Solution assignment 09 Linear inequalities

Solve:

$3(x-1)+2x(x-3)\leq{x(x+1)+x(x-1)}$

We write this inequality as:

$3x-3+2x^2-6x\leq{x^2+x+x^2-x}$

$3x-3-6x\leq0$

$-3x\leq3$

We divide both sides of this inequality by the negative number $-3$ and thus the 'less than' becomes a 'greater than' sign:

$x\geq-1$

Note that the inequality seemed to be a quadratic inequality but this did not appear the case.