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select the correct answer. melanie used cross multiplication to correct…

Question

select the correct answer.

melanie used cross multiplication to correctly solve a rational equation. she found one valid solution and one extraneous solution. if 3 is the extraneous solution, which equation could she have solved?

  • the equation is \\(\frac{4}{x-3} = \frac{x}{10}\\) because 3 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying.
  • the equation is \\(\frac{x-3}{4} = \frac{2x-6}{4x}\\) because 3 makes a numerator equal zero and is a solution of the equation derived from cross multiplying.
  • the equation is \\(\frac{8}{x^2-9} = \frac{5}{2x-6}\\) because 3 makes a denominator equal zero and is a solution of the equation derived from cross multiplying.
  • the equation is \\(\frac{8}{x+3} = \frac{12}{4x-3}\\) because 3 is a solution of both the original equation and the equation derived from cross multiplying.

Explanation:

Analyze the definition of an extraneous solution

An extraneous solution of a rational equation is a value that:

  1. Is a solution to the polynomial equation obtained after cross-multiplying (or clearing denominators).
  2. Makes at least one denominator in the original rational equation equal to zero, meaning it is excluded from the domain of the original equation.

Evaluate each option for \(x = 3\)

  • Option 1: \(\frac{4}{x-3} = \frac{x}{10}\)
  • Cross-multiplying: \(40 = x(x-3) \implies x^2 - 3x - 40 = 0 \implies (x-8)(x+5) = 0 \implies x = 8, -5\).
  • \(x = 3\) is not a solution to the cross-multiplied equation.
  • Option 2: \(\frac{x-3}{4} = \frac{2x-6}{4x}\)
  • At \(x = 3\), the numerators are zero, but the denominators are \(4\) and \(12\) (not zero). Thus, \(3\) does not make a denominator zero.
  • Option 3: \(\frac{8}{x^2-9} = \frac{5}{2x-6}\)
  • Denominators are \(x^2-9\) and \(2x-6\). At \(x = 3\), both denominators are zero: \(3^2-9 = 0\) and \(2(3)-6 = 0\).
  • Cross-multiplying: \(8(2x-6) = 5(x^2-9) \implies 16x - 48 = 5x^2 - 45 \implies 5x^2 - 16x + 3 = 0\).
  • Factoring: \((5x-1)(x-3) = 0 \implies x = \frac{1}{5}, 3\).
  • Thus, \(3\) is a solution to the cross-multiplied equation and makes the original denominators zero.
  • Option 4: \(\frac{8}{x+3} = \frac{12}{4x-3}\)
  • At \(x = 3\), the denominators are \(6\) and \(9\) (not zero).

Identify the correct choice

Option 3 correctly states that the equation is \(\frac{8}{x^2-9} = \frac{5}{2x-6}\) because \(3\) makes a denominator equal to zero and is a solution of the equation derived from cross-multiplying.

Answer:

  • (A) The equation is \(\frac{4}{x-3} = \frac{x}{10}\) because 3 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying.
  • (B) The equation is \(\frac{x-3}{4} = \frac{2x-6}{4x}\) because 3 makes a numerator equal zero and is a solution of the equation derived from cross multiplying.
  • (C) The equation is \(\frac{8}{x^2-9} = \frac{5}{2x-6}\) because 3 makes a denominator equal zero and is a solution of the equation derived from cross multiplying. (Correct answer)
  • (D) The equation is \(\frac{8}{x+3} = \frac{12}{4x-3}\) because 3 is a solution of both the original equation and the equation derived from cross multiplying.