Problem 88 - Elementary and Intermediate Algebra

Question

Evaluate each expression, for \(x=3, y=-2,\) and \(z=-4\) See Example 10. $$ -\frac{2 z^{2}-x}{2 x-y^{2}} $$

Step-by-Step Solution

Answer

The evaluated expression is \(\frac{35}{2}.\)

Step 1: Substitute the Given Values

Replace every instance of the variables in the expression with their given numerical values: Replace \(x\) with 3, \(y\) with -2, and \(z\) with -4.

Step 2: Calculate the Numerator

Calculate the value of the numerator \(-2z^2-x\) by first squaring \(z\), then substituting and simplifying as follows:\[ (-2(-4)^2 - 3) = (-2\times16 - 3) = -32 - 3 = -35.\]

Step 3: Calculate the Denominator

Calculate the value of the denominator \(2x-y^2\) by first squaring \(y\), then substituting and simplifying as follows:\[ 2(3) - (-2)^2 = 6 - 4 = 2. \]

Step 4: Evaluate the Expression

Substitute the values from Steps 2 and 3 back into the fraction:\[ -\frac{-35}{2} = \frac{35}{2}. \]

Key Concepts

Substitution

Substitution is a key concept when evaluating algebraic expressions. It involves replacing variables with their specified numerical values. In the given exercise, we're provided specific numbers for each variable: \( x = 3 \), \( y = -2 \), and \( z = -4 \). Replacing these variables in the expression is the first step to find the numerical value. You just need to go through the expression and substitute the variables systematically:

• Where you see \( x \), place \( 3 \).
• Where you see \( y \), place \( -2 \).
• Where you see \( z \), place \( -4 \).

Substitution lays the foundation for simplifying expressions and ensures you're working with a numerical equation rather than variable terms. This step is integral before moving on to actual calculations.

Numerical Evaluation

Numerical evaluation refers to calculating the value of an expression once all variables have been replaced with their numerical counterparts. After substitution, the expression becomes a numerical problem that needs solving. In the step-by-step solution, one begins by focusing on the numerator, \(-2z^2-x\). This involves a series of calculations: 1. Square the number \( z \), i.e., \((-4)^2 = 16\). 2. Multiply the squared value by \(-2\) to get \(-32\). 3. Finally, subtract \(x\) from your result, which yields \(-35\). By breaking down the expression into smaller parts, you simplify the process, and numerical evaluation becomes a straightforward task. This practice helps in managing complex expressions by tackling one operation at a time.

Fractions

Fractions represent the division of two values, and they often appear in algebraic expressions. Here's how to approach them:In the exercise, after substituting and simplifying, we end up with a fraction, \(-\frac{-35}{2}\). The top part is called the numerator, and the bottom is the denominator.

• Numerator: The result of your calculations for \(-2z^2-x\), which is \(-35\).
• Denominator: The outcome of simplifying \(2x-y^2\), which is \(2\).

To simplify the fraction: 1. Handle the negative sign in front, \(-(-35)\), which makes the fraction positive since a double negative equals a positive. 2. Your fraction simplifies to \(\frac{35}{2}\), a simplified form that shows the results of the entire calculation clearly.

Order of Operations

When tackling any algebraic expression, it's crucial to follow the order of operations to ensure accuracy. This commonly follows the rule known as PEMDAS/BODMAS:

• P/B for Parentheses/Brackets
• E/O for Exponents/Orders like squares and powers
• MD for Multiplication and Division (from left to right)
• AS for Addition and Subtraction (from left to right)

In the context of the original exercise:- We first square \( z \) and \( y \), since exponents take precedence.- Next, multiply the squared value of \(z\) by \(-2\).- After managing exponents and multiplication, proceed with subtracting and adding as outlined.- Finally, division is handled inherently within the fraction. Correctly applying the order of operations prevents common errors, ensuring each calculation step is done accurately. This principle is crucial not just for this problem, but for all mathematical problem-solving.