## College Algebra (10th Edition)

$\left\{x\ |\ x \lt -11 \right\}$ or $\left(-\infty,-11\right)$
$(x-1)(x+1) \gt (x-3)(x+4)$ $x^{2}-1 \gt x^{2}+x-12 \qquad$ ... add $1-x-x^{2}$ $-x \gt -11\qquad$ ... multiply with $-1$ (negative) ... the inequality changes direction ... $x \lt -11$ Solution set: $\left\{x\ | x \lt -11 \right\}$ or $\left(-\infty,-11\right)$