The value of [tex]y[/tex] is required from the given conditions.
The value of [tex]y=6[/tex]
Fractions and Algebra
The fractions are
[tex]\dfrac{6}{y-4}[/tex] and [tex]\dfrac{y}{y+2}[/tex]
The difference between the given fractions is equal to the product of the fractions
[tex]\dfrac{6}{y-4}-\dfrac{y}{y+2}=\dfrac{6}{y-4}\times\dfrac{y}{y+2}\\\Rightarrow \dfrac{6y+12-y^2+4y}{(y-4)(y+2)}=\dfrac{6y}{(y-4)(y+2)}\\\Rightarrow 6y+12-y^2+4y=6y\\\Rightarrow 12-y^2+4y=0\\\Rightarrow y^2-4y-12=0\\\Rightarrow y=\dfrac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\times 1\times \left(-12\right)}}{2\times1}\\\Rightarrow y=6,-2[/tex]
Substituting [tex]-2[/tex] in the left hand side
[tex]\dfrac{6}{-2-4}-\dfrac{-2}{-2+2}=-1-\dfrac{-2}{0}[/tex]
This is undefined.
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