What is the solution of"

Question

asked Feb 2, 2020 46.0k views

1 Answer

Saeven · Answer 1 · 2020-02-06T19:31:27+0000

Answer:

Hence final answer is
$x \leq -3$ or
$2 \leq x<7$

correct choice is A because both ends are open circles.

Explanation:

Given inequality is
$(x^2+x-6)/(x-7)\leq 0$

Setting both numerator and denominator =0 gives:

x^2+x-6=0 , x-7=0

or
(x+3)(x-2)=0 , 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:

$(-\infty,-3)$ , (-3,2), (2,7),
$(7,\infty)$

We pick one number from each interval and plug into original inequality to see if that number satisfies the inequality or not.

Test for
$(-\infty,-3)$ .

Clearly x=-4 belongs to
$(-\infty,-3)$ interval then plug x=-4 into
$(x^2+x-6)/(x-7)\leq 0$

$((-4)^2+(-4)-6)/((-4)-7)\leq 0$

$(6)/(-11)\leq 0$

Which is TRUE.

Hence
$(-\infty,-3)$ belongs to the answer.

Similarly testing other intervals, we get that only
$(-\infty,-3)$ and
(2,7) satisfies the original inequality.

Hence final answer is
$x \leq -3$ or
$2 \leq x<7$

correct choice is A because both ends are open circles.