Question

A5 jos 2 sus - P3. Sa se rezolve inecuatia 5x-7<5 pe 9+ ​

Answer 1

 

$\displaystyle\\5x-7<\frac{5}{9}\\\\\\5x<\frac{5}{9}+7\\\\\\5x<\frac{5}{9}+\frac{63}{9}\\\\\\5x<\frac{5+63}{9}\\\\\\5x<\frac{68}{9}\\\\\\x<\frac{~~\dfrac{68}{9}~~}{5}\\\\\\x<\dfrac{68}{9\cdot5}\\\\\\x<\frac{68}{45}\\\\\\\boxed{\bf~x\in\left(-\infty,~~\frac{68}{45}\right)}$