Problem 52
Question
Solve each formula for the specified variable. $$ \frac{2}{c}+\frac{2}{d}=\frac{1}{h} \text { for } c $$
Step-by-Step Solution
Verified Answer
\( c = \frac{2hd}{d - 2h} \)
1Step 1: Understand the Equation
We are given the equation \( \frac{2}{c}+\frac{2}{d}=\frac{1}{h} \) and are asked to solve it for \( c \). The goal is to isolate \( c \) on one side of the equation.
2Step 2: Subtract \( \frac{2}{d} \) from Both Sides
To isolate \( \frac{2}{c} \), subtract \( \frac{2}{d} \) from both sides:\[ \frac{2}{c} = \frac{1}{h} - \frac{2}{d} \]
3Step 3: Find a Common Denominator
The right side has two fractions, so we need a common denominator. The common denominator of \( h \) and \( d \) is \( hd \). Rewrite the right side:\[ \frac{2}{c} = \frac{d}{hd} - \frac{2h}{hd} = \frac{d - 2h}{hd} \]
4Step 4: Cross-Multiply to Solve for \( c \)
Cross-multiply to isolate \( c \): \[ 2hd = c(d - 2h) \]
5Step 5: Solve for \( c \)
Finally, solve for \( c \) by dividing both sides by \( d - 2h \):\[ c = \frac{2hd}{d - 2h} \]
Key Concepts
Algebraic ManipulationIsolating VariablesFraction Operations
Algebraic Manipulation
Algebraic manipulation involves rearranging and simplifying equations to find the value of the variable you are solving for. It is like puzzle solving, where you move pieces around until they fit the way you need them to. Here, you use various strategies such as adding, subtracting, multiplying, and dividing while maintaining the equality of the equation.
In the given exercise, we start with the equation \( \frac{2}{c} + \frac{2}{d} = \frac{1}{h} \). Our goal is presented clearly: to solve it for \( c \). The very first step in algebraic manipulation is to understand the structure of your equation. Once understood, we can change its form without changing its meaning.
To manipulate the equation for solving, the key strategies include:
In the given exercise, we start with the equation \( \frac{2}{c} + \frac{2}{d} = \frac{1}{h} \). Our goal is presented clearly: to solve it for \( c \). The very first step in algebraic manipulation is to understand the structure of your equation. Once understood, we can change its form without changing its meaning.
To manipulate the equation for solving, the key strategies include:
- Combining like terms
- Finding a common denominator to simplify additions or subtractions of fractions
- Using inverse operations, such as subtracting when addition is present, or dividing when multiplication is present
Isolating Variables
Isolating variables is a core part of solving equations and involves getting your variable of interest on one side of the equation, all by itself. Doing so allows you to find the value of the variable directly.
In this exercise, we are focusing on solving for \( c \). Initially, \( c \) is tangled with other terms in \( \frac{2}{c} + \frac{2}{d} = \frac{1}{h} \). The first step to isolating \( c \) is to free \( \frac{2}{c} \) from the term \( \frac{2}{d} \) by subtracting \( \frac{2}{d} \) from both sides. This operation leaves \( \frac{2}{c} \) on its own on one side of the equation:
\[ \frac{2}{c} = \frac{1}{h} - \frac{2}{d} \]
Once this is achieved, the focus shifts to further simplifying the right side to maintain consistency with our next operation—solving for \( c \). By cross-multiplying, we set up a scenario where terms involving \( c \) can be isolated simply by division, resulting in \( c = \frac{2hd}{d - 2h} \). Hence, isolating a variable often works stage by stage, freeing \( c \) step by step from other mathematical entities.
In this exercise, we are focusing on solving for \( c \). Initially, \( c \) is tangled with other terms in \( \frac{2}{c} + \frac{2}{d} = \frac{1}{h} \). The first step to isolating \( c \) is to free \( \frac{2}{c} \) from the term \( \frac{2}{d} \) by subtracting \( \frac{2}{d} \) from both sides. This operation leaves \( \frac{2}{c} \) on its own on one side of the equation:
\[ \frac{2}{c} = \frac{1}{h} - \frac{2}{d} \]
Once this is achieved, the focus shifts to further simplifying the right side to maintain consistency with our next operation—solving for \( c \). By cross-multiplying, we set up a scenario where terms involving \( c \) can be isolated simply by division, resulting in \( c = \frac{2hd}{d - 2h} \). Hence, isolating a variable often works stage by stage, freeing \( c \) step by step from other mathematical entities.
Fraction Operations
Working with fractions involves several rules and strategies to simplify and manipulate them within equations. Often, fractions can seem daunting, but understanding their operations makes them much more straightforward.
In this problem, fractions are present on both sides of the equation. The solution requires operations that simplify handling these fractions:
1. **Subtract Fractions:** Begin by subtracting \( \frac{2}{d} \) to isolate \( \frac{2}{c} \). This operation relies on the concept that fractions with different denominators require a common denominator to subtract them directly.2. **Find a Common Denominator:** In this context, the common denominator is \( hd \) which allows us to combine the fractions \( \frac{1}{h} \) and \( - \frac{2}{d} \):
\[ \frac{1}{h} = \frac{d}{hd} \quad \text{and} \quad \frac{2}{d} = \frac{2h}{hd} \] 3. **Simplify:** With a common denominator, express the equation as \[ \frac{2}{c} = \frac{d - 2h}{hd} \]. Here, simplified fractions make further algebraic manipulations feasible.
Fraction operations are integral to solving equations like the given one. They ensure the clarity needed when dealing with parts instead of wholes.
In this problem, fractions are present on both sides of the equation. The solution requires operations that simplify handling these fractions:
1. **Subtract Fractions:** Begin by subtracting \( \frac{2}{d} \) to isolate \( \frac{2}{c} \). This operation relies on the concept that fractions with different denominators require a common denominator to subtract them directly.2. **Find a Common Denominator:** In this context, the common denominator is \( hd \) which allows us to combine the fractions \( \frac{1}{h} \) and \( - \frac{2}{d} \):
\[ \frac{1}{h} = \frac{d}{hd} \quad \text{and} \quad \frac{2}{d} = \frac{2h}{hd} \] 3. **Simplify:** With a common denominator, express the equation as \[ \frac{2}{c} = \frac{d - 2h}{hd} \]. Here, simplified fractions make further algebraic manipulations feasible.
Fraction operations are integral to solving equations like the given one. They ensure the clarity needed when dealing with parts instead of wholes.
Other exercises in this chapter
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