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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?

 Mar 17, 2021

Best Answer 

 #1
avatar+148 
+1

let's define two variables: 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 smiley

 Mar 17, 2021
 #1
avatar+148 
+1
Best Answer

let's define two variables: 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 smiley

idyllic Mar 17, 2021

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