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) $\frac{y^{2}+4}{3 y^{2}+7}=\frac{1}{2}$
$($ i $) \frac{x^{2}-9}{5+x^{2}}=\frac{-5}{9}$
or $9 \mathrm{x}^{2}-81=-25-5 \mathrm{x}^{2}[$ After c ross multiplication $]$
or $9 \mathrm{x}^{2}+5 \mathrm{x}^{2}=-25+81$
or $14 \mathrm{x}^{2}=56$
or $\mathrm{x}^{2}=\frac{56}{14}$
or $x^{2}=4=2^{2}$
or $\mathrm{x}=2$
Thus, $x=2$ is the solution of the given equation.
Check:
Substituting $x=2$ in the given equation, we get:
L.H.S. $=\frac{2^{2}-9}{5+2^{2}}=\frac{4-9}{5+4}=\frac{-5}{9}$
R.H.S. $=\frac{-5}{9}$
$\therefore$ L.H.S. $=$ R.H.S. for $\mathrm{x}=2$.
(ii) $\frac{\mathrm{y}^{2}+4}{3 \mathrm{y}^{2}+7}=\frac{1}{2}$
or $3 \mathrm{y}^{2}+7=2 \mathrm{y}^{2}+8[$ After $c$ ross multipl ication $]$
or $3 y^{2}-2 y^{2}=8-7$
or $\mathrm{y}^{2}=1$
or $\mathrm{y}=1$
Thus, $\mathrm{y}=1$ is the solution of the given equation.
Check :
Substituting $\mathrm{y}=1$ in the given equation, we get:
L.H.S. $=\frac{1^{2}+4}{3(1)^{2}+7}=\frac{5}{10}=\frac{1}{2}$
R.H.S. $=\frac{1}{2}$
$\therefore$ L.H.S. $=$ R.H.S. for $\mathrm{y}=1$.
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