Question

angie drew a rectangle. the length of a rectangles she drew is 2 less than three times the width. find the dimensions of the rectangle if the rectangle if the area is 65 square meters

Answer 1

The width of the rectangle is 5

The length of the rectangle is 13

Step-by-step explanation:

Let's call x the length of the rectangle and y the width of the rectangle.

The length is 2 less than 3 times the width, so

x = 3y - 2

And the area is 65 square meters. Since the area is length times width, we get:

xy = 65

Now, we can replace the first equation x = 3y - 2 on the second one to get

(3y - 2)y = 65

3y(y) - 2y = 65

3y² - 2y = 65

3y² - 2y - 65 = 0

So, using the quadratic equation, we get that the solutions to 3y² - 2y - 65 = 0 are

`\begin{gathered} y=\frac{-(-2)\pm\sqrt[]{(-2)^2-4(3)(-65)}_{}}{2(3)} \\ y=\frac{2\pm\sqrt[]{784}}{6} \\ y=(2\pm28)/(6) \\ \text{Then} \\ y=(2+28)/(6)=(30)/(6)=5 \\ or \\ y=(2-28)/(6)=(-26)/(6)=-(13)/(3) \end{gathered}`

The solution is y = 5 because the width can't have a negative length.

Then, replacing y = 5 on the first equation, we get:

x = 3y - 2

x = 3(5) - 2

x = 15 - 2

x = 13

Therefore, the length of the rectangle is 13 meters and the width of the rectangle is 5 meters