Question

Find the value of the variable y, for which:

Answer 1

The value of the variable y is $y=6$

Explanation:

The given equation is $\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)$

Taking LCM on LHS of the equation, we get,

$\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)$

Simplifying the term on RHS of the equation, we get,

$\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\frac{6 y}{(y-4)(y+2)}$

Since, both the sides of the equation have the same denominator, we can cancel them.

Thus, we have,

$6(y+2)-y(y-4)=6 y$

Multiplying the terms within the bracket, we get,

$6y+12-y^2-4y=6 y$

Adding the like terms, we have,

$2y+12-y^2=6 y$

Subtracting both sides by $6 y$, we have,

$-4y+12-y^2=0$

Factoring the equation, we get,

$(y+2)(y-6)=0$

$y=-2, y=6$

The values of the variable y are $y=-2, y=6$

At the point $y=-2$, the equation $\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)$ becomes undefined.

Thus, the value of the variable y is $y=6$