• anonymous
$\frac{ 3 }{ x-3}=\frac{ x }{ x-3 }-\frac{ 3 }{ 2 }x$ $\frac{ 3 }{ x-3 }=\frac{ 2x-3x(x-3) }{ 2\left( x-3 \right) }$ multiply both sides by 2(x-3) $6=2x-3x^2+9x,3x^2-11x+6=0$ 3x^2-9x-2x+6=0 3x(x-3)-2(x-3)=0 (x-3)(3x-2)=0 $x-3=0 ~or~ x=3~is~an~extraneous~ root.$ or 3x-2=0 calculate x
Mathematics

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