Question

Solve for y: [2y/ y-3 ] + [ 1/2y-3] = -9-3y/(y-3)(2y-3).

Answer 1

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Answer 2

Answer:

The values of y are $\frac{1+i\sqrt{23}}{4}$ and $\frac{1-i\sqrt{23}}{4}$.

Step-by-step explanation:

The given equation is

$\frac{2y}{y-3}+\frac{1}{2y-3}=\frac{-9-3y}{(y-3)(2y-3)}$

Multiply both sides by (y-3)(2y-3).

$2y(2y-3)+(y-3)=-9-3y$

On simplification we get

$4y^2-2y+6=0$

Divide both sides by 2.

$2y^2-y+3=0$

Using quadratic formula

$y=\frac{-(-1)\pm \sqrt{(-1)^2-4(2)(3)}}{2(2)}$

$y=\frac{1\pm \sqrt{-23}}{4}$

$y=\frac{1\pm i\sqrt{23}}{4}$

Therefore the values of y are $\frac{1+i\sqrt{23}}{4}$ and $\frac{1-i\sqrt{23}}{4}$.