+14965 3/4x+1/8=3/4(x+8)
\(\frac {3}{4}x+ \frac {1}{8}=\frac {3}{4} (x+8)\)
\(\frac {3}{4}x+ \frac {1}{8}=\frac {3}{4}x+\frac {24}{4}\)
\({\color{red} \frac {1}{8}\neq \frac {24}{4}}\)
so
\(\frac{3}{4x}+\frac{1}{8}=\frac{3}{4\left(x+8\right) }\)
\(\frac{6+x}{8x} =\frac{3}{4\left(x+8\right) } \)
\(4(6+x)(x+8)= 24x\)
\((6+x)(x+8)= 6x\)
\(6x + 48 + x² + 8x = 6x\)
\(x^2+8x+48=0\)
No real solution.
asinus :- ) 
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