Question

posttest: rational relationships
14
select the correct answer.
sean used cross multiplication to correctly solve a rational equation. he found one valid solution and one extraneous solution. if 1 is the extraneous solution, which equation could he have solved?
a. the equation is \\(\frac{10}{x^2 - 1} = \frac{5}{3x - 3}\\) because 1 makes a denominator equal zero and is a solution of the equation derived from cross multiplying.
b. the equation is \\(\frac{x + 2}{x + 3} = \frac{6x}{8}\\) because 1 is a solution of both the original equation and the equation derived from cross multiplying.
c. the equation is \\(\frac{4x - 4}{x + 6} = \frac{x - 1}{10}\\) is because 1 makes a numerator equal zero and is a solution of the equation derived from cross multiplying.
d. the equation is \\(\frac{4}{x - 1} = \frac{x + 2}{10}\\) because 1 makes a denominator equal zero and is not a solution of the equation derived from cross multiplying.

Explanation:

Step1: Recall Extraneous Solution Definition

An extraneous solution makes the original rational equation's denominator zero (so it's not in the domain) and satisfies the cross - multiplied (or transformed) equation.

Step2: Analyze Option A

For the equation \(\frac{10}{x^{2}-1}=\frac{5}{3x - 3}\), first factor denominators: \(x^{2}-1=(x + 1)(x - 1)\) and \(3x-3 = 3(x - 1)\). The denominator is zero when \(x = 1\) or \(x=-1\). Cross - multiply: \(10\times(3x - 3)=5\times(x^{2}-1)\), \(30x-30 = 5x^{2}-5\), \(5x^{2}-30x + 25 = 0\), \(x^{2}-6x + 5=0\), \((x - 1)(x - 5)=0\), solutions \(x = 1\) and \(x = 5\). \(x = 1\) makes the original denominator zero (extraneous) and is a solution of the cross - multiplied equation.

Step3: Analyze Option B

For \(\frac{x + 2}{x + 3}=\frac{6x}{8}\), the denominator \(x+3 = 0\) when \(x=-3\). Substitute \(x = 1\) into original equation: \(\frac{1 + 2}{1+3}=\frac{3}{4}\), \(\frac{6\times1}{8}=\frac{3}{4}\). So \(x = 1\) is a valid solution, not extraneous.

Step4: Analyze Option C

For \(\frac{4x-4}{x + 6}=\frac{x - 1}{10}\), numerator \(4x - 4=0\) when \(x = 1\) (not a denominator zero issue). Cross - multiply: \(10(4x - 4)=(x - 1)(x + 6)\), \(40x-40=x^{2}+5x - 6\), \(x^{2}-35x + 34 = 0\), \((x - 1)(x - 34)=0\). \(x = 1\) makes numerator zero, but extraneous solutions come from denominator zero, so this is incorrect.

Step5: Analyze Option D

For \(\frac{4}{x - 1}=\frac{x + 2}{10}\), denominator zero when \(x = 1\). Cross - multiply: \(4\times10=(x - 1)(x + 2)\), \(40=x^{2}+x - 2\), \(x^{2}+x - 42 = 0\), \((x + 7)(x - 6)=0\), solutions \(x=-7\) and \(x = 6\). \(x = 1\) is not a solution of the cross - multiplied equation, so it doesn't fit the extraneous solution definition (extraneous should satisfy the cross - multiplied equation).

Answer:

A. The equation is \(\frac{10}{x^{2}-1}=\frac{5}{3x - 3}\) because 1 makes a denominator equal zero and is a solution of the equation derived from cross multiplying.