Modulus Inequality

Question

Solve the inequality $$2|x-3| > |3x+1|$$ Is sketching the only way I can solve ALL modulus equations and inequalities? Does an algebraic technique work for all modulus equations and inequalities?

Answer 1

For generic absolute value inequalities, especially those with multiple summations, you can always split it up into regions where you know the value of the absolute value.

As an example, using the above equality, the regions are where $ x \leq -\frac{1}{3}$, $- \frac{1}{3} \leq x \leq 3$ and $ 3 \leq x $.

Case 1: $ x \leq -\frac{1}{3}$
We have $-2 ( x-3) > -(3x+1) \Rightarrow x> -7$.
Combined with the region that we're in, we have $ -7 < x \leq - \frac{1}{3} $.

Case 2: $- \frac{1}{3} \leq x \leq 3$
We have $-2(x-3) > (3x+1) \Rightarrow 5 > 5x \Rightarrow 1 > x$
Combined with the region that we're in, we have $ - \frac{1}{3} \leq x < 1$.

Case 3: $ 3 \leq x $.
We have $2(x-3) > (3x+1) \Rightarrow -7 > x$.
Combined with the region that we're in, there are no solution.

Combining all the regions, we get $ -7 < x < 1$.

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Alternative approach:

I would not advise this approach, unless you absolutely understand absolute values and WHY this works. In this special inequality, we can simply square everything, since

$$2|x-3| > |3x+1| \Leftrightarrow [2(x-3)]^2 > (3x+1)^2$$

This is equivalent to $0 > 5(x-1)(x+7) \Rightarrow -7 < x < 1$.