Question

a) 2x (x-1)+(x-3)(x-2)≥(x-3)(x+5)-6+2x²

Answer 1

$2x\left(x-1\right)+\left(x-3\right)\left(x-2\right)\ge \left(x-3\right)\left(x+5\right)-6+2x^2$
$\mathrm{Restar\:}\left(x-3\right)\left(x+5\right)-6+2x^2\mathrm{\:de\:ambos\:lados}:$
$2x\left(x-1\right)+\left(x-3\right)\left(x-2\right)-(\left(x-3\right)\left(x+5\right)-6+2x^2)\ge \left(x-3\right)\left(x+5\right)-6+2x^2-(\left(x-3\right)\left(x+5\right)-6+2x^2)$
$-2x^2+2\left(x-1\right)x+\left(x-3\right)\left(x-2\right)-\left(x-3\right)\left(x+5\right)+6\ge \:0$
$-2x^2+2x^2-2x+x^2-5x+6-x^2-2x+15+6\ge \:0$
$-9x+27\ge \:0$
$\mathrm{Restar\:}27\mathrm{\:de\:ambos\:lados}:$
$-9x+27-27\ge \:0-27$
$-9x\ge \:-27$
$\mathrm{Multiplicar\:ambos\:lados\:por\:-1\:\left(invierte\:la\:desigualdad\right)}:$
$\left(-9x\right)\left(-1\right)\le \left(-27\right)\left(-1\right)$
$9x\le \:27$
$\mathrm{Dividir\:ambos\:lados\:entre\:}9:$
$\frac{9x}{9}\le \frac{27}{9}$
$x\le \:3$
$\begin{bmatrix}\mathrm{Solucion:}\:&\:x\le \:3\:\\ \:\mathrm{Notacion\:intervalo}&\:(-\infty \:,\:3]\end{bmatrix}$