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Let \(a=3x^2+x\), then the equation becomes:

\[ \dfrac{a-1}{1}+\dfrac{1}{a-3}=0 \]

Multiply all terms by the LCD which is \(a-3\)

\[(a-3)(a-1)+\dfrac{1}{a-3}\cdot (a-3)=0\]

removing the brackets gives:

\[ a^2-a-3a+3a+1=0\]

\[a^2-4a+4=0\]

\[a^-2a-2a+4=0\]

\[a(a-2)-2(a-2)=0\]

\[(a-2)^2=0\]

thus

\[a=2 \quad \text{twice}\]

Now remember that

\[3x^2+x=2\]

\[3x^2+x-2=0\]

\[3x^2+3x-2x-2=0\]

\[3x(x+1)-2(x+1)=0\]

\[(x+1)(3x-2)=0\]

\[\therefore x=-1 \quad OR \quad x=\dfrac{2}{3}\]
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