Question

$\frac{x+2}{5} +\frac{x+3}{6} +\frac{x+4}{7} \leq \frac{x-1}{3} -\frac{x+1}{15}$

Answer 1

Respuesta:

$x\leq -\dfrac{393}{51}$

Explicación paso a paso:

$\dfrac{x+2}{5}+\dfrac{x+3}{6} +\dfrac{x+4}{7}-\dfrac{x-1}{3}+\dfrac{x+1}{15}\leq 0$

$\dfrac{42(x+2)+35(x+3)+30(x+4)-70(x-1)+14(x+1)}{210}\leq 0$

$\dfrac{42x+84+35x+105+30x+120-70x+70+14x+14}{210}\leq 0$

$\dfrac{51x+393}{210}\leq 0$

$51x+393\leq 0$

$51x\leq -393$

$x\leq -\dfrac{393}{51}$