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# so confused

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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

#1
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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

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

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

$$x = \frac{-7 \pm \sqrt{169}}{10}$$(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 Mar 17, 2021

#1
+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

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

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

$$x = \frac{-7 \pm \sqrt{169}}{10}$$(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