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Chapter 6: Problem 59

Simplify. $$\left(x^{-1}+y^{-1}\right)^{-1}$$

Short Answer

Expert verified
\(\frac{xy}{x+y}\)

Step by step solution

01

Rewrite the expression with common denominator

The given expression is \(\left(x^{-1} + y^{-1}\right)^{-1}. \) First, rewrite each term inside the parentheses with a common denominator: \( x^{-1} = \frac{1}{x} \text{ and } y^{-1} = \frac{1}{y}.\)
02

Combine the fractions

Next, combine the two fractions inside the parentheses using a common denominator: \( \frac{1}{x} + \frac{1}{y} = \frac{y}{xy} + \frac{x}{xy} = \frac{x+y}{xy}. \)
03

Take the reciprocal

Now, take the reciprocal of the combined fraction: \( \left( \frac{x+y}{xy} \right)^{-1} = \frac{xy}{x+y}. \)

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Key Concepts

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

Reciprocal
Reciprocals are essential in algebra and are the 'flips' of fractions and numbers.
A reciprocal of a number is one divided by that number.
For example, the reciprocal of 3 is \(\frac{1}{3}\). Similarly, the reciprocal of \(\frac{a}{b}\) is \(\frac{b}{a}\).
When we have negative exponents, they naturally lead us to find reciprocals.
For example, \(x^{-1}\) means \(\frac{1}{x}\).
Understanding reciprocals makes it easier to deal with algebraic expressions where terms are raised to the power of -1.
Common Denominator
Finding a common denominator is crucial when adding or subtracting fractions.
A common denominator is a shared multiple of the denominators of two or more fractions.
For instance, for the fractions \(\frac{1}{x}\) and \(\frac{1}{y}\), we seek a common multiple of the denominators \(x\) and \(y\).
The least common denominator here is \(xy\).
By rewriting the fractions as \(\frac{y}{xy}\) and \(\frac{x}{xy}\), it becomes easy to add them:
  • The sum is \(\frac{y}{xy} + \frac{x}{xy} = \frac{x+y}{xy}\).

This process simplifies the subsequent steps in solving algebraic problems.
Algebraic Fractions
Algebraic fractions are fractions that contain algebraic expressions in their numerator or denominator.
They often require techniques like simplifying, finding common denominators, and taking reciprocals to solve them.
Let's take the expression \(\frac{1}{x} + \frac{1}{y}\):
  • First, find a common denominator: \(xy\).
  • Rewrite the fractions: \(\frac{1}{x} = \frac{y}{xy}\) and \(\frac{1}{y} = \frac{x}{xy}\).
  • Add them: \(\frac{y}{xy} + \frac{x}{xy} = \frac{x+y}{xy}\).

Taking the reciprocal of \(\frac{x+y}{xy}\) gives us the simplified form: \(\frac{xy}{x+y}\).
These methods illustrate how handling algebraic fractions helps in solving complex algebraic problems.

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Most popular questions from this chapter

Perform the indicated operations. $$\frac{(r t)^{3}}{r t^{4}} \div \frac{\left(r t^{2}\right)^{3}}{r^{2} t^{3}}$$

Find the solution set to each equation. $$\frac{a-5}{a+6}=\frac{a-7}{a+8}$$

Perform the indicated operations. Variables in exponents represent integers. $$\frac{m^{3 k}-1}{m^{3 k}+1} \cdot \frac{m^{2 k+1}-m^{k+1}+m}{m^{3 k}+m^{2 k}+m^{k}}$$

Find the solution set to each equation. $$-\frac{5}{7}=\frac{2}{x}$$

Solve each equation. Identify each equation as a conditional equation, an inconsistent equation, or an identity. State the solution sets to the identities using interval notation. $$\frac{1}{x}+\frac{1}{2 x}=\frac{5}{4 x}$$
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