Question

please answer the questions that I have answered​

Answer 1

Answer:

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Answer 2

$\frac{1}{y - 1} - \frac{2}{y + 3} = 0 \\ taking \: lcm \\ \frac{1 \times (y + 3)}{(y - 1) \times (y + 3)} - \frac{2 \times (y - 1)}{(y + 3) \times (y - 1)} = 0 \\ \frac{(y + 3 )-2 \times (y - 1) }{(y - 1) \times (y + 3)} = 0 \\ \frac{y + 3 - 2y + 1}{ {y}^{2} + 3y - y - 3 } = 0 \\ \frac{(y - 2y) + 3 + 1}{ {y}^{2} + 2y - 3} = 0 \\ \frac{ - 2y + 4}{ {y}^{2} + 2y - 3 } = 0 \\ - 2y + 4 = 0 \times ( {y}^{2} + 2y - 3) \\ - 2y + 4 = 0 \\ - 2y = - 4 \\ y = \frac{ - 4}{ - 2} \\ y = 2$