Answer
Explanation
Variables
• x: width of the rectangle, in ft
,
• y: length of the rectangle, in ft
Given that the length, y, is 30 feet less than 6 times the width, x, then:
![y=6x-30\text{ \lparen eq. 1\rparen}](https://img.qammunity.org/2023/formulas/mathematics/college/1t5qmv5a5ngdc6j97fpfegi4lozhxa1xza.png)
The area of a rectangle is calculated as follows:
![Area=length{}t* width](https://img.qammunity.org/2023/formulas/mathematics/college/sacqopq3fa16sshc9mk1hqx645u5t86e1w.png)
In this case, the area is 4836 square ft. Substituting this value and using the before defined variables, we get:
![4836=yx\text{ \lparen eq. 2\rparen}](https://img.qammunity.org/2023/formulas/mathematics/college/940lqqh6w9xk8ab85iy88873ja1962f3vo.png)
Substituting equation 1 into equation 2:
![\begin{gathered} 4836=(6x-30)x \\ 4,836=6x^2-30x \\ 0=6x^2-30x-4836 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/tpesuv83t263l2o3n5pq4g1k12hh2igfey.png)
We can solve this equation with the help of the quadratic formula, as follows:
![\begin{gathered} x_(1,2)=(-b\pm√(b^2-4ac))/(2a) \\ x_(1,2)=(3^0\pm√((-30)^2-4\cdot6\cdot(-4836)))/(2\cdot6)\frac{}{} \\ x_(1,2)=(30\pm√(116964))/(12) \\ x_(1,2)=(30\pm342)/(12) \\ x_1=(30+342)/(12)=31 \\ x_2=(30-342)/(12)=-26 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/eeha74y0z2waixd4zkf8eis9ij4btgz9qk.png)
Given that the width cannot be negative, then the second solution is discarded.
Substituting x = 31 ft into equation 1: