(y + 30)=0

→ y -25 =0  ∧ y + 30=0

y= 25 ∧ y = -30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

2. y² - 5 y =750

→y² - 5 y - 750=0

→y² - 30 y + 25 y - 750=0

→y× (y -30) +25 × (y -30)=0

→ (y + 25)(y - 30)=0

→ y +25 =0  ∧ y - 30=0

y= -25 ∧ y = 30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

3. 750 - y×(y-5)=0

⇒750 - y² +  5 y=0

⇒-1× (y² - 5 y - 750)=0

→y² - 5 y - 750=0

→y² - 30 y + 25 y - 750=0

→y× (y -30) +25 × (y -30)=0

→ (y + 25)(y - 30)=0

→ y +25 =0  ∧ y - 30=0

y= -25 ∧ y = 30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

4. y × (y -5) + 750 =0

y² - 5 y + 750=0

you will not get real root,so this equation can't be used  to solve for y, the length of the room.

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

→y+ 25 =0 ∧→ y -30=0

y= -25 ∧→ y =30

It gives a Positive value.So, this equation can be used to solve for y, the length of the room.

→→Equation 1,2,3,and 5 that is apart from equation 4, all of equation can  be used to solve for y, the length of the room.