Patricia has 3030 meters of fencing to make a rectangular garden in her backyard. She wants the length to be 55 meters more than the width. Complete steps 11 & 22 of the word problem solving process to set up an equation Patricia could use to find the width of her rectangular fence.

30 = 2 ( w + 5 ) + 2 w 30=2\left(w+5\right)+2w 30 = 2 ( w + 5 ) + 2 w

Patricia has 30 30 meters of fencing to make a rectangular garden in her backyard. She wants the length to be 5 5 meters more than the width. Complete steps 1 1 & 2 2 of the word problem solving process to set up an equation Patricia could use to find the width of her rectangular fence.

For this problem, Patricia has 30 meters of fencing to make a rectangular garden in her backyard. She wants the length to be five meters more than the width. It wants us to complete steps one and two of the word problem solving process to set up an equation that Patricia can use to find the width of her rectangular fence. So steps one and two are to first understand the problem, and then second set up an equation to solve the problem. So let's first understand this problem. To do help us do this, let's draw a picture. So she has a rectangular fence in her garden, so we'll draw a rectangle. And what do we know about this rectangle? We know that in total there is 30 meters of fencing, so we can say that the perimeter of this rectangle is 30 meters. So let's let p be our perimeter, and we'll say p equals 30. We also know that this rectangle has some kind of length and some kind of width, so we'll call those l and w respectively. So now to build an equation that sets up this problem for us, we need to find a way to relate P, L, and W, and we can do that with the perimeter formula. We know that to find perimeter, we simply add up our lengths, L plus L, and add up our widths, W plus W. Combining like terms, this simplifies to P equals two L plus two W. We have a bit of a problem with this equation. There's three variables and we only know how to solve equations when there's one variable. So we need to either plug in values for some of these variables or find ways to rewrite one of these variables in terms of another. And I'll show you what I mean by both of these. So first, we know that P equals 30, so we can plug that value in for P in our equation. And second, we know that the length is to be five meters more than the width. So we can translate this as L is equal to W plus five. Our length is five more than the width. And now we can substitute in w plus five for l in our equation. Doing both of these, we get 30 equals two times w plus five plus two w. So now we have an equation with only one variable and we'd be ready to solve if that's what we needed to do in our next step. But here we're only doing till step two. So this is the answer to our problem.

Patricia has 30 30 meters of fencing to make a rectangular garden in her backyard. She wants the length to be 5 5 meters more than the width. Complete steps 1 1 & 2 2 of the word problem solving process to set up an equation Patricia could use to find the width of her rectangular fence.

30 = 2 ( w + 5 ) + 2 w 30=2\left(w+5\right)+2w 30 = 2 ( w + 5 ) + 2 w

30 = 2 ( w + 5 ) + w 30=2\left(w+5\right)+w 30 = 2 ( w + 5 ) + w

30 = ( w + 5 ) + 2 w 30=\left(w+5\right)+2w 30 = ( w + 5 ) + 2 w

30 = 2 ( w − 5 ) + 2 w 30=2\left(w-5\right)+2w 30 = 2 ( w − 5 ) + 2 w

3. Solving Word Problems

Introduction to Problem Solving

Intermediate Algebra

3. Solving Word Problems

Introduction to Problem Solving

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Multiple Choice

Patricia has $30$ meters of fencing to make a rectangular garden in her backyard. She wants the length to be $5$ meters more than the width. Complete steps $1$ & $2$ of the word problem solving process to set up an equation Patricia could use to find the width of her rectangular fence.

$30=2$\left$(w+5$\right$)+2w$

$30=2$\left$(w+5$\right$)+w$

$30=$\left$(w+5$\right$)+2w$

$30=2$\left$(w-5$\right$)+2w$

Verified step by step guidance

Identify the variables: Let $ w $ represent the width of the rectangular garden in meters.

Express the length in terms of the width: Since the length is 5 meters more than the width, write the length as $ w + 5 $.

Recall the formula for the perimeter of a rectangle: $ P = 2 \times \text{length} + 2 \times \text{width} $.

Substitute the known perimeter and expressions for length and width into the perimeter formula: $ 30 = 2(w + 5) + 2w $.

This equation can now be used to solve for the width $ w $.

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