Problem 63 - Algebra for College Students

Question

For Problems 59-68, simplify each rational expression. You may want to refer to Example 12 of this section. \(\frac{2 y-2 x y}{x^{2} y-y}\)

Step-by-Step Solution

Answer

The simplified expression is \(\frac{-2}{x+1}\).

Step 1: Factor Out Common Terms in the Numerator

First, identify any common terms in the numerator, which is \(2y - 2xy\). Notice that \(2y\) is a common factor. Factor it out: \[2y - 2xy = 2y(1-x)\].

Step 2: Factor Out Common Terms in the Denominator

Next, consider the denominator \(x^2y - y\). Observe that \(y\) is a common factor here. Factor \(y\) out:\[x^2y - y = y(x^2 - 1)\].

Step 3: Simplify Using the Difference of Squares

The expression \(x^2 - 1\) in the denominator is a difference of squares, which can be factored into \[(x - 1)(x + 1)\]. Thus, the denominator becomes:\[y(x-1)(x+1)\].

Step 4: Cancel Common Factors

Now, write the rational expression with the factored numerator and denominator:\[\frac{2y (1-x)}{y(x-1)(x+1)}\].Notice that \(y\) is a common factor in both numerator and denominator, which can be canceled out:\[\frac{2(1-x)}{(x-1)(x+1)}\].Also recognize that \(1-x\) is equivalent to \[-(x-1)\]. Substituting for simplification, we get:\[\frac{2(-1)(x-1)}{(x-1)(x+1)} = \frac{-2(x-1)}{(x-1)(x+1)}\].Now cancel the \((x-1)\) terms, giving us:\[\frac{-2}{x+1}\].

Key Concepts

Factoring

Factoring is like taking things apart to see what they’re made of. It’s a technique used to break down algebraic expressions into simpler parts, known as factors, which when multiplied together give back the original expression.
For instance, if you look at the original numerator of the rational expression, which is \(2y - 2xy\), you can notice a common factor, which is \(2y\). By factoring out \(2y\), we turn \(2y - 2xy\) into \(2y(1-x)\).
This step is crucial because it simplifies the expression and reveals hidden common factors which we can later cancel. Factoring is a helpful first step in making complex math problems easier to handle.
When dealing with polynomials, always check for common factors. It's like simplifying a fraction: simple factors in the numerator and denominator can often cancel out.

Cancel Common Factors

Once we have factored both the numerator and the denominator, our next move is to look for common factors that can be "canceled out". In the given problem, the rational expression becomes \(\frac{2y(1-x)}{y(x-1)(x+1)}\). Both the numerator and denominator have \(y\) as a common factor.
Canceling these common factors makes the expression cleaner, resulting in \(\frac{2(1-x)}{(x-1)(x+1)}\). This does not change the value of the expression because we're merely dividing by terms that appear both on top and bottom.

• Identify the common factors in both numerator and denominator.
• Cancel them by dividing them out.

Cancelling helps in both simplifying the expressions and significantly reducing error potential when solving problems.

Difference of Squares

When you come across an expression like \(x^2 - 1\), you’re dealing with a difference of squares. This is a key pattern where \(a^2 - b^2\) factors into \((a-b)(a+b)\).
In the denominator \(x^2 - 1\), recognize that it can be rewritten as \((x-1)(x+1)\).
So always remember:

• A difference of squares can be split into two simpler binomials.
• It simplifies faster calculations in complex expressions.

Understanding this pattern is an essential skill in algebra for both solving equations and simplifying expressions efficiently.

Simplifying Algebraic Expressions

Simplifying algebraic expressions means making them as straightforward as they can be. It involves using all the techniques we've discussed: factoring, canceling common factors, and recognizing special patterns like the difference of squares.
For the given expression \(\frac{2(-1)(x-1)}{(x-1)(x+1)}\), observe that \(2(-1)(x-1)\) simplifies to \(-2(x-1)\). Then, cancelling out \((x-1)\), we arrive at \(\frac{-2}{x+1}\).
Remember, simplification:

• Involves reducing expressions to their simplest form.
• Makes calculating and understanding easier.

Simplified expressions are tidier, more manageable, and help reveal relationships and properties of algebraic forms.