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Chapter 0: Problem 43

Evaluate each expression if \(x=-2\) and \(y=3\). \(5 x y+2\)

Short Answer

Expert verified
-28

Step by step solution

01

Substitute the values of x and y

Replace the variables in the expression with the given values: If \(x = -2\) and \(y = 3\), substitute these into the expression: \(5 x y + 2\) becomes \(5 (-2) (3) + 2\).
02

Perform multiplication

First, multiply the values inside the parentheses: \(5 \cdot (-2) \cdot 3 = -30\).
03

Add the constant term

Finally, add the constant term to the product: \(-30 + 2 = -28\).

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Key Concepts

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

Substitution
Substitution is a critical concept when evaluating algebraic expressions. It involves replacing variables with their given values. For our exercise, we are given the expression: \(5xy + 2\) and the values: \(x = -2\) and \(y = 3\). By substituting these values, the expression transforms from: \(5xy + 2\) to \(5(-2)(3) + 2\). Substitution makes complex expressions easier by converting them into numerical calculations.
Order of Operations
Understanding the order of operations is essential when evaluating expressions. The correct order is often remembered by the acronym PEMDAS:
  • P: Parentheses
  • E: Exponents
  • M: Multiplication
  • D: Division
  • A: Addition
  • S: Subtraction
For our given expression \(5(-2)(3) + 2\), we first handle the multiplication before applying the addition. This ensures that each step follows logically and avoids mistakes.
Multiplication
Multiplication is a fundamental arithmetic operation. When we see an expression like \(5xy + 2\), we must consecutively multiply the constants and variables. In the step \(5(-2)(3)\), we calculate as follows: First, compute \(5 \times -2\) to get \(-10\). Then, multiply \(-10 \times 3\) yielding \(-30\). Proper handling of multiplication ensures that the product correctly influences any subsequent operations.
Constants in Algebra
Constants are numbers that do not change. In our expression, \(2\) is a constant. Unlike variables, constants remain the same regardless of the values substituted for the variables. After simplifying the terms involving variables, you can straightforwardly manage constants in the final step. For instance, after reaching \(-30\) from the multiplication, you simply add the constant \(2\) as specified in the expression: \(-30 + 2 = -28\). Constants help simplify expressions and bring them to a final, manageable result.

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Most popular questions from this chapter

Expressions that occur in calculus are given. Write each expression as a single quotient in which only positive exponents and radicals appear. $$2 x\left(x^{2}+1\right)^{1 / 2}+x^{2} \cdot \frac{1}{2}\left(x^{2}+1\right)^{-1 / 2} \cdot 2 x$$

Simplify each expression. Assume that all variables are positive when they appear. $$\sqrt[4]{32 x}+\sqrt[4]{2 x^{5}}$$

The period \(T\), in seconds, of a pendulum of length \(l,\) in feet, may be approximated using the formula $$T=2 \pi \sqrt{\frac{l}{32}}$$ Express your answer both as a square root and as a decimal approximation. Find the period \(T\) of a pendulum whose length is 16 feet.

Expressions that occur in calculus are given. Factor each expression. Express your answer so that only positive exponents occur. $$6(6 x+1)^{1 / 3}(4 x-3)^{3 / 2}+6(6 x+1)^{4 / 3}(4 x-3)^{1 / 2} \quad x \geq \frac{3}{4}$$

Expressions that occur in calculus are given. Factor each expression. Express your answer so that only positive exponents occur. $$6 x^{1 / 2}\left(x^{2}+x\right)-8 x^{3 / 2}-8 x^{1 / 2} \quad x \geq 0$$
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