The length of a rectangle is 7 inches more than 5 times its width. The area of the rectangle is 6 square inches. What is the width?

Is this even possible?

Guest Mar 17, 2021

#1**+1 **

let's define two variables: *l *and *w*, to represent the length and width of this rectangle.

we'll have a system of equations.

l = 7 + 5w

w * l = 6

now let's use substitution.

(7 + 5w) * w = 6

5w^2 + 7w = 6

5w^2 + 7w - 6 = 0

now let's solve the quadratic by using the quadratic formula.

\(x = {-7 \pm \sqrt{7^2-4 \cdot 5 \cdot -6} \over 2 \cdot 5} \)

\(x = \frac{-7 \pm \sqrt{49 + 120}}{10}\)

\(x = \frac{-7 \pm \sqrt{169}}{10}\)(oops, i meant to do w, not x for the variable in these three; sorry about that!)

\(w = \frac{-7 + 30}{10}, \frac{-7-30}{10}\)

\(w = \frac{23}{10}, \frac{-37}{10}\)

i doubt that the width can be negative, so i believe your answer is 23/10.

hope this helped! please let me know if you are confused about anything i did

idyllic Mar 17, 2021

#1**+1 **

Best Answer

let's define two variables: *l *and *w*, to represent the length and width of this rectangle.

we'll have a system of equations.

l = 7 + 5w

w * l = 6

now let's use substitution.

(7 + 5w) * w = 6

5w^2 + 7w = 6

5w^2 + 7w - 6 = 0

now let's solve the quadratic by using the quadratic formula.

\(x = {-7 \pm \sqrt{7^2-4 \cdot 5 \cdot -6} \over 2 \cdot 5} \)

\(x = \frac{-7 \pm \sqrt{49 + 120}}{10}\)

\(x = \frac{-7 \pm \sqrt{169}}{10}\)(oops, i meant to do w, not x for the variable in these three; sorry about that!)

\(w = \frac{-7 + 30}{10}, \frac{-7-30}{10}\)

\(w = \frac{23}{10}, \frac{-37}{10}\)

i doubt that the width can be negative, so i believe your answer is 23/10.

hope this helped! please let me know if you are confused about anything i did

idyllic Mar 17, 2021