Problem 35 - Elementary Algebra

Question

Perform the indicated multiplications and divisions and express your answers in simplest form. $$\frac{2 x^{2}-x y-3 y^{2}}{(x+y)^{2}} \div \frac{4 x^{2}-12 x y+9 y^{2}}{10 x-15 y}$$

Step-by-Step Solution

Answer

\(\frac{5(2x + 3y)(x - y)}{(x+y)^2(2x - 3y)}\) in simplest form.

Step 1: Write the Division as Multiplication

To divide fractions, we multiply by the reciprocal. Therefore, rewrite the division problem as a multiplication by flipping the second fraction. This gives us:\[\frac{2x^2 - xy - 3y^2}{(x+y)^2} \times \frac{10x - 15y}{4x^2 - 12xy + 9y^2}\]

Step 2: Factor the Numerators and Denominators

Next, we factor both the numerators and denominators.- For the first numerator, \(2x^2 - xy - 3y^2\), we can factor as \((2x + 3y)(x - y)\).- The first denominator \((x+y)^2\) is already in factored form.- For the second numerator \(10x - 15y\), factor out the greatest common factor, which is 5: \(5(2x - 3y)\).- The second denominator \(4x^2 - 12xy + 9y^2\) can be factored as \((2x - 3y)^2\).

Step 3: Substitute the Factored Expressions

Substituting the factored expressions into our multiplication gives:\[\frac{(2x + 3y)(x - y)}{(x+y)^2} \times \frac{5(2x - 3y)}{(2x - 3y)^2}\]

Step 4: Simplify the Expression

Now, simplify the expression by canceling out common factors in the numerators and denominators.- The factor \((2x - 3y)\) appears in both the numerator and denominator, so one \((2x - 3y)\) cancels out with the squared term in the denominator.This results in:\[\frac{5(2x + 3y)(x - y)}{(x+y)^2(2x - 3y)}\]

Step 5: Further Simplification

Check if any further simplification is possible. In this case, there are no more common factors to cancel, which means the expression is already in its simplest form.

Key Concepts

Factoring Polynomials

Factoring polynomials is one of the essential skills in algebra. It involves breaking down a complex polynomial, such as \[2x^2 - xy - 3y^2\] into simpler terms or products of simpler polynomials. The factored form helps in solving equations and simplifying expressions. To factor a polynomial:

• Look for the greatest common factor (GCF). For instance, in \[10x - 15y,\] 5 is the GCF, giving us \[5(2x-3y).\]
• For quadratic expressions like \[2x^2 - xy - 3y^2,\] think about finding two binomials that multiply to give the original expression (e.g., \[(2x + 3y)(x - y)\]).
• When a polynomial is a perfect square, it can be expressed as a binomial square, such as \[4x^2 - 12xy + 9y^2 = (2x - 3y)^2.\]

Understanding factoring helps make expressions manageable, allowing for easier manipulation in calculations and simplifications.

Simplifying Expressions

Simplifying expressions is all about reducing them to their simplest form. This can make solving mathematical problems much less complex. Here's how to simplify an expression, as seen in our exercise:

• First, factor expressions when possible, as we did with each polynomial in the numerators and denominators. This step reveals common terms that might cancel.
• Next, rewrite the expression by substituting the factored terms.
• Finally, cancel out any common factors in both numerators and denominators, simplifying your expression.

In our exercise, simplifying included canceling the \[(2x - 3y)\] factor from both the numerator and denominator. Simplifying not only beautifies an expression but is also crucial for finding the most direct path to the solution, as it eliminates unnecessary complexity.

Fraction Division

Dividing fractions might feel a bit tricky at first, but a simple rule can make it straightforward: multiply by the reciprocal. To divide one fraction by another, here's what you do:

• Reverse the second fraction. In the given problem, the initial division \[\frac{2x^2 - xy - 3y^2}{(x+y)^2} \div \frac{4x^2 - 12xy + 9y^2}{10x - 15y}\] changes to a multiplication:\[\frac{2x^2 - xy - 3y^2}{(x+y)^2} \times \frac{10x - 15y}{4x^2 - 12xy + 9y^2}.\]
• Then, treat it like a multiplication problem. Once you're in multiplication mode, follow the steps for simplifying and factoring.

Breaking dividing tasks into multiplication after inverting the second fraction turns a potentially head-scratching problem into a more straightforward multiplication problem, allowing you to easily factor and simplify.