## Precalculus: Mathematics for Calculus, 7th Edition

SOLUTION $\frac{(2+x)(3-x)}{x(x-1)} \geq0$ INTERVAL [-2,0) (1,3] GRAPH in the real number line put four numbers. Put an open circle in 0 and 1. Put a close circle in -2 and 3. Connect the -2 and 0 and connect the 1 and three.
$\frac{6}{x-1}-\frac{6}{x} \geq1$ $\frac{6}{x-1}-\frac{6}{x}-1 \geq0$ $\frac{6(x)-6(x-1)-1(x^{2}-1)}{x(x-1)} \geq0$ $\frac{6x-6x+6-x^{2}+x}{x(x-1)} \geq0$ $\frac{x-x^{2}+6}{x(x-1)} \geq0$ $\frac{(2+x)(3-x)}{x(x-1)} \geq0$ You need to see what are the solutions of the inequality. Key numbers: NUMERATOR x=-2 x=3 DENOMINATOR x=0 x=1 INTERVALS $(- \infty,-2] [-2,0) (0,1) (1,3] [3,\infty)$ test the values to know if there is a solution. $(- \infty,-2]$ $y(-4)=\frac{(2-4)(3+4)}{-4(-4-1)} \geq0= y(-4)=-0.7$ No solution. $[-2,0)$ $y(-1)=\frac{(2-1)(3+1)}{-1(-1-1)} \geq0= y(-1)=2$ Solution. $(0,1)$ $y(0.5)=\frac{(2+0.5)(3-0.5)}{0.5(0.5-1)} \geq0= y(0.5)=-25$ No solution $(1,3]$ $y(2)=\frac{(2+2)(3-2)}{2(2-1)} \geq0= y(2)=2$ Solution. $[3,\infty)$ $y(6)=\frac{(2+6)(3-6)}{6(6-1)} \geq0= y(6)=-0.8$ No solution. CORRECT [-2,0) (1,3] GRAPH in the real number line put four numbers. Put an open circle in 0 and 1. Put a close circle in -2 and 3. Connect the -2 and 0 and connect the 1 and three.