Problem 60 - Introductory Algebra for College Students

Question

denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{2 x-y}{x-y}+\frac{x-2 y}{y-x}$$

Step-by-Step Solution

Answer

The simplification of the expression \(\frac{2x - y}{x - y} + \frac{x - 2y}{y - x}\) results to \((x + y) / (x - y)\).

Step 1: Identify the inverse relationship

See that the two denominators \(x - y\) and \(y - x\) are negatives of each other. This means \((x - y) = - (y - x)\).

Step 2: Simplify the second fraction

Knowing that a fraction divided by its negative equals -1, change the denominator \(y - x\) to its negative \(x - y\) and add a negative sign to the fraction to give \(- (x - 2y) / (x - y)\). This then changes the whole expression to \((2x - y) / (x - y) - (x - 2y) / (x - y)\)

Step 3: Add the fractions

Since the fractions now have the same denominator \((x - y)\), add the numerators: \(2x - y - x + 2y\) to get \(x + y\) as the new numerator. The whole fraction now is \((x + y) / (x - y)\).

Key Concepts

Simplify Fractions

Understanding how to simplify fractions is essential when working with mathematical expressions, especially when they involve variables. When simplifying fractions, our goal is to reduce them to their simplest form - a form where the numerator and the denominator have no common factors except for 1. This process can make calculations easier and results more interpretable.

To simplify a fraction, you should:

• Factor both the numerator and the denominator to their prime components.
• Divide both by any common factors.
• Rewrite the fraction using the reduced numerator and denominator.

When we apply simplification to algebraic expressions, we look out for like terms in the numerator and the denominator, which can be simplified. For instance, if we encounter \(\frac{x^2 - y^2}{x-y}\), we can factor the numerator as \(\frac{(x+y)(x-y)}{x-y}\) and simplify to \(x+y\) since \(x-y\) in the numerator and denominator cancel each other out, being the same factor. In this exercise, we see a similar situation, though it involves understanding additive inverses to truly simplify the expressions involved.

Additive Inverses

The concept of additive inverses is critical in understanding how to work with fractions that have opposite denominators. An additive inverse is simply a number that, when added to the original number, yields zero. In other words, the additive inverse of a number 'a' is '-a', and vice versa. The additive inverse of \(5\) is \( -5\) and for \(x\) it is \( -x\).

In our given exercise, the denominators \(x-y\) and \(y-x\) are additive inverses. This is because if we add \(x-y\) to \(y-x\), we obtain zero: \(x-y + y-x = x - y + (-x + y) = x - x + y - y = 0\). Recognizing this relationship allows us to simplify the expression, by realizing one of the denominators can be transformed into the other by multiplying by -1, which does not change the value of the expression but simplifies the addition process.

Inverse Relationship

The inverse relationship in mathematics refers to a pair of elements that, when combined using a specific operation, bring us back to the identity element for that operation. For addition, the identity element is zero as already mentioned in the context of additive inverses.

In our exercise, the inverse relationship between the denominators \(x - y\) and \(y - x\) informs us that they can nullify each other. This understanding allows us to manipulate the equation for easier simplification. By adjusting the sign of one denominator and its corresponding numerator, we align the fractions to have a uniform denominator, making it possible to carry out the addition.

Recognizing and using inverse relationships enables us to overcome obstacles that might seem complex at an initial glance. For example, using inverses can help solve equations, simplify algebraic expressions, and understand the behavior of functions, which is a valuable strategy for working with algebraic fractions.