Question

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

\[\frac{y^2 + 4}{3 y^2 + 7} = \frac{1}{2}\]

Answer 1

\[\frac{y^2 + 4}{3 y^2 + 7} = \frac{1}{2}\]
\[\text{ or }3 y^2 + 7 = 2 y^2 + 8 [\text{ After cross multiplication }]\]
\[\text{ or }3 y^2 - 2 y^2 = 8 - 7\]
\[\text{ or }y^2 = 1\]
\[\text{ or }y = 1\]
\[\text{ Thus, }y = 1\text{ is the solution of the given equation . }\]
\[\text{ Check: }\]
\[\text{ Substituting }y = 1 \text{ in the given equation, we get: }\]
\[\text{ L . H . S .} = \frac{1^2 + 4}{3(1 )^2 + 7} = \frac{5}{10} = \frac{1}{2}\]
\[\text{ R . H . S . }= \frac{1}{2}\]
\[ \therefore\text{ L . H . S . = R . H . S . for }y = 1 .\]