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...
 A: Separate the cases. You can notice that the value in absolutes reaches a zero at $x=\frac97.$
Case 1: If $x\geq \frac97$, then the inequality becomes equivalent to $$7x-9\geq x+3$$ which can be simplified to $$6x\geq 12\\ x\geq 2$$
This means that, if $x\geq \frac97$, then the inequality holds if and only if $x\geq 2$. Therefore, $x\geq 2$ is part of the solution.
Case 2: If $x<\frac97$, the inequality becomes
$$9-7x\geq x+3\\
8x\leq 6\\
x\leq \frac34$$
So the whole solution is $x\in (-\infty, \frac34]\cup[2,\infty)$
A: It is equivalent to:
$$7x-9 \geq x+3 \text{ or } 7x-9 \leq -x-3 \\ \Rightarrow 6x \geq 12 \text{ or } 8x \leq 6 \\ \Rightarrow x \geq 2 \text{ or } x \leq \frac{6}{8} $$
A: HINT: Look towards a graphical solution by plotting both sides of the inequality as standalone objects on the same axes and then look for the domains that satisfy the inequality. 
This is just my opinion, I find it much more beautiful than jumping straight to the case. 
A: 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 
$$S = \left(-\infty, \frac{3}{4}\right] \cup [2, \infty)$$
