Question

(x-7)(x-15)<0,(x-2)(x-8)(x-12)>=0 find common solution from it

Answer 1

(x - 7)(x - 15) ≤ 0 , (x - 2)(x -8)(x - 12) ≥ 0

find solution for (x - 7)(x - 15) < 0
first replace inequality sign by equal
e.g., (x - 7)(x - 15) = 0 ,
x = 7, 15
now, put 7 and 15 in number line .
And apply the condition of inequality ,
we get 7 < x < 15
Similarly find the solutions of (x - 2)(x - 8)(x - 12) ≥ 0
replace inequality by equal
And we get x = 2, 8, 12 now put it in number line.
apply condition we get , 2 ≤ x ≤ 8 or, x ≥ 12

Now, take common solution of it using all numbers { e.g., 2, 7, 8, 12, 15} in number line .
see attachment , we get the solution x ∈(7, 8] ∪ [12, 15)

Answer 2

Any question asking for a common solution requires solving it using wavy curves, a tool, to make stuff easier.

For example,

Equation: $(x- \alpha )(x- \beta ) \leq 0$ will have a wavy curve like the first attachment below.
The convention is that the upper loop in a wave of the wavy curve represents where the equation will generate a positive result and vice versa.
Remember, the wavy curve is a number line representation of values, it's NOT a graph of the given equation.

Now, let's draw the wavy curves for the given equations.

$(x-7)(x-15)\ \textless \ 0$ will have the wavy curve as attached in the second attachment below.
Let's take a sample value >15, let's take 16.
The given equation with x=16 will yield (16-7)(16-15) = 9 * 1 = 9, positive.
In 7<x<15, let's take x=10. The equation will yield (10-7)(10-15) = 3 * -5 = -15, negative.
For values less than x=7, let's take x=6, the equation is yielding (6-7)(6-15) = -1* -9 = 9, positive.

Clearly, we need to take the <0 part for the required range of the inequality.
According to the first equation, x ∈ (7, 15)                                                        ...(i)

The second equation $(x-2)(x-8)(x-12) \geq 0$ will give a wavy curve as represented by the third attachment.

One can take sample values to check a wavy curve.

Clearly, x∈ [2, 8] ∪ [12, ∞)                                                                                   ...(ii)

The region covered by both (i) and (ii) shall be,
x ∈ (7, 8] ∪ [12, 15)

And that, shall be the answer!