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How would you go about solving this equation for x? (In comments)
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\[\frac{ x }{ x-3 } = \frac{ 3 }{ x-3 }+9 \]
\[\frac{x}{x-3}=\frac{3}{x-3}+9\\\]first note that x cant be 3 now lets go \[\frac{x}{x-3}-\frac{3}{x-3}=9\\\frac{x-3}{x-3}=9\\x \neq 3\\we \\can\\simplify\\1=9\\\] and it has no roots
another method multiply by (x-3) \[(x-3)(\frac{x}{x-3}=\frac{3}{x-3}+9)\\x=3+9(x-3)\\x-9x=-27+3\\-8x=-24\\x=3\] but \[x \neq 3\\\] because denominator goes to zero
the equation has no root
or no solution
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it's simply like that 2=2+1, which is false equation. above is conditional equation with no solution.
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