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$\displaystyle |x| < 5$
$ -5 < x < 5 $
$\displaystyle |x| \geq 3$
$x\leq -3 \quad \mathrm{or} \quad x\geq 3$
$\displaystyle \frac{3x^2+x-7}{x-1} > x+1$
$-2 < x < 1 \quad \mathrm{or} \quad x > \frac{3}{2}$
$\displaystyle \frac{x^2+2x+3}{2x^2+x-6} < 0$
$-2 < x < \frac{3}{2}$
$\displaystyle \frac{x^2+2|x|+3}{2x^2+|x|-6} < 0$
$x<-\frac{3}{2} \quad \mathrm{or} \quad -1 < x < 1 \quad \mathrm{or} \quad x>\frac{3}{2}$
$\displaystyle \mathrm{e}^x = 5-x^2$
$x = -2.211 \quad \mathrm{or} \quad x=1.241$
$\displaystyle \mathrm{e}^x \leq 5-x^2$
$-2.211 \leq x \leq 1.241$
$\displaystyle 2^n - 1 > 5n^2 - 4n$
$n \geq 9, \; n \in \mathbb{Z}$
$x^3 + x^2 - 3x + 1 = 0$
$x=-2.414 \quad \mathrm{or} \quad x=0.414 \quad \mathrm{or} \quad x=1 $
$\displaystyle \begin{align} x+y-z&= -0.8 \\ 2x-y+2z&=9.4 \\ x-5z &= -5 \end{align}$
$x=2.375, \; y=-1.7, \; z=1.475$
$\displaystyle y=\frac{x-1}{x^2+1}.$
Find, algebraically, the set of values that $y$ can take. $\{ y \in \mathbb{R}: \frac{-1-\sqrt{2}}{2} \leq y \leq \frac{-1+\sqrt{2}}{2} \}$
Find, algebraically, the set of values that $y$ can take. $\{ y \in \mathbb{R}: \frac{-1-\sqrt{2}}{2} \leq y \leq \frac{-1+\sqrt{2}}{2} \}$