Question

which of the following statements are true about the given rational equation? choose two correct answers.
\\(\frac{4}{x+6} + \frac{1}{x^2} = \frac{x+10}{x^3+6x^2}\\)

• \\(x = 1\\) is a solution.
• \\(x = -1\\) is a solution.
• \\(x = 0\\) is a solution.
• \\(x = -6\\) is a solution.

Explanation:

Step1: Find the domain

First, determine the domain of the rational equation. The denominators are \(x + 6\), \(x^2\), and \(x^3+6x^2=x^2(x + 6)\). So, \(x
eq0\) and \(x
eq - 6\) (since these values make the denominators zero). So \(x = 0\) and \(x=-6\) are excluded from the domain, so they can't be solutions.

Step2: Test \(x = 1\)

Substitute \(x = 1\) into the left - hand side (LHS) and right - hand side (RHS) of the equation \(\frac{4}{x + 6}+\frac{1}{x^2}=\frac{x + 10}{x^3+6x^2}\)

LHS: \(\frac{4}{1+6}+\frac{1}{1^2}=\frac{4}{7}+1=\frac{4 + 7}{7}=\frac{11}{7}\)

RHS: \(\frac{1+10}{1^3+6\times1^2}=\frac{11}{1 + 6}=\frac{11}{7}\)

Since LHS = RHS when \(x = 1\), \(x = 1\) is a solution.

Step3: Test \(x=-1\)

Substitute \(x=-1\) into the equation.

LHS: \(\frac{4}{-1 + 6}+\frac{1}{(-1)^2}=\frac{4}{5}+1=\frac{4+5}{5}=\frac{9}{5}\)

RHS: \(\frac{-1 + 10}{(-1)^3+6\times(-1)^2}=\frac{9}{-1 + 6}=\frac{9}{5}\)

Since LHS = RHS when \(x=-1\), \(x=-1\) is a solution.

Answer:

A. \(x = 1\) is a solution.
B. \(x=-1\) is a solution.