Problem 33 - Algebra 2

Question

Simplify each expression. $$ \frac{(x+y)\left(\frac{1}{x}-\frac{1}{y}\right)}{(x-y)\left(\frac{1}{x}+\frac{1}{y}\right)} $$

Step-by-Step Solution

Answer

The expression simplifies to \(-1\).

Step 1: Distribute and simplify numerators and denominators

Rewrite the expression by expanding the numerator and the denominator. For the numerator, use the distributive property: \((x+y)\left(\frac{1}{x}-\frac{1}{y}\right) = (x+y)\cdot \frac{1}{x} - (x+y)\cdot \frac{1}{y}\). Similarly, for the denominator: \((x-y)\left(\frac{1}{x}+\frac{1}{y}\right) = (x-y)\cdot \frac{1}{x} + (x-y)\cdot \frac{1}{y}\)."With these, the initial expression becomes: \[\frac{\left(\frac{x+y}{x} - \frac{x+y}{y}\right)}{\left(\frac{x-y}{x} + \frac{x-y}{y}\right)}\]".

Step 2: Find Common Denominators for Fractions

For the numerator, simplify \(\frac{x+y}{x} - \frac{x+y}{y}\). The common denominator for this is \(xy\). So, \(\frac{x+y}{x} = \frac{(x+y)\cdot y}{xy}\) and \(\frac{x+y}{y} = \frac{(x+y)\cdot x}{xy}\). This gives: \(\frac{(x+y)\cdot y - (x+y)\cdot x}{xy}\)."Perform the same for the denominator, \(\frac{x-y}{x} + \frac{x-y}{y}\), with a common denominator of \(xy\): \(\frac{(x-y)\cdot y + (x-y)\cdot x}{xy}\)".

Step 3: Simplify Numerator and Denominator Terms

In each of the above simplified fractions, factor out the common terms. For the numerator, \(\frac{(x+y)\cdot (y - x)}{xy}\) and for the denominator, \(\frac{(x-y)\cdot (y + x)}{xy}\)."Notice that \(y - x = -(x - y)\), allowing the rewritten form as \(\frac{(x+y)\cdot -(x-y)}{xy}\)" and \(\frac{(x-y)\cdot (y + x)}{xy}\)".

Step 4: Cancel out and finalize expression

At this stage, simplify by canceling out the common terms. In both the simplified numerator \((x+y)\cdot -(x-y)\) and denominator \((x-y)\cdot (y + x)\), \(x-y\) can be canceled out. The expression simplifies to: \(-\frac{(x+y)}{(y+x)}\)."Remembering \(x+y = y+x\), finally you have: \(-1\).".

Key Concepts

Distributive Property

The distributive property plays a central role in algebra and simplifies complex expressions. It essentially states that for any numbers or variables, the product of a term with a sum (or difference) can be distributed over the addition (or subtraction). In mathematical terms, this is expressed as: \(a(b+c) = ab + ac\).
Applying this to our exercise, we used the distributive property in both the numerator and the denominator.

• For the numerator: \((x+y)\left(\frac{1}{x} - \frac{1}{y}\right)\). This means we distribute \((x+y)\) to each fraction inside the brackets: \((x+y) \cdot \frac{1}{x} - (x+y) \cdot \frac{1}{y}\).
• For the denominator: \((x-y)\left(\frac{1}{x} + \frac{1}{y}\right)\) works similarly, distributing the \((x-y)\) term: \((x-y) \cdot \frac{1}{x} + (x-y) \cdot \frac{1}{y}\).

Understanding this property helps in breaking down and solving complex expressions efficiently.

Common Denominator

When working with fractions, identifying and using a common denominator simplifies the process of adding or subtracting them. The common denominator is essentially the least common multiple (LCM) of the denominators.
In the given exercise, finding the common denominator allowed us to simplify both the numerator and the denominator expressions further:

• For example, in the numerator, \(\frac{x+y}{x} - \frac{x+y}{y}\), both fractions need to be expressed in terms of the same denominator \(xy\) to allow subtraction.
• This results in: \(\frac{(x+y)\cdot y}{xy} - \frac{(x+y)\cdot x}{xy}\).
• The same technique is mirrored in the denominator: \(\frac{x-y}{x} + \frac{x-y}{y}\) becomes \(\frac{(x-y)\cdot y + (x-y)\cdot x}{xy}\).

Common denominators ease several operations on fractions, making complex expressions much simpler.

Factorization

Factorization is a powerful algebraic tool, allowing you to rewrite expressions as the product of their factors. This can simplify complex expressions and make calculations more manageable.
In our step-by-step solution, factorization played a key role:

• We factor out common terms in both the numerator and the denominator. Consider the numerator \((x+y)\cdot (y-x)\), here \((y-x)\) can be rewritten as \(-(x-y)\).
• The denominator \((x-y)\cdot (y+x)\) does not require much change initially, but noting like terms can simplify further cancellations.
• This step facilitates the cancellation process in the following stage, making factorization quite useful.

By recognizing common terms, the entire algebraic dance becomes simpler and elegant.

Fraction Simplification

Simplifying fractions means writing them in their simplest form. This often involves canceling common factors from the numerator and the denominator after factorization.
In our example, after rewriting all the terms from the previous steps, you get:

• The simplified numerator is \((x+y)\cdot -(x-y)\).
• The denominator becomes \((x-y)\cdot (y+x)\).
• Since \(x-y\) is a common factor in both, it cancels out, leaving \(-\frac{x+y}{y+x}\).

By remembering \(x+y = y+x\), the expression simplifies to \(-1\).
The journey to simplify a fraction often ends in unveiling much simpler results, emphasizing the beauty and utility of mathematical reduction.