Solve each equation. | (6x + 1)/ (x - 1) | = 3

검증된 단계별 안내

Recognize that the equation involves an absolute value expression: \(\left| \frac{6x + 1}{x - 1} \right| = 3\). Recall that \(|A| = B\) means \(A = B\) or \(A = -B\).

Set up two separate equations to solve: \(\frac{6x + 1}{x - 1} = 3\) and \(\frac{6x + 1}{x - 1} = -3\).

Solve the first equation \(\frac{6x + 1}{x - 1} = 3\) by multiplying both sides by \((x - 1)\) to eliminate the denominator, giving \(6x + 1 = 3(x - 1)\).

Solve the second equation \(\frac{6x + 1}{x - 1} = -3\) similarly by multiplying both sides by \((x - 1)\), resulting in \(6x + 1 = -3(x - 1)\).

For each equation, expand the right side, collect like terms, and solve for \(x\). Finally, check for any restrictions such as \(x \neq 1\) because the denominator cannot be zero.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Absolute Value Equations

An absolute value equation involves expressions within absolute value bars, which represent the distance from zero on the number line. To solve, set up two separate equations: one where the expression equals the positive value, and one where it equals the negative value. This accounts for both possible cases of the absolute value.

Solving Rational Equations

A rational equation contains variables in the denominator. To solve, first identify restrictions where the denominator is zero, then multiply both sides by the least common denominator to eliminate fractions. Afterward, solve the resulting equation while checking for extraneous solutions.

Checking for Extraneous Solutions

Extraneous solutions arise when solving equations involving absolute values or rational expressions, especially after multiplying both sides by expressions containing variables. Always substitute solutions back into the original equation to verify their validity and discard any that do not satisfy the original equation.