Solve:
$\frac{y^{2} + 4}{3 y^{2} + 7} = \frac{1}{2}$

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Solution

Given,
$\frac{y^{2} + 4}{3 y^{2} + 7} = \frac{1}{2}$
or, $2 y^{2} + 8 = 3 y^{2} + 7$
or, $y^{2} = 1$
or, $y = \pm 1$ .

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Similar questions

Solve the following equations and verify your answer :

Find a positive value of x for which the given equation is satisfied :

(i) $\frac{x^{2} - 9}{5 + x^{2}} = - \frac{5}{9}$

(ii) $\frac{y^{2} + 4}{3 y^{2} + 7} = \frac{1}{2}$

\frac{2}{3} y^{3} - \frac{1}{3} y^{2} + \frac{5}{6} y + 7

\frac{4}{3} y^{3} - \frac{2}{3} y^{2} + \frac{1}{y} + 3

7 y = - 3 y^{2} - 4

3 y^{2} - 118 y + 16 = 3 y^{2} + 2 y - 4

12 y

Question 1 (vi)

Simplify combining like terms:

$(3 y^{2} + 5 y - 4) - (8 y - y^{2} - 4)$

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