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Chapter 8: Problem 45

Evaluate the given expression for \(x=-3\) and \(y=5\) $$(x y)^{-1}$$

Short Answer

Expert verified
\(-\frac{1}{15}\)

Step by step solution

01

Substitute the Values

First, substitute the given values of the variables into the expression. Here, substitute \(x = -3\) and \(y = 5\) into \((xy)^{-1}\). This gives us:\[(-3 \times 5)^{-1}\]
02

Compute the Product

Next, compute the product inside the parentheses. Here, \(-3 \times 5 = -15\). So the expression is now:\[(-15)^{-1}\]
03

Apply the Negative Exponent Rule

Finally, apply the rule of negative exponents, which states that \(a^{-1} = \frac{1}{a}\). Therefore, \[(-15)^{-1} = \frac{1}{-15}\]

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Key Concepts

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

Evaluating Expressions
In algebra, evaluating an expression means finding the numerical value of the expression by substituting its variables with given numbers. For instance, if you are given the expression \((xy)^{-1} \) and specific values for \(x\) and \(y\), such as \(x = -3\) and \(y = 5\), you will replace \(x\) with -3 and \(y\) with 5. This step makes it easier to simplify and get the final result.

To start evaluating:
  • First, replace each variable in the expression with its given value.
  • Next, perform any operations like multiplication or addition.
  • Ensure you follow the order of operations (PEMDAS/BODMAS).
This process helps break down complex expressions into simpler numerical forms that are easy to interpret and solve.
Substitution
Substitution is a key concept in algebra where you replace variables with actual numbers. It helps transform algebraic expressions into numerical ones, making them easier to solve. For instance, given \(x = -3\) and \(y = 5\) with the expression \((xy)^{-1}\), you substitute x and y:
  • Take the expression \((xy)^{-1}\).
  • Replace \(x\) with -3 and \(y\) with 5.
  • The expression now looks like \[(-3 \times 5)^{-1}\].
After substitution, the expression turns into a numerical form that is easier to simplify and solve. Substitution is particularly useful because it converts abstract variables into concrete numbers, simplifying the entire mathematical process.
Negative Exponents
Understanding negative exponents is crucial in algebra. A negative exponent indicates that the base should be reciprocated. For example, if we have \(a^{-1}\), it simplifies to \(\frac{1}{a}\). Combining this with the solution from the problem:
  • After substituting values and multiplying, we get \[(-3 \times 5) = -15\].
  • Thus, the expression becomes \[(-15)^{-1}\].
  • Applying the negative exponent rule, \[(-15)^{-1} \text{is equivalent to} \frac{1}{-15} \].
The rule for negative exponents is simple: take the reciprocal of the base to turn the exponent positive. For \((xy)^{-1}\), it ensures that no matter how complex the initial expression looks, you can break it down and apply basic arithmetic rules to find the solution.

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Most popular questions from this chapter

Factor each of the following expressions as completely as possible. If an expression is not factorable, say so. $$x^{2}-8 x-20$$

Factor each of the following as completely as possible. If the expression is not factorable, say so. Try factoring by grouping where it might help. $$x(x-5)+4(x-5)$$

Divide each of the following. Use the long division process where necessary. $$\frac{2 x^{3}-3 x^{2}-x+12}{2 x+3}$$

Explain how you would generalize the long division process to handle $$\frac{5 x^{4}-8 x^{3}+7 x^{2}-2 x+1}{x^{2}+x-1}$$

Divide each of the following. Use the long division process where necessary. $$\frac{3 x^{2}-4 x+7}{x-3}$$
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