Question

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. $y(y + 5) = 750$ $y^2 - 5y = 750$ $750 - y(y - 5) = 0$ $y(y - 5) + 750 = 0$ $(y + 25)(y - 30) = 0$

Explanation:

Step1: Define variables

Let \( y \) be the length of the room. Then the width is \( y - 5 \) (since width is 5 feet less than length). The area of a rectangle is \( \text{length} \times \text{width} \), so the area equation is \( y(y - 5)=750 \).

Step2: Analyze each option

• Option 1: \( y(y + 5)=750 \). Here width is \( y + 5 \), which is incorrect (width should be \( y - 5 \)). So this is wrong.
• Option 2: Start with \( y(y - 5)=750 \), expand the left - hand side: \( y^{2}-5y = 750 \). This is correct.
• Option 3: Start with \( y(y - 5)=750 \), rearrange to get \( 750 - y(y - 5)=0 \). This is correct.
• Option 4: From \( y(y - 5)=750 \), we get \( y(y - 5)-750 = 0 \), not \( y(y - 5)+750 = 0 \). So this is wrong.
• Option 5: Start with \( y^{2}-5y - 750=0 \). Factor the quadratic: \( y^{2}-5y - 750=(y + 25)(y - 30)=0 \) (since \( 25\times(- 30)=-750 \) and \( 25+( - 30)=-5 \)). This is correct.

Answer:

B. \( y^{2}-5y = 750 \), C. \( 750 - y(y - 5)=0 \), E. \( (y + 25)(y - 30)=0 \)