Problem 44
Question
Simplify each algebraic fraction. $$\frac{x^{2}-3 x y+2 y^{2}}{x^{2}-4 y^{2}}$$
Step-by-Step Solution
Verified Answer
The simplified form is \(\frac{x-y}{x+2y}\).
1Step 1: Factor the Numerator
To simplify the algebraic fraction, start with factoring both the numerator and the denominator. The numerator \( x^2 - 3xy + 2y^2 \) is a quadratic expression. We need numbers that multiply to \( 2y^2 \) and add to \(-3xy\). These numbers are \(-xy\) and \(-2xy\). The factorization is \((x - 2y)(x - y)\).
2Step 2: Factor the Denominator
Next, factor the denominator \( x^2 - 4y^2 \). This is a difference of squares, which can be factored as \((x - 2y)(x + 2y)\).
3Step 3: Simplify the Fraction
Now that both the numerator and the denominator are factored, the expression to simplify is \(\frac{(x - 2y)(x - y)}{(x - 2y)(x + 2y)}\). Notice that \((x - 2y)\) is a common factor in both the numerator and the denominator. Cancel it out to simplify the fraction, resulting in \(\frac{x - y}{x + 2y}\).
Key Concepts
Factoring QuadraticsDifference of SquaresAlgebraic Expressions
Factoring Quadratics
Factoring quadratics is an important skill in algebra and involves expressing a quadratic expression as a product of two binomials. A quadratic expression generally takes the form \(ax^2 + bx + c\). Factoring involves finding numbers that multiply to \(ac\) and add to \(b\).
In the exercise, the numerator is \(x^2 - 3xy + 2y^2\). We need numbers that multiply to \(2y^2\) (the product of the leading coefficient \(1\) and the constant term \(2y^2\)) and add to \(-3xy\).
These numbers are \(-xy\) and \(-2xy\). Therefore, the factorization of the quadratic expression becomes \((x - 2y)(x - y)\).
In the exercise, the numerator is \(x^2 - 3xy + 2y^2\). We need numbers that multiply to \(2y^2\) (the product of the leading coefficient \(1\) and the constant term \(2y^2\)) and add to \(-3xy\).
These numbers are \(-xy\) and \(-2xy\). Therefore, the factorization of the quadratic expression becomes \((x - 2y)(x - y)\).
- Multiply to \(2y^2\)
- Add to \(-3xy\)
- Factored form: \((x - 2y)(x - y)\)
Difference of Squares
The difference of squares is a special case in algebra where you have two perfect squares separated by a subtraction sign. It follows the identity \(a^2 - b^2 = (a - b)(a + b)\). This formula helps in easily factoring expressions like \(x^2 - 4y^2\) as seen in the denominator of the algebraic fraction.
To apply the difference of squares, recognize that \(x^2\) is \(x\) squared and \(4y^2\) is \((2y)^2\). Therefore, the expression \(x^2 - 4y^2\) can be rewritten using the difference of squares formula:
To apply the difference of squares, recognize that \(x^2\) is \(x\) squared and \(4y^2\) is \((2y)^2\). Therefore, the expression \(x^2 - 4y^2\) can be rewritten using the difference of squares formula:
- Identify \(a = x\) and \(b = 2y\)
- Apply formula: \((x - 2y)(x + 2y)\)
Algebraic Expressions
Algebraic expressions are mathematical phrases that can include numbers, variables, and operation symbols. Simplifying algebraic expressions helps in making them easier to work with. In algebra, this often involves factoring, canceling common factors, and performing arithmetic operations.
In this exercise, we worked with the fraction \(\frac{(x - 2y)(x - y)}{(x - 2y)(x + 2y)}\). Simplification involves:
In this exercise, we worked with the fraction \(\frac{(x - 2y)(x - y)}{(x - 2y)(x + 2y)}\). Simplification involves:
- Identifying common factors in numerator and denominator
- Cancelling out the common factor \((x - 2y)\)
- Resulting in a simplified form: \(\frac{x - y}{x + 2y}\)
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Problem 44
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