Question

A rectangle has a length of 30 feet less than five times its width if the area of the rectangle is 360 ft.² find the length of the rectangle

Answer 1

`l = 30` OR
`l=-60`

Explanation:

We can solve this problem with a system of equations by creating 2 equations from the information given.

1. "a length of 30 feet less than 5 times its width"

`l = 5w - 30`

2. "the area of the rectangle is 360 ft²"

`l\cdot w = 360`

To solve for the rectangle's length, we can create a substitution for w by dividing both sides of the second equation by length (
`l`).

`l\cdot w = 360\\\overline{\ \ l\ \ } \ \ \ \: \ \overline{\ l \ \ }`

`w = (360)/(l)`

Then, we can substitute this value back into the first equation and solve for
`l`.

`l = 5((360)/(l)) - 30`

↓ distribute the 5

`l(l) = \left((1800)/(l) - 30\right)l`

↓ multiply both sides by
`l`

`l^2 = 1800 - 30l`

↓ move all
`l`'s to one side

`l^2 + 30l = 1800`

To solve this quadratic, we can complete the square.

↓ add (30/2)² to both sides

`l^2 + 30l + \left((30)/(2)\right)^2 = 1800 + 15^2`

↓ simplify 15²

`l^2 + 30l + 225 = 1800 + 225`

↓ factor the left side as a perfect square

`(l + 15)^2 = 2025`

↓ take the square root of both sides

`l + 15 = \pm √(2025)`

`l + 15 = \pm 45`

↓ split into 2 equations

`l + 15 = 45` OR
`l + 15 = -45`

`\boxed{l = 30}` OR
`\boxed{l=-60}`