Question

select the correct answer.
which statement describes the solutions of this equation?
\\(\frac{x}{x + 2} + \frac{1}{x} = 1\\)
\\(\bigcirc\\) a. the equation has two valid solutions and no extraneous solutions.
\\(\bigcirc\\) b. the equation has no valid solutions and two extraneous solutions.
\\(\bigcirc\\) c. the equation has one valid solution and no extraneous solutions.
\\(\bigcirc\\) d. the equation has one valid solution and one extraneous solution.

Explanation:

Step1: Find the domain

The denominators are \(x + 2\) and \(x\), so \(x
eq - 2\) and \(x
eq0\).

Step2: Solve the equation

Multiply both sides by \(x(x + 2)\) (the least common denominator) to get:

$$ x\cdot x+(x + 2)=x(x + 2) $$

Simplify the left - hand side: \(x^{2}+x + 2\)
Simplify the right - hand side: \(x^{2}+2x\)
So the equation becomes:

$$ x^{2}+x + 2=x^{2}+2x $$

Subtract \(x^{2}\) from both sides: \(x + 2=2x\)
Subtract \(x\) from both sides: \(2=x\) or \(x = 2\)

Step3: Check for extraneous solutions

We check if \(x = 2\) is in the domain. Since \(2
eq-2\) and \(2
eq0\), \(x = 2\) is a valid solution.
There is only one solution (\(x = 2\)) and no extraneous solutions because when we solved the equation, we only got one solution and it is in the domain of the original equation.

Answer:

C. The equation has one valid solution and no extraneous solutions.