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

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).

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

$4y^2-6y+y-3+9+3y=0$

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

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

Using quadratic formula:

$y=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

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

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

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

Therefore the values of y are $y=\frac{1\pm i\sqrt{23}}{4}$.