Question Number 880 by arnav last updated on 11/Apr/15 | ||
$${Solve} \\ $$ $$\frac{{x}+\mathrm{3}}{{x}−\mathrm{3}}<\mathrm{1} \\ $$ | ||
Answered by 123456 last updated on 11/Apr/15 | ||
$$\frac{{x}+\mathrm{3}}{{x}−\mathrm{3}}−\mathrm{1}<\mathrm{0} \\ $$ $$\frac{{x}+\mathrm{3}−{x}+\mathrm{3}}{{x}−\mathrm{3}}<\mathrm{0} \\ $$ $$\frac{\mathrm{6}}{{x}−\mathrm{3}}<\mathrm{0} \\ $$ $${x}−\mathrm{3}<\mathrm{0} \\ $$ $${x}<\mathrm{3} \\ $$ | ||