Can someone please help me with this problem, I can't seem to figure it out. Thank you in advance.
8/x +1/(x+1) < 9/(x+2)
8/x +1/(x+1) -9/(x+2) < 0
Combine the 3 fractions into one fraction only,
the common denominator is the product of the 3 denominators,
[8(x+1)(x+2) +1(x)(x+2) -9(x)(x+1)] / [x(x+1)(x+2)] < 0
[8(x^2 +3x +2) +x^2 +2x -9(x^2 +x)] / [x(x+1)(x+2)] < 0
[8x^2 +24x +16 +x^2 +2x -9x^2 -9x] / [x(x+1)(x+2)] < 0
[17x +16] / [x(x+1)(x+2)] < 0
Therefore,
A = 17
B = 16
The interval where the inequality is defined:
Solve for x.
17x +16 < 0
17x < -16
x < -16/17 ---no reversing of sense because it dividing by positive 17 only.
Therefore, the interval is (-infinity,-16/17)
So far so good, but ticbol missed a few intervals in the solution.
You want to find "critical points" for the solution. These are points where the numerator or denominator are 0. So we have x = -2, -1, -16/17, and 0.
Now you want to break the real line into intervals according to these critical points and test each interval in the inequality:
$\displaystyle ( -\infty, -2);~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} > 0$ No.
$\displaystyle ( -2, -1);~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} < 0$ Yes.
$\displaystyle \left ( -1, -\frac{16}{17} \right );~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} > 0$ No.
$\displaystyle \left ( -\frac{16}{17}, 0 \right );~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} < 0$ Yes.
$\displaystyle ( 0, \infty );~ \implies ~ \frac{17x + 16}{x(x + 1)(x + 2)} > 0$ No.
So the solution set is the union of the two intervals:
$\displaystyle ( -2, -1) , ~ \left ( -\frac{16}{17}, 0 \right )$
-Dan