Question

Resolver las siguientes inecuaciones​

Answer 1

Respuesta:

$2x-1\leq x-3(x-1)\\2x-1\leq x-3x+3\\2x-x+3x\leq 3+1\\4x\leq 4\\x\leq 4/4\\x=1$                                $5x-7\leq 2x+\frac{1}{2}\\5x-2x\leq \frac{1}{2} +7\\3x\leq \frac{15}{2} \\x\leq \frac{15}{2}/3\\x\leq \frac{5}{2}$

$x-2(x-3)<-2\\x-2x+6<-2\\-x<-8\\x<8$                                    $2-\frac{3x+4}{5}\leq 4x-\frac{1+2x}{3} \\2-\frac{3}{5}x-\frac{4}{5}\leq 4x-\frac{1}{3}-\frac{2}{3}x\\-\frac{3}{5}x-4x+\frac{2}{3}x \leq -\frac{1}{3}-2+\frac{4}{5} \\\frac{59}{15}x\leq -\frac{23}{15}\\x\leq\frac{23}{15}/-\frac{59}{15}\\x\leq \frac{23}{59}$

$3x+2<10+3(x-2)\\3x<10+3x-6-2\\3x-3x<2\\0x<2\\x<2/0\\x<No hay c.s$                            $\frac{x}{2} -\frac{x-3}{3}\geq 1\\ \frac{x}{2} -\frac{x}{3}+1\geq 1\\\frac{x}{6} \geq 1-1\\x\geq 0*6\\x\geq 0$

$3(2x+5)-4(x-1)>0\\6x+15-4x+4>0\\2x>-19\\x>-19/2$                      $\frac{x-4}{4}+1<\frac{x+4}{8} \\\frac{x}{4} -1+1<\frac{x}{8}+\frac{1}{2}\\\frac{x}{4} -\frac{x}{8} <\frac{1}{2}+0\\\frac{x}{8}<\frac{1}{2}\\x<\frac{1}{2}(8)\\x<4$

$6x-3(x+5)>7x+4\\6x-3x-15-7x>4\\-4x>19\\x>-19/4$

Espero te sirva. Saludos