The true statement is (c) the equation has one valid solution and no extraneous solutions
The equation is given as:
[tex]\frac{x}{x + 2} + \frac 1x = 1[/tex]
Take LCM
[tex]\frac{x^2 + x + 2}{x(x + 2)} = 1[/tex]
Expand the numerator
[tex]\frac{x^2 + x + 2}{x^2 + 2x} = 1[/tex]
Cross multiply
[tex]x^2 + x + 2 = x^2 + 2x[/tex]
Subtract x^2 from both sides
[tex]x + 2 = 2x[/tex]
Collect like terms
[tex]2x -x = 2[/tex]
[tex]x = 2[/tex]
Hence, the equation has one valid solution and no extraneous solutions
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