Question - Solving Absolute Value Inequalities
mathsolving absolute value inequalitiesinequality casesnon-negative expressionsabsolute value brackets
Solution:
To solve the inequality $$|2x - 3| > |x + 3|$$, you need to consider different cases based on the values of $$x$$ which change the expressions inside the absolute value brackets.There are two main cases to consider for each absolute value expression:1. When the expression inside the absolute value brackets is non-negative;2. When the expression inside the absolute value brackets is negative.Let's deal with these step by step:For $$|2x - 3|$$:- Case 1: $$2x - 3 \geq 0$$, which simplifies to $$x \geq \frac{3}{2}$$.- Case 2: $$2x - 3 < 0$$, which simplifies to $$x < \frac{3}{2}$$.For $$|x + 3|$$:- Case 1: $$x + 3 \geq 0$$, which simplifies to $$x \geq -3$$.- Case 2: $$x + 3 < 0$$, which simplifies to $$x < -3$$.Combining these cases, you have to solve the inequality under the following scenarios:1. $$x \geq \frac{3}{2}$$ and $$x \geq -3$$ (since $$x \geq \frac{3}{2}$$ covers $$x \geq -3$$). This simplifies the inequality to $$2x - 3 > x + 3$$: $$x > 6$$.2. $$x < \frac{3}{2}$$ and $$x \geq -3$$: Here we have to take two subcases because of the range of $$x$$: a. $$ -3 \leq x < \frac{3}{2} $$, therefore the inequality becomes $$3 - 2x > x + 3$$: $$-3x > 0$$, which simplifies to $$x < 0$$. b. $$ x < -3 $$, therefore the inequality becomes $$3 - 2x > 3 - x$$: $$x < 0$$, which always holds true for $$x < -3$$.Putting it all together, there are two intervals for $$x$$ that solve the inequality:- $$x < 0$$ for the range between $$-3$$ and $$\frac{3}{2}$$,- $$x > 6$$.The final solution to the inequality $$|2x - 3| > |x + 3|$$ is:\[x < 0 \text{ or } x > 6.\]
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com