Answer:
Hence final answer is [tex]x \leq -3[/tex] or [tex]2 \leq x<7[/tex]
correct choice is A because both ends are open circles.
Step-by-step explanation:
Given inequality is [tex]\frac{x^2+x-6}{x-7}\leq 0[/tex]
Setting both numerator and denominator =0 gives:
[tex]x^2+x-6=0[/tex], x-7=0
or [tex](x+3)(x-2)=0[/tex], x-7=0
or x+3=0, x-2=0, x-7=0
or x=-3, x=2, x=7
Using these critical points, we can divide number line into four sets:
[tex](-\infty,-3)[/tex], (-3,2), (2,7), [tex](7,\infty)[/tex]
We pick one number from each interval and plug into original inequality to see if that number satisfies the inequality or not.
Test for [tex](-\infty,-3)[/tex].
Clearly x=-4 belongs to [tex](-\infty,-3)[/tex] interval then plug x=-4 into [tex]\frac{x^2+x-6}{x-7}\leq 0[/tex]
[tex]\frac{(-4)^2+(-4)-6}{(-4)-7}\leq 0[/tex]
[tex]\frac{6}{-11}\leq 0[/tex]
Which is TRUE.
Hence [tex](-\infty,-3)[/tex] belongs to the answer.
Similarly testing other intervals, we get that only [tex](-\infty,-3)[/tex] and [tex](2,7)[/tex] satisfies the original inequality.
Hence final answer is [tex]x \leq -3[/tex] or [tex]2 \leq x<7[/tex]
correct choice is A because both ends are open circles.
