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?
+152 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 
+152 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
