In Exercises 1–34, solve each rational equation. If an equation has no solution, so state. (x−6)/(x+5) = (x−3)/(x+1)

Verified step by step guidance

1. Identify the rational equation: $\frac{x-6}{x+5}=\frac{x-3}{x+1}$.
2. Cross-multiply to eliminate the fractions: $(x-6)(x+1)=(x-3)(x+5)$.
3. Expand both sides of the equation: $x^2+x-6x-6=x^2+5x-3x-15$.
4. Simplify the expanded equation: $x^2-5x-6=x^2+2x-15$.
5. Subtract $x^2$ from both sides and solve the resulting linear equation: $-5x-6=2x-15$.

Key Concepts

Rational Equations

Rational equations are equations that involve fractions with polynomials in the numerator and denominator. To solve these equations, one typically finds a common denominator to eliminate the fractions, allowing for easier manipulation and simplification. Understanding how to work with rational expressions is crucial for solving these types of equations.

Cross Multiplication

Cross multiplication is a technique used to solve rational equations where two fractions are set equal to each other. By multiplying the numerator of one fraction by the denominator of the other, and vice versa, one can create a simpler equation without fractions. This method is particularly useful in rational equations, as it helps to eliminate the denominators and simplify the solving process.

Extraneous Solutions

Extraneous solutions are solutions that emerge from the algebraic manipulation of an equation but do not satisfy the original equation. When solving rational equations, it is essential to check each potential solution by substituting it back into the original equation to ensure it does not result in division by zero or an invalid statement. Recognizing and discarding extraneous solutions is a key step in the solution process.