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)+2w30=2\left(w+5\right)+2w

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 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)+2w30=2\left(w+5\right)+2w

30=2(w+5)+w30=2\left(w+5\right)+w

30=(w+5)+2w30=\left(w+5\right)+2w

30=2(w−5)+2w30=2\left(w-5\right)+2w

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1. Review of Real Numbers2h 24m

Simplifying Fractions
30m

Evaluating Exponents
14m

Evaluating Expressions
15m

Translating Phrases to Expressions
16m

Translating Sentences to Equations
9m

The Distributive Property
17m

Simplifying Expressions
40m

2. Linear Equations and Inequalities3h 42m

The Addition and Subtraction Properties of Equality
47m

The Multiplication and Division Properties of Equality
30m

Solving Linear Equations
1h 14m

Solving Linear Inequalities
1h 8m

3. Solving Word Problems2h 48m

Introduction to Problem Solving
37m

Formulas
21m

Percent Problem Solving
1h 4m

Mixture Problem Solving
43m

4. Graphing4h 42m

The Rectangular Coordinate System
44m

Graph Linear Equations in Two Variables
24m

Graph Linear Equations Using Intercepts
23m

Slope of a Line
44m

Slope-Intercept Form
38m

Point Slope Form
22m

Linear Inequalities in Two Variables
28m

Introduction to Relations and Functions
40m

Function Notation
15m

5. Systems of Linear Equations2h 6m

Solving Systems of Linear Equations by Graphing
41m

Solving Systems of Linear Equations by Substitution
15m

Solving Systems of Linear Equations by Elimination
45m

Systems of Linear Inequalities
23m

6. Exponents and Polynomials3h 25m

The Product Rule
10m

The Quotient Rule
13m

Intro to the Power Rules
18m

The Power of a Quotient Rule
18m

Negative Exponents
28m

Simplifying Exponential Expressions Using All Exponent Rules
6m

Intro to Polynomials
21m

Adding and Subtracting Polynomials
20m

Multiplying Polynomials
33m

Special Products
34m

7. Factoring2h 36m

Factoring by Greatest Common Factor & Grouping
37m

Factoring Trinomials of the Form x² + bx + c
11m

Factoring Trinomials of the Form ax² + bx + c
37m

Factoring Special Products
33m

Quadratic Equations & Applications
35m

8. Rational Expressions and Equations3h 51m

Simplifying Rational Expressions
39m

Multiplying and Dividing Rational Expressions
25m

Adding and Subtracting Rational Expressions with Common Denominators
24m

Least Common Denominators
32m

Adding and Subtracting Rational Expressions with Different Denominators
39m

Rational Equations
44m

Direct & Inverse Variation
27m

9. Roots and Radicals2h 46m

Introduction to Radical Expressions
47m

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46m

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19m

Multiplying, Dividing, and Rationalizing Radicals
53m

10. Quadratic Equations3h 2m

The Square Root Property
17m

Completing the Square
24m

The Quadratic Formula
29m

Introduction to Complex Numbers
18m

Multiplying and Dividing Complex Numbers
51m

Complex Solutions of Quadratic Equations
10m

Graphing Quadratic Equations
30m

3. Solving Word Problems

Introduction to Problem Solving

Beginning 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.

$30 = 2 (w + 5) + 2 w$

$30 = 2 (w + 5) + w$

$30 = (w + 5) + 2 w$

$30 = 2 (w - 5) + 2 w$

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Verified step by step guidance

Identify the variables: Let the width of the garden be $w$ meters. Since the length is 5 meters more than the width, express the length as $w + 5$ meters.

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

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

This equation represents the total fencing Patricia has, set equal to the perimeter of the garden with the given dimensions.

Now the equation \$30 = 2(w + 5) + 2w$ can be used to solve for the width $w$.

2:11m

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