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Chapter 2: Problem 60

Solve \(A=\frac{x_{1}+x_{2}+x_{3}}{3}\) for \(x_{3}\).

Short Answer

Expert verified
x_{3} = 3A - x_{1} - x_{2}

Step by step solution

01

Multiply both sides by 3

To isolate the terms involving the variables, multiply both sides of the equation by 3: 3A = x_{1} + x_{2} + x_{3}
02

Subtract x_{1} and x_{2} from both sides

Next, subtract both x_{1} and x_{2} from both sides of the equation to solve for x_{3}: 3A - x_{1} - x_{2} = x_{3}
03

Simplify the equation

The equation is now solved for x_{3} and can be simplified as: x_{3} = 3A - x_{1} - x_{2}

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Key Concepts

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

headline of the respective core concept
Isolating variables is a fundamental step when solving equations. It involves rearranging the equation in a way that the variable we are solving for is alone on one side of the equation.
In this exercise, we started with the equation:
\[A=\frac{x_{1}+x_{2}+x_{3}}{3}\].
The goal was to isolate \(x_{3}\).
Here鈥檚 how we did it:
  • We first multiplied both sides by 3 to get rid of the denominator. This gave us:
    \[3A = x_{1} + x_{2} + x_{3}\].

  • Next, we subtracted \(x_{1}\) and \(x_{2}\) from both sides to isolate \(x_{3}\):
    \[3A - x_{1} - x_{2} = x_{3}\].
A variable is isolated when it stands alone on one side of the equation, like we have for \(x_{3}\) here. This process allows us to find the value of the unknown variable easily.
headline of the respective core concept
Algebraic manipulation refers to the various operations we use to solve equations, which include addition, subtraction, multiplication, and division. Understanding how and when to apply these operations is critical.
Let's break down the algebraic manipulation in our example:
  • We began with the equation
    \[A=\frac{x_{1}+x_{2}+x_{3}}{3}\]. To clear the fraction, we multiplied both sides by 3:
    \[3A = x_{1} + x_{2} + x_{3}\].

  • To isolate \(x_{3}\), we needed to isolate it on one side of the equation. We achieved this by subtracting \(x_{1}\) and \(x_{2}\) from both sides:
    \[3A - x_{1} - x_{2} = x_{3}\].
This example shows the importance of performing algebraic operations on both sides of the equation to maintain equality and progressively isolate the variable of interest.
headline of the respective core concept
Equation simplification is the process of making an equation easier to work with. It involves combining like terms and performing operations that reduce the equation to its simplest form.
In our example, once we had:
\[3A = x_{1} + x_{2} + x_{3}\],
we needed to solve for \(x_{3}\). This required simplifying the equation.
We did this by:
  • Subtracting \(x_{1}\) and \(x_{2}\) from both sides:
    \[3A - x_{1} - x_{2} = x_{3}\].

  • After this step, the equation is already in a simplified form where \(x_{3}\) is isolated. Further simplification isn't necessary because the equation is clear and straightforward.
This illustrates how simplification can both make an equation easier to understand and help to obtain the final solution more efficiently.

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Most popular questions from this chapter

For exercises 9-36, (a) solve. (b) check the direction of the inequality sign. $$ 20>-2 u+40 $$

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

For exercises 53-62, (a) clear the fractions or decimals and solve. (b) check the direction of the inequality sign. $$ \frac{1}{4} m+\frac{1}{6}>1 $$

For exercises 53-62, (a) clear the fractions or decimals and solve. (b) check the direction of the inequality sign. $$ \frac{3}{4}(2 u+8)<\frac{5}{6}(3 u-12) $$

For exercises 89-92, solve. Use a calculator to do the arithmetic. $$ -\frac{3}{8}=\frac{b}{160} $$
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