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Olivia has a $20$ -meter-long fence that she plans to use to enclose a rectangular garden of width $w$ . The fencing will be placed around all four sides of the garden so that its area is $18.75$ square meters. $\newline$ Write an equation in terms of $w$ that models the situation.

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Evaluate two-variable equations: word problems

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Q. Olivia has a

20

-meter-long fence that she plans to use to enclose a rectangular garden of width

w

. The fencing will be placed around all four sides of the garden so that its area is

18.75

square meters.

\newline

Write an equation in terms of

w

that models the situation.

Perimeter Equation: The perimeter of the rectangular garden is equal to the length of the fence, which is $20$ meters. The perimeter $P$ of a rectangle is given by $P = 2l + 2w$ , where $l$ is the length and $w$ is the width.
Total Perimeter: Since we know the total perimeter is $20$ meters, we can write the equation as $20 = 2l + 2w$ .
Area Equation: We also know the area $A$ of the rectangle is given by $A = l \times w$ , and the area is $18.75$ square meters. So we have $18.75 = l \times w$ .
Expressing Length in Terms of Width: To write an equation in terms of $w$ , we need to express $l$ in terms of $w$ using the area equation. From the area equation, we get $l = \frac{18.75}{w}$ .
Substitution into Perimeter Equation: Now we substitute $l = \frac{18.75}{w}$ into the perimeter equation: $20 = 2\left(\frac{18.75}{w}\right) + 2w$ .
Simplify Equation: Simplify the equation: $20 = \frac{37.5}{w} + 2w$ .
Eliminating Fraction: To make it look neater, we can multiply every term by $w$ to get rid of the fraction: $20w = 37.5 + 2w^2$ .
Quadratic Equation: Rearrange the terms to get a quadratic equation in standard form: $2w^2 - 20w + 37.5 = 0$ .

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