Here's the question you clicked on:
Envii
1/2 + x/6 = 18/x
Get the common denominator, and cross multiply...\[\frac{ 1 }{ 2 }+\frac{ x }{ 6 }=\frac{ 18 }{ x } \rightarrow \frac{ 3+x }{ 6 }=\frac{ 18 }{ x }\]\[x(3+x)=6(18) \rightarrow 3x+x^2-108=0 \rightarrow x^2+3x-108=0\]\[(x+12)(x-9)=0\]From here we got two values of x...\[x+12=0 \rightarrow x=-12\]\[x-9=0 \rightarrow x=9\]
The book says no solution. Does that have no solution?
What's the question? Do they ask for the values of x?
Oh, I see now. Thank you so much. :)