Solve algebraically using only variable: The length of a rectangle is two less than three times
its width. If the area of the rectangle is 65, find the length and the width. (The area of a rectangle is equal to its length times its width.
L = 3W-2
L*W = 65
(3W-2)W = 65
3W^2 - 2W - 65 = 0
Factor to solve for W, then insert W value in the first equation to find L.
To solve this problem algebraically, let's assign variables to the length and width of the rectangle.
Let's say the width of the rectangle is 'w.'
According to the problem, the length of the rectangle is two less than three times its width.
So we can represent the length as '3w - 2.'
The area of a rectangle is equal to its length times its width, so we can set up the equation:
Area = Length × Width
65 = (3w - 2) × w
Now, we can solve this equation to find the value of 'w' (width) and then use it to find the value of '3w - 2' (length).
Expanding the equation, we get:
65 = 3w^2 - 2w
Rearranging the equation, we have:
3w^2 - 2w - 65 = 0
Now, we can solve this quadratic equation for 'w' using factoring, completing the square, or using the quadratic formula.
Let's use factoring to solve it.
The factored form of the equation is:
(3w + 13)(w - 5) = 0
Setting each factor equal to zero, we get two possible solutions:
3w + 13 = 0 or w - 5 = 0
Solving each equation separately:
For 3w + 13 = 0:
3w = -13
w = -13/3
For w - 5 = 0:
w = 5
Since the width cannot be negative, we disregard the solution w = -13/3.
Therefore, the width of the rectangle is 5.
Now, we can find the length (3w - 2):
Length = 3w - 2
Length = 3(5) - 2
Length = 15 - 2
Length = 13
So, the length of the rectangle is 13 units and the width is 5 units.