Problem 15 - Algebra 2

Question

Divide. State any restrictions on the variables. $$ \frac{6 x+6 y}{x-y} \div \frac{18}{5 x-5 y} $$

Step-by-Step Solution

Answer

The result of the division is \(\frac{5(x+y)(x-y)}{3}\). The restriction on the variables is that \(x\) must not be equal to \(y\).

Step 1: Reverse the Second Fraction and Turn Division into Multiplication

Change the division problem into a multiplication problem by taking the reciprocal of the second fraction. So, \(\frac{6x+6y}{x-y} \div \frac{18}{5x-5y}\) becomes \(\frac{6x+6y}{x-y} * \frac{5x-5y}{18}\).

Step 2: Simplify the Fractions

Factor out the common factors in all the numerators and denominators to simplify the fractions. \[\frac{6(x+y)}{x-y} * \frac{5(x-y)}{18}\]Further simplify to\[\frac{5*6(x+y)}{18} * \frac{(x-y)}{1}\], which simplifies to \[\frac{5(x+y)(x-y)}{3}\].

Step 3: Determine Restrictions on Variables

At this point, the division is complete. However, the task requires us to identify any restrictions on the variables. These restrictions come from the issue that a fraction cannot have a denominator of zero. This means that \(x\) cannot equal \(y\), since this would cause the denominators in the original fractions to be zero. So the restriction is \(x \neq y\).

Key Concepts

Fraction Division

In algebra, dividing fractions is a common operation that is crucial to master. Instead of directly dividing one fraction by another, we utilize the reciprocal of the second fraction. The reciprocal of a fraction simply involves flipping its numerator and denominator.
By doing this, we transform a division operation into multiplication, which is often simpler to handle. For instance, \( \frac{a}{b} \div \frac{c}{d} \) changes into \( \frac{a}{b} \times \frac{d}{c} \).
This method is handy because multiplication of fractions is straightforward — you multiply the numerators together and the denominators together.

• Always remember: Flip the second fraction and change the division sign to multiplication.
• This step sets the stage for easy simplification in later steps.

Variable Restrictions

When dealing with algebraic fractions, it's essential to consider variable restrictions. These restrictions occur due to the mathematical rule that denominators cannot be zero — having a zero in the denominator would make the expression undefined.
To find variable restrictions, set each denominator expression equal to zero and solve for the variable. In the example exercise, the restrictions arise from the original denominators \(x-y\) and \(5x-5y\) being zero.

• Set \(x-y = 0\), resulting in \(x = y\).
• This means \(x eq y\) to ensure a valid fraction.

Hence, the restriction tells us the values that the variables cannot take, ensuring the fractions remain defined, maintaining the integrity of the solution.

Simplifying Expressions

Simplifying algebraic expressions involves reducing them to their simplest form while preserving equality. This usually means factoring out common terms from the numerators and denominators.
In each fraction, look for terms that can be factored out or canceled. Let's take the expression from the exercise, \( \frac{6(x+y)}{x-y} \times \frac{5(x-y)}{18} \).

• First, recognize \(x-y\) is a common factor which can be canceled from the numerator and the denominator.
• Next, multiply the remaining terms \(6 \times 5(x+y)\) divided by \(18\).
• Simplify the numeric coefficients: \( \frac{30(x+y)}{18}\), which simplifies further to \( \frac{5(x+y)}{3}\).

This simplification process trims the expression to its cleanest form, making it easier to analyze or use in further calculations.