Problem 106 - Elementary Algebra

Question

For Problems \(93-106\), evaluate each algebraic expression for the given values of the variables. $$ x^{2}+5 x y \quad \text { for } x=-0.2 \text { and } y=-0.6 $$

Step-by-Step Solution

Answer

The expression evaluates to 0.64.

Step 1: Substitute Values Into Expression

We want to evaluate the expression \( x^2 + 5xy \) for the given values of \( x = -0.2 \) and \( y = -0.6 \). First, substitute these values into the expression: \((-0.2)^2 + 5(-0.2)(-0.6)\).

Step 2: Calculate the Square Term

Calculate \((-0.2)^2\). The square of a number is the number multiplied by itself. So, \((-0.2)^2 = 0.04\).

Step 3: Calculate the Product in the Second Term

Now calculate \(5(-0.2)(-0.6)\). First, multiply \(-0.2\) and \(-0.6\), which equals \(0.12\) because a negative times a negative is positive. Then multiply the result by 5: \(5 \times 0.12 = 0.6\).

Step 4: Combine the Two Terms

Add the results from Step 2 and Step 3: \(0.04 + 0.6 = 0.64\). This is the value of the expression when the given values of \(x\) and \(y\) are substituted.

Key Concepts

Substitution of Values

In algebra, substitution of values is a technique used to replace variables with numbers to simplify an expression or equation. It's like swapping a placeholder with an actual number. To substitute values into an expression:

• Identify the variable and the corresponding value given in the problem.
• Replace each instance of the variable with the given number.

In our exercise, we start with the expression \( x^2 + 5xy \) and substitute \( x = -0.2 \) and \( y = -0.6 \). So, wherever we see \( x \) in the expression, we replace it with \(-0.2\), and wherever we see \( y \), we replace it with \(-0.6\).
This process allows us to transform the expression with variables into a simple arithmetic expression: \((-0.2)^2 + 5(-0.2)(-0.6)\). Once substitution is complete, we evaluate the arithmetic operations to find the final result.

Basic Arithmetic Operations

Basic arithmetic operations form the foundation of algebraic calculations. These include addition, subtraction, multiplication, and division. They are the tools we use to manipulate and simplify expressions.
In the given exercise, we focus mainly on multiplication and addition:

• Multiplication: This is used when you see two numbers or a number and a variable beside each other, often indicated with parentheses. For example, in \((-0.2) \times (-0.6)\).
• Square: Taking the square of a number means multiplying it by itself. For instance, \((-0.2)^2 = (-0.2) \times (-0.2) = 0.04\).
• Addition: We add numbers together to combine results, like \(0.04 + 0.6 = 0.64\), which gives the total value of the expression after performing all operations.

These operations are fundamental for solving the expression step by step, leading to the correct evaluation of the algebraic expression.

Negative Numbers in Algebra

Working with negative numbers in algebra requires understanding how they interact with various mathematical operations. Negative numbers have unique properties that can affect the outcome of an expression when used in calculations.
Here are key points to remember:

• Negative times Negative: When multiplying two negative numbers, the result is positive. For example, \(-0.2 \times -0.6 = 0.12\).
• Negative times Positive: The result of multiplying a negative and a positive number is negative. However, in our problem, we multiply two negatives first, resulting in a positive number.
• Negative Squared: Squaring a negative number (multiplying it by itself) will yield a positive result, such as \((-0.2)^2 = 0.04\), because a negative times a negative results in a positive.

Understanding these rules is critical for correctly evaluating expressions involving negative numbers, ensuring each operation is accurately computed to maintain the integrity of the solution.