Problem 50 - Algebra and Trigonometry with Analytic Geometry

Question

Show that the equation is an identity. $$\frac{49 x^{2}-25}{7 x-5}=7 x+5$$

Step-by-Step Solution

Answer

The equation \(\frac{49x^{2}-25}{7x-5}=7x+5\) is an identity.

Step 1: Understand the Problem

We need to prove that the given equation \(\frac{49x^2-25}{7x-5}=7x+5\) is an identity. This means that the equation must hold true for all values of \(x\) except where the expressions are undefined.

Step 2: Factorize the Numerator

Notice that \(49x^2-25\) is a difference of squares, which can be factorized using the formula \(a^2-b^2 = (a-b)(a+b)\). Here, \(a^2=49x^2\) and \(b^2=25\). Thus, we have:\[49x^2 - 25 = (7x)^2 - 5^2 = (7x-5)(7x+5)\]

Step 3: Simplify the Left-Hand Side (LHS)

Substitute the factorized form back into the equation:\[\frac{49x^2-25}{7x-5} = \frac{(7x-5)(7x+5)}{7x-5}\]Since the numerator and denominator have \(7x-5\) as a common factor, we can simplify the fraction by canceling \(7x-5\):\[= 7x+5\]

Step 4: Compare Both Sides

After simplification, the left-hand side becomes \(7x + 5\). Compare it with the right-hand side of the original equation:\[7x + 5 = 7x + 5\]Since both sides are equal, the equation is an identity.

Step 5: Consider Domain Restrictions

Remember that the original equation had the denominator \(7x-5\). To prevent division by zero, \(x\) cannot be \(\frac{5}{7}\). However, this doesn't affect the identity since we're only excluding the undefined point.

Key Concepts

Difference of Squares

The 'difference of squares' is a useful algebraic identity that allows us to simplify math expressions. This identity states that for any two numbers \(a\) and \(b\), the difference of their squares can be expressed as

• \(a^2 - b^2 = (a-b)(a+b)\).

This formula is incredibly helpful in both simplifying expressions and solving equations. To apply this, identify square terms in your equation.
In the given problem, the expression \(49x^2 - 25\) is a difference of squares. Recognizing this, we let \(a = 7x\) and \(b = 5\), leading to the factorization:

• \(49x^2 - 25 = (7x-5)(7x+5)\).

Understanding this identity helps to break down complex expressions into simpler factors. It's a fundamental building block in algebra.

Factoring Polynomials

Factoring polynomials involves breaking down a polynomial into the product of simpler polynomials or factors. By recognizing patterns such as the difference of squares, we can greatly simplify this process.
In our exercise, the polynomial \(49x^2 - 25\) has been identified as a difference of squares, leading to the factorization \((7x-5)(7x+5)\).

• This allows us to express the polynomial in a more manageable form.

The factorization process is essential in solving algebraic equations, simplifying rational expressions, and understanding what values make the polynomial zero.
The key to factoring successfully is recognizing these common patterns and applying them correctly. By practicing factoring, students become more adept at identifying and simplifying expressions, making more complex algebraic operations easier to handle.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Simplifying rational expressions often involves factoring one or both parts of the fraction, enabling cancellation of common factors.
In the problem we have, the rational expression was \(\frac{49x^2-25}{7x-5}\). After factoring the numerator to obtain \((7x-5)(7x+5)\), we notice a common factor, \(7x-5\), in both the numerator and the denominator.

• This allows us to simplify the expression by canceling out the common term, leading to \(7x + 5\).

Simplifying rational expressions correctly can significantly reduce the complexity of algebraic problems. It highlights the power of understanding and applying algebraic identities. However, it's important to consider restrictions, such as avoiding division by zero. So always look out for values that could make the denominator zero and handle these with care.