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Chapter 2: Problem 68

Solve \(R=\frac{E+C}{L}\) for \(E\)

Short Answer

Expert verified
E = RL - C

Step by step solution

01

Understand the Given Equation

We start with the equation: \( R = \frac{E + C}{L} \)
02

Multiply Both Sides by L

To isolate \(E\), first eliminate the fraction by multiplying both sides of the equation by \(L\): \( R \cdot L = \frac{E + C}{L} \cdot L \) This simplifies to: \( RL = E + C \)
03

Isolate E

Now, isolate \( E \) by subtracting \( C \) from both sides: \( RL - C = E + C - C \) This simplifies to: \( E = RL - C \)

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Key Concepts

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

Isolation of Variables
Often, equations involve multiple variables. To solve for one variable, we need to isolate it, a fundamental concept in algebra.

In the given problem: \( R = \frac{E + C}{L} \),
the goal is to solve for \( E \). This means we need to get \( E \) by itself on one side of the equation. We do this through specific steps:
  • First, eliminate any fractions or groupings by performing the same operation on both sides.
  • Next, move any terms involving the variable of interest to one side.
  • Finally, isolate the variable by simplifying.
Ultimately, isolation of variables helps make complex problems simpler and prepares the equation for solving. Understanding this approach is key to mastering algebraic equations.
Basic Algebra
Basic algebra is about understanding and using the rules of arithmetic with variables. It lays the groundwork for more advanced math topics.

In our exercise, we see these rules in action:

We start with the equation:
\( R = \frac{E + C}{L} \).

First, we need to eliminate the fraction to simplify our equation.

By multiplying both sides by \( L \):
\( R \cdot L = \frac{E + C}{L} \cdot L \),

we get \( RL = E + C \).

We see here how basic operations鈥攎ultiplication and subtraction鈥攁re applied to isolate \( E \). Mastering these operations and understanding their use in different situations is essential for solving equations.
Equation Manipulation
Equation manipulation involves changing the form of an equation to make it easier to solve. This often includes operations like addition, subtraction, multiplication, and division.

In this exercise, we manipulate the equation to solve for \( E \) by:
  • Multiplying both sides by \( L \) to get rid of the fraction. This gives us \( RL = E + C \).

  • Next, we subtract \( C \) from both sides to isolate \( E \): \( RL - C = E \).
Each step simplifies the equation, bringing us closer to the solution.

These techniques are not only useful in solving linear equations but are also foundational for more complex mathematical tasks. Being adept in equation manipulation makes tackling advanced math problems much more manageable.

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Most popular questions from this chapter

For problems \(89-92\), do the arithmetic with a calculator. The area \(A\) of a rectangular playground is \(28,800 \mathrm{ft}^{2}\). The length \(L\) is \(180 \mathrm{ft}\). Solve the formula \(A=L W\) for \(L\), and use it to find the width \(W\) of the playground.

For exercises 1-8, use a number line graph to represent the inequality. $$ -7

Solve: \(2(n+3)=4 n+6\)

For exercises \(65-72\), assign a variable, and write an inequality that represents the constraint. The maximum fuel economy of a car is \(\frac{32 \mathrm{mi}}{1 \mathrm{gal}}\). Find the maximum distance it can travel on a full tank of \(12 \mathrm{gal}\) of gasoline.

An estimated \(3.5\) million people in southeast Asia are infected with HIV. More than two out of three in need of antiretroviral treatment in this region do not receive it. Find the minimum number of people who need and do not receive the antiretroviral treatment. Round to the nearest tenth of a million. (Source: www.searo.who.int, Sept. 8, 2011)
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