Answer
SOLUTION
$\frac{2x}{3-x}\geq0$
INTERVAL
[0,3)
GRAPH
In a real number line write 0 and 3. The 0 should be with a close circle and the 3 with an open circle
Work Step by Step
SOLUTION
$\frac{3+x}{3-x}\geq1$
$\frac{3+x}{3-x}-1\geq0$
$\frac{3+x}{3-x}-\frac{1(3-x)}{3-x}\geq0$
$\frac{3+x}{3-x}-\frac{(3-x)}{3-x}\geq0$
$\frac{3+x-3+x}{3-x}\geq0$
$\frac{2x}{3-x}\geq0$
Solutions from the inequality.
For the numerator to be 0 x=0.
For the denominator to be 0 it has to be x=3. In this case this number is not allowed because it will make the fraction undefined. But we can use it to construct the intervals.
INTERVALS
$(-\infty,0] [0,3) (3,\infty)$
In order to know whihc one is the soltion you have to test the values.
$(-\infty,0]$ = $y(-2)=\frac{2(-2)}{3-(-2)}\geq0$ = $y(-2)=0.8$
This interval is not the solution because the result is less than 0 and it should be greater than or equal to 0.
[0,3) = $y(1)=\frac{2(1)}{3-1}\geq0$ = $y(1)= 1$
This interval is a solution because the result is greater or equal to 0.
$(3,\infty)$ = $y(4)\frac{2(4)}{3-4}$ = $y(4)=-8$
This interval is not a solution because the result is less than 0.
The correct interval is [0,3)
GRAPH
In a real number line write 0 and 3. The 0 should be with a close circle and the 3 with an open circle. Then connect them. Indicating that 3 should not be used.
