Problem 44 - Intermediate Algebra

Question

Evaluate the algebraic expressions for the given values of the variables. $$ x^{2} y^{3}-2 x y+x^{2} y^{2}, \quad x=-1 \text { and } y=-3 $$

Step-by-Step Solution

Answer

The evaluated expression is -24.

Step 1: Substitute the variable values

We need to substitute \(x = -1\) and \(y = -3\) into the expression \(x^{2}y^{3}-2xy+x^{2}y^{2}\). This means replacing every \(x\) with \(-1\) and every \(y\) with \(-3\).

Step 2: Evaluate each term separately

Now evaluate the terms one by one:1. \((x^2)(y^3) = ((-1)^2)((-3)^3) = 1(-27) = -27\)2. \(-2xy = -2(-1)(-3) = 6\)3. \(x^2y^2 = ((-1)^2)((-3)^2) = 1(9) = 9\)

Step 3: Combine the terms

Now add the results of each term from Step 2:\(-27 - 6 + 9 = -24\)

Key Concepts

Substitution Method

When working with algebraic expressions, the substitution method is a crucial technique. It involves replacing variables within an expression with given numerical values to simplify and evaluate the expression.
Here’s how it works:

• Identify the variables in the expression.
• Replace each variable with its corresponding given value.

This method helps make abstract expressions tangible by turning them into numerical calculations. For example, if an expression is given as \(x^{2}y^{3}-2xy+x^{2}y^{2}\) and you need to evaluate it for \(x = -1\) and \(y = -3\), you substitute each occurrence of \(x\) with \(-1\) and \(y\) with \(-3\). This straightforward approach simplifies complex expressions quickly and accurately.

Polynomial Evaluation

Evaluating polynomials at specific values is like peeling away layers to see what the expression amounts to in numbers. For a polynomial, consider these steps:

• After substituting the given values for variables, focus on each term separately.
• Compute powers first, as they set the basis for the remaining operations.
• Solve multiplications and products within each term.

In our previous example, once we substitute \(x = -1\) and \(y = -3\) into the expression, we evaluate each part individually, like \((x^2)(y^3)\) turning into \(1\times(-27)\). Performing calculations one term at a time ensures accuracy and helps you spot any errors. After evaluating each term separately, you then add or subtract them according to the expression's structure to get the final result.

Negative Numbers in Algebra

Understanding how negative numbers behave in algebra is essential for correct computations. Here's a quick guide:

• When you square a negative number, it becomes positive; e.g., \((-1)^2 = 1\).
• Multiplying two negatives results in a positive; e.g., \(-1 \times -3 = 3\).
• Multiplying a negative and a positive number results in a negative.

Always pay attention to the signs during calculations, especially in complex expressions involving several operations. In our example, the term \(-2xy\) becomes 6 after resolving the signs. Such operations show why understanding the mechanics of negative numbers helps maintain precision and accuracy in your solutions.