The length of a rectangle is 2 less than 5 times the width. The area is 39 square inches. What is the length and the width?

I did this: Length = 5x-2; width = 5x
Then I solved the question and came up with x=3

i think x=1.5

Thank you ChrismB

To solve this problem, you set up two equations based on the given information. Let's assign variables to the length and width of the rectangle.

Let's assume the width of the rectangle is 'x'. According to the problem, the length is 2 less than 5 times the width, so the length can be represented as (5x - 2).

Now, we know that the formula for the area of a rectangle is length times width. In this case, the area is given as 39 square inches. So, our equation for the area is:

Length × Width = Area
(5x - 2) × x = 39

Next, we can simplify the equation by expanding the brackets:

5x^2 - 2x = 39

Now, rearrange the equation to solve for x:

5x^2 - 2x - 39 = 0

This is a quadratic equation. To solve it, you can either factorize or use the quadratic formula. In this case, it is not easily factorizable, so let's use the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / (2a)

The coefficients a, b, and c in our quadratic equation are:
a = 5, b = -2, c = -39

Now, plug in the values into the quadratic formula and solve for x:

x = (-(-2) ± sqrt((-2)^2 - 4*5*(-39))) / (2*5)

Simplifying further:

x = (2 ± sqrt(4 + 780)) / 10
x = (2 ± sqrt(784)) / 10
x = (2 ± 28) / 10

This gives us two possible solutions for x:
x₁ = (2 + 28) / 10 = 30 / 10 = 3
x₂ = (2 - 28) / 10 = -26 / 10 = -2.6

Since we're dealing with length and width measurements, a negative width doesn't make sense in this context. Therefore, we discard the negative solution.

So, the width of the rectangle is x = 3 inches. Now, substitute this value back into the equation for length:

Length = 5x - 2
Length = 5(3) - 2
Length = 15 - 2
Length = 13 inches

Therefore, the length is 13 inches and the width is 3 inches.