Question

$y { }^{2} + 4 \div 3y {}^{2} + 7 = 1 \div 2$
Find y.

Answer 1

hope this will help you

Answer 2

$\frac{ {y}^{2} + 4}{ {3y}^{2} + 7} = \frac{1}{2} \\ = > 2 {y}^{2} + 8 = 3 {y}^{2} + 7 \\ = > - {y}^{2} = - 1 \\ = > y = + - \sqrt{1}$
y= ± 1

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Amrit⭐