Problem 64 - Beginning and Intermediate Algebra

Question

Evaluate each expression when \(x=-5, y=4,\) and \(t=10 .\) See Example 6. $$ \frac{15-x}{y+2} $$

Step-by-Step Solution

Answer

The expression evaluates to \( \frac{10}{3} \).

Step 1: Substitute the Values into the Expression

We are given the expression \( \frac{15-x}{y+2} \) and need to evaluate it for \( x = -5 \), \( y = 4 \), and \( t = 10 \). Substitute the values of \( x \) and \( y \) into the expression. The expression becomes \( \frac{15 - (-5)}{4 + 2} \).

Step 2: Simplify the Numerator

In the expression \( \frac{15 - (-5)}{4 + 2} \), simplify the numerator by adding 5 to 15, since subtracting a negative number is the same as adding. This gives you \( 15 + 5 = 20 \). The expression now looks like \( \frac{20}{4 + 2} \).

Step 3: Simplify the Denominator

Simplify the denominator of the expression \( \frac{20}{4 + 2} \) by adding 4 and 2, which gives \( 6 \). The expression becomes \( \frac{20}{6} \).

Step 4: Simplify the Fraction

Now, simplify the fraction \( \frac{20}{6} \). To do this, find the greatest common divisor (GCD) of 20 and 6, which is 2. Divide both the numerator and the denominator by 2: \( \frac{20 \div 2}{6 \div 2} = \frac{10}{3} \). This is the simplified form of the expression.

Key Concepts

Substitution in Algebra

Substitution in algebra is about replacing variables with specific values. It often makes complex expressions easier to solve. In our example, the expression is \( \frac{15-x}{y+2} \). To evaluate it, you substitute the values of \( x \) and \( y \), which are \( -5 \) and \( 4 \) respectively. Doing so involves replacing \( x \) with \( -5 \) and \( y \) with \( 4 \) in the expression. So, the expression becomes \( \frac{15 - (-5)}{4 + 2} \).

Substitution often involves:

• Identifying which values to substitute for each variable.
• Carefully doing the replacement without altering the structure of the expression.
• Checking your work to ensure values were substituted correctly.

This process is essential in algebra because it helps convert abstract expressions into something tangible, allowing one to perform arithmetic operations to find a solution.

Fraction Simplification

Fraction simplification is a process that makes fractions easier to work with by reducing them to their simplest form. In this case, once we substituted \( x \) and \( y \) into the expression, it evaluated to \( \frac{20}{6} \). The goal here is to simplify it as much as possible, which is particularly useful in broader algebraic problems.

To simplify fractions:

• Find the greatest common divisor (GCD) of the numerator and denominator.
• Divide both the numerator and denominator by this GCD.
• Check to make sure all values are integers after division.

For \( \frac{20}{6} \), the GCD is 2. Dividing the numerator and denominator by their GCD results in \( \frac{10}{3} \). This fraction cannot be simplified further, making it the simplest form.

Negative Number Operations

Working with negative numbers in arithmetic operations can be tricky, but with practice it becomes straightforward. In our expression \( \frac{15 - (-5)}{4+2} \), the key operation involves subtracting \(-5\).

When subtracting a negative number:

• Remember that subtracting a negative is equivalent to adding its positive counterpart.
• \( 15 - (-5) \) thus becomes \( 15 + 5 \).
• This principle applies in any arithmetic operation involving negative numbers.

Understanding this concept ensures correct evaluations and helps avoid common pitfalls in algebra.

Greatest Common Divisor

The greatest common divisor (GCD) is the largest number that can divide two integers without leaving a remainder. Calculating the GCD is crucial in reducing fractions to their simplest form. In our example, we needed the GCD of 20 and 6.

To find the GCD:

• Identify the factors of both numbers.
• List these factors and find the largest number they have in common.
• Alternatively, use the Euclidean algorithm, which involves dividing the larger number by the smaller number and repeating the process with remainders.

For 20 and 6, the common factors are 1 and 2, with 2 being the greatest. This means that dividing both 20 and 6 by 2 results in the fraction \( \frac{10}{3} \), which is fully simplified.