Answer:
C. The equation has one valid solution and no extraneous solutions.
Step-by-step explanation:
Given the equation;
[tex]\frac{x}{x+2} + \frac{1}{x} = 1[/tex]
First, take note of the domain. The domain of the equation is the values for which the equation holds true. As of now, one doesn't have the actual value of the unknown, therefore, one will have to take not of solutions which the variable cannot equal. Since it is impossible for the denominator (part of a fraction underneath the fraction bar) of a fraction to be zero, one will have to set each of the denominators equal to zero and solve to find the value of (x). One will get that;
{ [tex]x[/tex] ∈ [tex]R[/tex] | [tex]x\neq -2, x\neq -2[/tex] }
Now one must solve the equation,
[tex]\frac{x}{x+2} + \frac{1}{x} = 1[/tex]
Multiply the equation by the LCD (least common denominator), this makes the equation easier to solve. The LCD of this equation is [tex](x+2)(x)[/tex]
[tex]((x+2)(x)\frac{x}{x+2}) + ((x+2)(x)\frac{1}{x}) = (x)(x+2)(1)\\[/tex]
Simplify
[tex]x^2 + x + 2 = x^2 + 2x[/tex]
Inverse operations,
[tex]x^2 + x +2 = x^2 + 2x\\-x^2\\\\x + 2 = 2x\\-x\\\\2 = x[/tex]
Since ([tex]x=2[/tex]) is not excluded by the domain, it is a real solution, and since it is the only solution, the correct answer is, (C. The equation has one valid solution and no extraneous solutions.).
