Answer
SOLUTION
$\frac{-x+16}{x-5}\leq0$
INTERVALS
$(-\infty,5)$$[16,\infty)$
GRAPH
the graph can be a real number line with the numbers 5 and 16. You should make an open circle in 5 and a line pointing to $-\infty$ and put a close circle in 16 and a line poinying to $\infty$.
Work Step by Step
SOLUTION
$\frac{2x+1}{x-5}\leq3$
$\frac{2x+1}{x-5}-3\leq0$
$\frac{2x+1}{x-5}\frac{-3(x-5)}{x-5}\leq0$
$\frac{2x+1}{x-5}-\frac{-3x+15}{x-5}\leq0$
$\frac{-x+16}{x-5}\leq0$
If we analyze the inequality we know that the answer has to be equal or less than 0.
First we should look at the solutions that the inequality is giving us.
If the numerator of an equation is 0 the result is 0 so we should solve the numerator to know one solution.
$$-x+16=0$$$$x=16$$
In this case we know 16 is a solution because it will give us 0 which the inequality allows
Also the denumerator can’t be equal to 0 becuase it would be undefined. So we have to solve it
$$x-5=0$$$$x=5$$
In this case we know that 5 can’t be a solution because it will make the inequality undefined.
INTERVAL
You should analyze what interval do you have and then test some values between that interval.
$(-\infty,5)$ $(5,16]$ $[16,\infty)$
$(-\infty,5)$=$y(3)=\frac{-3+16}{3-5}\leq0$ $y(3)=-6,5$
This interval is a solution because the reslut is less than 0.
$(5,16]$=$y(8)=\frac{-8+16}{8-5}\leq0$=$y(8)=2.6$
This interval is not a solution because the result is greater tha 0.
$[16,\infty)$= $y(20)=\frac{-20+16}{20-5}\leq0$ = $y(20)=-0.27$
This interval is a solution because the result is less than or equal 0.
the solution is
$(-\infty,5)$$[16,\infty)$
GRAPH
the graph can be a real number line with the numbers 5 and 16. You should make an open circle in 5 and a line pointing to $-\infty$ and put a close circle in 16 and a line poinying to $\infty$.
