Respuesta :
The value of the variable y is [tex]y=6[/tex]
Explanation:
The given equation is [tex]\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)[/tex]
Taking LCM on LHS of the equation, we get,
[tex]\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)[/tex]
Simplifying the term on RHS of the equation, we get,
[tex]\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\frac{6 y}{(y-4)(y+2)}[/tex]
Since, both the sides of the equation have the same denominator, we can cancel them.
Thus, we have,
[tex]6(y+2)-y(y-4)=6 y[/tex]
Multiplying the terms within the bracket, we get,
[tex]6y+12-y^2-4y=6 y[/tex]
Adding the like terms, we have,
[tex]2y+12-y^2=6 y[/tex]
Subtracting both sides by [tex]6 y[/tex], we have,
[tex]-4y+12-y^2=0[/tex]
Factoring the equation, we get,
[tex](y+2)(y-6)=0[/tex]
[tex]y=-2, y=6[/tex]
The values of the variable y are [tex]y=-2, y=6[/tex]
At the point [tex]y=-2[/tex], the equation [tex]\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)[/tex] becomes undefined.
Thus, the value of the variable y is [tex]y=6[/tex]
