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

Question:

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

Solution:

$($ 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 $\mathrm{y}=1$.