Question

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

Answer 1

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

By cross multiplication :

$9(x^2-9)=-5(5+x^2)\Rightarrow 9x^2-81=-25-5x^2\Rightarrow 9x^2+5x^2=-25+81\Rightarrow 14x^2=56\Rightarrow x^2=\frac{56}{14}=4$
$\therefore x=\pm \sqrt{4}=\pm 2$
$\because$ We have to take only positive value of x
$\therefore x=2$

(ii) $\frac{y^2+4}{3y^2+7}=\frac{1}{2}$
By cross multiplication
$2(y^2+4)=3y^2+7\Rightarrow 2y^2+8=3y^2+7\Rightarrow 3y^2-2y^2=8-7\Rightarrow y^2=1\therefore y=\pm \sqrt{1}=\pm 1$
$\because$ We have to take only positive value of y
$\therefore y=1$