Problem 55
Question
Solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? $$A=lw \quad \text{for} \quad w$$
Step-by-Step Solution
Verified Answer
The solution to the formula \(A=lw\) solved for \(w\) is \(w=\frac{A}{l}\).
1Step 1: Understanding the Formula
The formula \(A=lw\) is an expression for the area of a rectangle, where \(A\) represents the area, \(l\) represents the length, and \(w\) represents the width. To solve for \(w\), you want \(w\) to be on one side of the equation.
2Step 2: Isolate Width \(w\)
To isolate \(w\), divide both sides of the equation by \(l\). This gives: \(\frac{A}{l}=w\). This operation canciles out \(l\) on the right side of the equation, leaving \(w\) alone. The right side becomes \(\frac{A}{l}\)
3Step 3: Check the Result
The equation has been successfully solved for \(w\) if it reads \(w=\frac{A}{l}\).
Key Concepts
Area of a RectangleAlgebraic ManipulationIsolating Variables
Area of a Rectangle
When it comes to geometry, understanding the concept of area is fundamental. Specifically, the area of a rectangle is determined by the simple formula:
\( A = l \times w \)
where \( A \) represents the area, \( l \) stands for the length, and \( w \) symbolizes the width of the rectangle. This formula is used to calculate how much surface an object has.
In real-life applications, knowing the area can be crucial -- whether you're wrapping a gift, covering a wall with paint, or planting grass seeds in your garden. Each of these tasks requires measuring the surface area so you can estimate the amount of material you need.
It's essential to comprehend that area is always measured in square units, such as square meters (\( m^2 \) ), square feet (\( ft^2 \) ), or another appropriate unit, depending on the scale of the rectangle. Whether you're dealing with small objects like books or larger spaces like rooms, mastering this formula is key to solving many practical problems.
\( A = l \times w \)
where \( A \) represents the area, \( l \) stands for the length, and \( w \) symbolizes the width of the rectangle. This formula is used to calculate how much surface an object has.
In real-life applications, knowing the area can be crucial -- whether you're wrapping a gift, covering a wall with paint, or planting grass seeds in your garden. Each of these tasks requires measuring the surface area so you can estimate the amount of material you need.
It's essential to comprehend that area is always measured in square units, such as square meters (\( m^2 \) ), square feet (\( ft^2 \) ), or another appropriate unit, depending on the scale of the rectangle. Whether you're dealing with small objects like books or larger spaces like rooms, mastering this formula is key to solving many practical problems.
Algebraic Manipulation
Algebraic manipulation refers to the process of rearranging and simplifying expressions and equations using algebraic rules. This is vital in solving for an unknown variable or simplifying expressions to make them easier to understand or use.
When you encounter an equation like \( A = lw \) , you can apply several techniques to manipulate it: adding or subtracting terms on both sides, multiplying or dividing both sides by a number, and using distributive, associative, or commutative properties.
In the context of our rectangle area problem, to solve for width \( w \), you would divide both sides of the equation by the length \( l \), a form of algebraic manipulation, yielding \( w = \frac{A}{l} \). This skill is not only essential in geometry but in all fields of mathematics and scientific investigation, as it provides a structured methodology for exploring relationships between variables and constants.
When you encounter an equation like \( A = lw \) , you can apply several techniques to manipulate it: adding or subtracting terms on both sides, multiplying or dividing both sides by a number, and using distributive, associative, or commutative properties.
In the context of our rectangle area problem, to solve for width \( w \), you would divide both sides of the equation by the length \( l \), a form of algebraic manipulation, yielding \( w = \frac{A}{l} \). This skill is not only essential in geometry but in all fields of mathematics and scientific investigation, as it provides a structured methodology for exploring relationships between variables and constants.
Isolating Variables
Isolating a variable is an algebraic technique used to find the value of that variable. It means rearranging an equation so that the target variable is on one side by itself. This is often a primary goal when solving algebraic equations, as it allows us to discover the value of an unknown.
To isolate a variable, we perform operations that will 'undo' the equation around the variable we are interested in. For the rectangle problem, where we are solving \( A = lw \) for \( w \) , we divide both sides of the equation by \( l \) because \( w \) is multiplied by \( l \) in the initial equation. What we are essentially doing is the inverse operation of multiplication – division, in this case – to both sides, ensuring that we keep the equation balanced.
Once the variable is isolated, as in \( w = \frac{A}{l} \), we have a formula that can be used directly to calculate its value given the other variables. Mastery of this process is incredibly valuable, enabling students to solve not only for geometrical problems but for any algebraic equation they encounter in mathematics.
To isolate a variable, we perform operations that will 'undo' the equation around the variable we are interested in. For the rectangle problem, where we are solving \( A = lw \) for \( w \) , we divide both sides of the equation by \( l \) because \( w \) is multiplied by \( l \) in the initial equation. What we are essentially doing is the inverse operation of multiplication – division, in this case – to both sides, ensuring that we keep the equation balanced.
Once the variable is isolated, as in \( w = \frac{A}{l} \), we have a formula that can be used directly to calculate its value given the other variables. Mastery of this process is incredibly valuable, enabling students to solve not only for geometrical problems but for any algebraic equation they encounter in mathematics.
Other exercises in this chapter
Problem 54
Graph each equation. $$ y=-\frac{1}{x}\left(\text { Let } x=-2,-1,-\frac{1}{2},-\frac{1}{3}, \frac{1}{3}, \frac{1}{2}, 1, \text { and } 2 .\right) $$
View solution Problem 54
Find all values of \(x\) satisfying the given conditions. \(y_{1}=\frac{x+1}{4}, y_{2}=\frac{x-2}{3},\) and \(y_{1}-y_{2}=-4\)
View solution Problem 55
Complex numbers are used in electronics to describe the current in an electric circuit. Ohm's law relates the current in a circuit, \(I\), in amperes, the volta
View solution Problem 55
Solve each equation in Exercises \(47-64\) by completing the square. $$ x^{2}-5 x+6=0 $$
View solution