Answer
SOLUTION
$\frac{(2+x)(2-x)}{x(x-1)} \geq0$
INTERVALS
$ [-2,0) (1,2] $
GRAPH
Put four numbers in the real number line -2, 0, 1 and 2. 0 and 1 should be with open circle. -2 and 2 with close circlel. Connect -2 and 0. Also, connect 1 and 2.
Work Step by Step
$\frac{3}{x-1}-\frac{4}{x}\geq1$
$\frac{3}{x-1}-\frac{4}{x}-1\geq0$
$\frac{3x-4(x-1)-1(x^{2}-x)}{x(x-1)}\geq0$
$\frac{3x-4x+4-x^{2}+x}{x(x-1)} \geq0$
$\frac{4-x^{2}}{x(x-1)}\geq0$
$\frac{(2+x)(2-x)}{x(x-1)} \geq0$
Key numbers
NUMERATOR: x=2 x=-2
DENOMINATOR x=0 x=1
INTERVALS
$(-\infty,-2] [-2,0) (0,1) (1,2] [2,\infty)$
test the values
$(-\infty,-2]= y(-3)=\frac{(2-3)(2+3)}{-3(-3-1)} \geq0= y(-3)=-0.41$
no solution
[-2,0)= $y(-1)=\frac{(2-1)(2+1)}{-1(-1-1)} \geq0= y(-1)=1.5$
Solution
(0,1)= y(0.5)=\frac{(2+0.5)(2-0.5)}{0.5(0.5-1)} \geq0= y(0.5)=-15$
No solution
(1,2]= y(1.5)=\frac{2+1.5)(2-1.5)}{1.5(1.5-1)} \geq0= y(1.5)=2.3$
Solution
$[2,\infty)$= $y(4)=\frac{(2+4)(2-4)}{2(2-1)} \geq0= Y(4)=-8$
No solution
SOLUTION
$ [-2,0) (1,2] $
GRAPH
Put four numbers in the real number line -2, 0, 1 and 2. 0 and 1 should be with open circle. -2 and 2 with close circlel. Connect -2 and 0. Also, connect 1 and 2.
