Question

|4x − 1| < 3x + 2

Решить неравенства

Answer 1

Ответ: x∈($-\frac{1}{7}$; 3)

Решение:

$|4x - 1| < 3x + 2\\\\\left \{ {{4x-1 < 3x + 2} \atop {4x - 1 > -3x - 2}} \right. \\\\\left \{ {{4x-3x < 2+1} \atop {4x + 3x > - 2 + 1}} \right.\\\\\left \{ {{x<3} \atop {7x > - 1}} \right.\\\\\left \{ {{x<3} \atop {x > -\frac{1}{7}}} \right.$

x∈($-\frac{1}{7}$; 3)

Answer 2

$|4x-1|<3x+2\\\\ \left \{ {{4x-1<3x+2} \atop {4x-1>-3x-2}} \right.\\\\\left \{ {{4x-3x<2+1} \atop {4x+3x>-2+1}} \right.\\\\\left \{ {{x<3} \atop {7x>-1}} \right.\\\\\left \{ {{x<3} \atop {x>-\frac{1}{7} }} \right. \\\\Otvet:\boxed{ x\in(-\frac{1}{7};3)}$