How to solve inequality for : $|7x - 9| \ge x +3$ How to solve inequality for : $|7x - 9| \ge x + 3$
There is a $x$ on both side that's make me confused...

We can solve the absolute value inequality by squaring both sides, then solving the resulting quadratic inequality.
\begin{align*}
|7x - 9| & \geq x + 3\\
|7x - 9|^2 & \geq (x + 3)^2\\
(7x - 9)^2 & \geq (x + 3)^2\\
49x^2 - 126x + 81 & \geq x^2 + 6x + 9\\
48x^2 - 132x + 72 & \geq 0\\
4x^2 - 11x + 6 & \geq 0\\
4x^2 - 8x - 3x + 6 & \geq 0\\
4x(x - 2) - 3(x - 2) & \geq 0\\
(4x - 3)(x - 2) & \geq 0
\end{align*}
Equality holds when $x = 3/4$ or $x = 2$.  Since $(4x - 3)(x - 2)$ is continuous, the sign of the product can only change at one of the roots.  We perform a line analysis. 

The sign of the product is the product of the signs of the factors.  Hence $(4x - 3)(x - 2) \geq 0$ when $x \leq 3/4$ or when $x \geq 2$.  Since $|7x - 9| = \sqrt{(7x - 9)^2}$, the steps are reversible.  Hence, the solution set of the inequality $(4x - 3)(x - 2) \geq 0$ is the solution set of the absolute value inequality $|7x - 9| \geq x + 3$.  Therefore, the solution set of the absolute value inequality is 
$$S = \left(-\infty, \frac{3}{4}\right] \cup [2, \infty)$$
