Question

What is the answer?
|3x-2|<=|2x-5|

Answer 1

Sayan, if you graph the two you will get two 'v' shapes, and you're looking for the area in which the |3x-2| line is below the |2x-5| line.

You can find where they intersect to find the bounds on x for which there are solutions

Answer 2

Remember that \[|x|=\sqrt{x^2}\]
so we have
\[\sqrt{(3x-2)^2} \le \sqrt{(2x-5)^2}\]
squaring both sides gives:
\[(3x-2)^2 \le (2x-5)^2\]
\[9x^2-12x+4 \le 4x^2-20x+25\]
\[5x^2+8x-21 \le 0\]
\[5x^2+15x-7x-21 \le 0\]
\[5x(x+3)-7(x+3) \le 0\]
\[(x+3)(5x-7) \le 0\]
so =0 when x=-3 and x=7/5

now i like to to make a number line and test where the inequality is true

----------|---------------------|--------------------------
-3 7/5

so choose a number before -3 like -4
plug in -4 to see if inequality holds

now plug in 0 to see if inequality holds (note you chould have chose any number btw -3 and 7/5)


now last interval: choose a number afterr 7/5 and see if the inequality holds

Answer 3

|3(-4)-2|<|2(-4)-5| does not hold so (-inf,-3) is not a part of our solution
|3(0)-2|<|2(0)-5| does hold so (-3,7/5) is a part of our solution
|3(3)-2|<|2(3)-5| does not hold so (7/5, inf) is not a part our solution
so the solution is the set: (-3,7/5)