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Chapter 1: Problem 34

Solving for a Variable Solve the equation for the indicated variable. $$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}} ; \quad \text { for } R_{1}$$

Short Answer

Expert verified
\( R_1 = \frac{R \cdot R_2}{R_2 - R} \)

Step by step solution

01

Introduce the given equation

The given equation to solve is \( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \). We need to solve for \( R_1 \).
02

Isolate the term

Subtract \( \frac{1}{R_2} \) from both sides to isolate the \( \frac{1}{R_1} \) term. This results in: \[ \frac{1}{R} - \frac{1}{R_2} = \frac{1}{R_1} \]
03

Find a common denominator

The left side of the equation currently has two fractions. To simplify it, find a common denominator for the fractions \( \frac{1}{R} \) and \( \frac{1}{R_2} \). The common denominator is \( R \cdot R_2 \).
04

Rewrite the left side with the common denominator

Multiply the fractions by suitable expressions to have the common denominator, resulting in:\[ \frac{R_2}{R \cdot R_2} - \frac{R}{R \cdot R_2} = \frac{1}{R_1} \] which simplifies to:\[ \frac{R_2 - R}{R \cdot R_2} = \frac{1}{R_1} \]
05

Take the reciprocal

Now, both sides of the equation can be written as the reciprocal since \( \frac{1}{R_1} \) is simply the reciprocal of \( R_1 \). Therefore, take the reciprocal of both sides:\[ R_1 = \frac{R \cdot R_2}{R_2 - R} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solving Equations
In algebra, solving an equation means finding the value of the variable that makes the equation true. The goal is to manipulate the equation until you isolate the desired variable on one side. It often starts with understanding the equation's structure and carefully choosing operations to simplify it. Let's go through a quick overview for solving equations.

  • Identify the variable you need to solve for.
  • Apply inverse operations to both sides of the equation to simplify it.
  • Use arithmetic operations such as addition, subtraction, multiplication, or division appropriately.
  • Check your solutions by substituting them back into the original equation.
  • Ensure each step is justified based on arithmetic principles.
In our example, we solved for \( R_1 \) in the given equation by rearranging terms and performing arithmetic operations, leading to the solution expressing \( R_1 \) in terms of \( R \) and \( R_2 \). These strategies are pieced together to achieve a final expression where \( R_1 \) is isolated.
Isolating Variables
Isolating a variable in an equation is a crucial concept in algebra. It involves rearranging the equation so that the variable you are solving for is by itself on one side of the equation. Let's dive into how exactly this process is carried out.

  • The primary objective is to perform operations that will remove other terms from the side of the equation where your variable resides.
  • You might need to add or subtract terms from both sides of the equation.
  • Multiplication or division can be helpful for eliminating coefficients attached to the variable.
  • In complex equations, this might involve dealing with fractions, like in our exercise.
When isolating \( R_1 \) in the provided equation, we first subtracted \( \frac{1}{R_2} \) from both sides to isolate the term \( \frac{1}{R_1} \). This clarified which operations were necessary in the following steps and ultimately led to isolating \( R_1 \) itself.
Fractions in Equations
Dealing with fractions in equations can sometimes seem tricky, but with practice, they become quite manageable. Here's how you can handle fractions effectively when solving algebraic equations.

  • Identify if the fractions share a common denominator; if not, find the least common denominator (LCD).
  • Rewrite fractions so they all have the same denominator, simplifying operations across the equation.
  • Use the property that multiplying both sides of an equation by the reciprocal of a fraction effectively eliminates it, aiding simplification.
  • When fractions are isolated, taking their reciprocal can often simplify the process of finding the value of your variable.
In solving the equation for \( R_1 \), we first found a common denominator for the fractions \( \frac{1}{R} \) and \( \frac{1}{R_2} \), which allowed us to combine them effectively. By rewriting the fractions and utilizing the reciprocal, \( R_1 \) was expressed elegantly without fractions.

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Most popular questions from this chapter

A sealed room in a hospital, measuring \(5 \mathrm{m}\) wide, \(10 \mathrm{m}\) long, and \(3 \mathrm{m}\) high, is filled with pure oxygen. One cubic meter contains \(1000 \mathrm{L},\) and \(22.4 \mathrm{L}\) of any gas contains \(6.02 \times 10^{23}\) molecules (Avogadro's number). How many molecules of oxygen are there in the room?

Solving an Equation in Different Ways We have learned several different ways to solve an equation in this section. Some equations can be tackled by more than one method. For example, the equation \(x-\sqrt{x}-2=0\) is of quadratic type. We can solve it by letting \(\sqrt{x}=u\) and \(x=u^{2},\) and factoring. Or we could solve for \(\sqrt{x},\) square each side, and then solve the resulting quadratic equation. Solve the following equations using both methods indicated, and show that you get the same final answers. (a) \(x-\sqrt{x}-2=0 \quad\) quadratic type; solve for the radical, and square (b) \(\frac{12}{(x-3)^{2}}+\frac{10}{x-3}+1=0 \quad \begin{array}{l}\text { quadratic type; multiply } \\ \text { by } \mathrm{LCD}\end{array}\)

Profit \(\quad\) A small-appliance manufacturer finds that the profit \(P\) (in dollars) generated by producing \(x\) microwave ovens per week is given by the formula \(P=\frac{1}{10} x(300-x)\) provided that \(0 \leq x \leq 200 .\) How many ovens must be manufactured in a given week to generate a profit of \(\$ 1250 ?\)

Solve the equation for the variable \(x\). The constants \(a\) and \(b\) represent positive real numbers. $$x^{4}-5 a x^{2}+4 a^{2}=0$$

A Family of Equations The equation $$3 x+k-5=k x-k+1$$ is really a family of equations, because for each value of \(k\) we get a different equation with the unknown \(x\). The letter \(k\) is called a parameter for this family. What value should we pick for \(k\) to make the given value of \(x\) a solution of the resulting equation? (a) \(x=0\) (b) \(x=1\) (c) \(x=2\)
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