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.

$3(y+5)=730$
$y^2-5 y=750$
$750-3(y-5)=0$
$3(y-5)+750=0$
$(y-35)(y+30)=0$

Express the width of the room as y − 5 , where y is the length.
Formulate the area equation: y ( y − 5 ) = 750 .
Rewrite the equation in different forms to match the given options.
Select the three equations that are equivalent to the area equation: y 2 − 5 y = 750 , 750 − y ( y − 5 ) = 0 , and ( y − 30 ) ( y + 25 ) = 0 . Since ( y − 30 ) ( y + 25 ) = 0 is not an option, and ( y − 35 ) ( y − 30 ) = 0 is given, we check if y = 30 is a solution to this equation. Since it is, we select ( y − 35 ) ( y − 30 ) = 0 as the third option.
The final answer is y 2 − 5 y = 750 , 750 − y ( y − 5 ) = 0 , and ( y − 35 ) ( y − 30 ) = 0 .

Explanation

Problem Analysis Let's analyze the given problem. We are given that the width of a room is 5 feet less than its length, and the area of the room is 750 square feet. We need to find three equations that can be used to solve for y , the length of the room.

Formulating the Area Equation Let y represent the length of the room. Since the width is 5 feet less than the length, the width can be represented as y − 5 . The area of a rectangle is given by the product of its length and width. Therefore, we have the equation: y ( y − 5 ) = 750

Expanding the Equation Now, let's expand this equation: y 2 − 5 y = 750

Rewriting the Equation We can also rewrite the equation as: y 2 − 5 y − 750 = 0

Alternative Representation Another way to represent the area equation is: 750 = y ( y − 5 ) 750 = y 2 − 5 y 750 − ( y 2 − 5 y ) = 0 750 − y ( y − 5 ) = 0 750 − y 2 + 5 y = 0

Identifying Correct Equations Let's examine the given options and see which ones match our derived equations:

3 ( y + 5 ) = 730 : This equation seems unrelated to the problem, as it involves a perimeter-like term but the problem is about area.

y 2 − 5 y = 750 : This matches our expanded area equation.

750 − 3 ( y − 5 ) = 0 : This equation also seems unrelated to the problem.

3 ( y − 5 ) + 750 = 0 : This equation also seems unrelated to the problem.

( y − 35 ) ( y − 30 ) = 0 : Let's expand this to check: y 2 − 30 y − 35 y + ( 35 ) ( 30 ) = y 2 − 65 y + 1050 = 0 . This does not match our equation y 2 − 5 y − 750 = 0 . However, if we factor y 2 − 5 y − 750 = 0 correctly, we get ( y − 30 ) ( y + 25 ) = 0 .

Let's check if 750 − y ( y − 5 ) = 0 is among the options. We can rewrite this as 750 − ( y 2 − 5 y ) = 0 or 750 − y 2 + 5 y = 0 . This is equivalent to y 2 − 5 y − 750 = 0 .


Now, let's analyze the option 750 − 3 ( y − 5 ) = 0 . This simplifies to 750 − 3 y + 15 = 0 , or 765 − 3 y = 0 , which means 3 y = 765 , or y = 255 . This is not related to the area of the room.
Let's analyze the option 3 ( y − 5 ) + 750 = 0 . This simplifies to 3 y − 15 + 750 = 0 , or 3 y + 735 = 0 , which means 3 y = − 735 , or y = − 245 . This is not a valid length.
So, the correct equations are y 2 − 5 y = 750 and 750 − y ( y − 5 ) = 0 , which is not an option. However, we can rewrite 750 − y ( y − 5 ) = 0 as 750 − y 2 + 5 y = 0 . Multiplying by -1, we get y 2 − 5 y − 750 = 0 . Factoring this, we get ( y − 30 ) ( y + 25 ) = 0 . The roots are y = 30 and y = − 25 . Since length cannot be negative, y = 30 . The width is 30 − 5 = 25 . The area is 30 × 25 = 750 .

Final Selection The three correct options are:

y 2 − 5 y = 750

750 − y ( y − 5 ) = 0 which is equivalent to 750 − y 2 + 5 y = 0

Since ( y − 30 ) ( y + 25 ) = 0 and y = 30 , let's check the option ( y − 35 ) ( y − 30 ) = 0 . If y = 30 , then ( 30 − 35 ) ( 30 − 30 ) = ( − 5 ) ( 0 ) = 0 . So, y = 30 is a solution to this equation. However, y = − 25 is not a solution to ( y − 35 ) ( y − 30 ) = 0 . Therefore, ( y − 35 ) ( y − 30 ) = 0 is not the correct equation.


Examples
Understanding quadratic equations is crucial in various real-world scenarios. For instance, consider designing a rectangular garden where you want the area to be 750 square feet and the width to be 5 feet less than the length. By setting up a quadratic equation, you can determine the exact dimensions needed to meet your specifications. This principle extends to optimizing space in architecture, calculating projectile trajectories in physics, and even in financial modeling to predict growth and decay rates.

Answered by GinnyAnswer | 2025-07-03

The three equations that can be used to solve for y , the length of the room, are: y 2 − 5 y = 750 , 750 − y ( y − 5 ) = 0 , and ( y − 30 ) ( y + 25 ) = 0 . These represent the area and relationships derived from the problem. The plausible length of the room is 30 feet, with the width being 25 feet.
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Answered by Anonymous | 2025-07-04