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Solve |2x+1| <= |x-3| for x

  1. Mar 8, 2007 #1


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    \left| {2x + 1} \right| \ge \left| {x - 3} \right|

    Now, i must solve for x. I could easily do it by graphing it on my calculator, and get an answer of [tex]x \le - 4,x \ge \frac{2}{3}[/tex].

    How can i do it algebraically?

    Thanks in advance,
    Last edited: Mar 8, 2007
  2. jcsd
  3. Mar 8, 2007 #2


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    You could start by finding the points where |2x+1| = |x+3|

    You can get rid of the absolute values by squaring each side.
  4. Mar 8, 2007 #3


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    Agreed. The best way to solve most inequalities is to solve the associated equation first. For continuous functions such as absolute value, the places where they are equal are the only places where the inequality can change so you need check only one point in each interval between solutions of the inequality to see whether the inequality is true or false there.

    Another way of solving an absolute value equation is to consider whether the quantity inside is positive or negative. If 2x+1 and x+3 are both negative or both positive then 2x+1= x+ 3. If they are of different sign then 2x+ 1= -(x_ 3). After solving check to make sure the signs are right for that point.
  5. Mar 8, 2007 #4


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    ok solving the equation, i get x=2/3, x=-4. So would i pick any 'test' value in between these points, and find weather the inequality is true or false?

    Using zero, the inequality is false. So that would mean that x is outside of the solutions, less than -4, but greater than 2/3? Is that the reasoning i should be using to attack these problems?
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