Question

The number of solution of integer which satisfies this inequality $\frac{x^2-1}{2x+5}<3$ is (x>0)

Answer 1

$\frac{x^2-1}{2x-3}<3$

$x^2-1<6x+15$

$x^2-6x-16<0$

$x^2-8x+2x-16<0$

$x(x-8)+2(x-8)<0$

$(x+2)(x-8)<0$

so, $x+2<0$

$x<-2$

or $x-8<0$

$x<8$

No. of solutions lies from $(-\infty ,-2)\cup (-\infty ,8)$

$⟹(-\infty ,8)$