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Math Help - Solving an inequality that has a single variable (i.e. just x, no y)

  1. #1
    Junior Member jonnygill's Avatar
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    Solving an inequality that has a single variable (i.e. just x, no y)

    the equation is...

    |3x-4|\leq{x}

    I first approached this problem by solving for x in both -(3x-4)\leq{x} and (3x-4)\leq{x}

    In both cases, I ended up with an inequality that still had x on one side and an expression that is probably equivalent to the original expression on the other side. This didn't really tell me anything.

    One thing that occurred to me was that it may be true in this case that y=x was one of the barriers.

    So i graphed |3x-4| myself (without the use of a graphing calculator). the parent function is y=|x|. so y=|3x-4| is y=|x| compressed horizontally by a factor of (1/3) and moved to the right 4 units. I do not know how to graph with latex (or if it is even possible), so maybe you could quickly graph this yourself with pen and paper, or just imagine it. basically its a narrower "V" shape moved over to the right 4 units. After graphing this, I then graphed y=x on the same plot and figured that the solution set existed between the two functions (i.e. < y=x & > y=|3x-4|).

    But then I graphed the two equations using a graphing calculator and saw that my graph of y=|3x-4| looked different from the graphing calculator's graph of the same function. Furthermore, the back of the book shows that the solution set is between and including 1 and 2, which is confirmed when looking at the graphing calculators graphs of y=x and y=|3x-4|.

    So i'm like, did i graph y=|3x-4| wrong? but i don't see how i did. Basically, the only difference between my graph of the function and the graphing calculator's is that it seems the calculator moved the graph over (4/3) units to the right and i moved the graph over 4 units to the right.

    This is the longest post, i just thought it would be helpful for you to know how i was thinking about this problem. So, how would i solve this inequality algebraically and why is my graph wrong?
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  2. #2
    MHF Contributor
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    Quote Originally Posted by jonnygill View Post
    the equation is...

    |3x-4|\leq{x}
    You can split into 2 parts :
    - if 3x-4 \leq 0 or x \leq \frac43 then |3x-4|=4-3x then solve 4-3x\leq{x} which gives 4\leq{4x} or x\geq{1} therefore 1 \leq x \leq \frac43

    - if 3x-4 \geq 0 ...

    Quote Originally Posted by jonnygill View Post
    So i graphed |3x-4| myself (without the use of a graphing calculator). the parent function is y=|x|. so y=|3x-4| is y=|x| compressed horizontally by a factor of (1/3) and moved to the right 4 units.
    What you are getting here is |3(x-4)| or |3x-12|
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  3. #3
    A riddle wrapped in an enigma
    masters's Avatar
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    Quote Originally Posted by jonnygill View Post
    the inequality is...

    |3x-4|\leq{x}
    Hi jonnygill,

    Algebraically, this is the solution:

    -x \leq 3x-4 \leq x

    3x-4 \ge -x \text{ and }3x-4 \le x

    4x \ge 4 \text{ and } 2x \le 4

    x \ge 1 \text { and } x \le 2

    Solution set: \{x|1 \le x \le 2\}

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  4. #4
    Junior Member jonnygill's Avatar
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    Quote Originally Posted by running-gag View Post
    You can split into 2 parts :
    - if 3x-4 \leq 0 or x \leq \frac43 then |3x-4|=4-3x then solve 4-3x\leq{x} which gives 4\leq{4x} or x\geq{1} therefore 1 \leq x \leq \frac43

    - if 3x-4 \geq 0 ...

    What you are getting here is |3(x-4)| or |3x-12|
    so i solved for 3x-4\geq{0} and i got x\leq{2}

    combined with your solution we arrive at the solution set. thank you.

    also, thanks for clarifying the mistake i made. In the future, if i'm given an equation like (3x-4)^2 i will factor out the coefficient of x to get it into a more recognizable form. Which, in this case would be [3(x-(4/3))]^2
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  5. #5
    Junior Member jonnygill's Avatar
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    Quote Originally Posted by masters View Post
    Hi jonnygill,

    Algebraically, this is the solution:

    -x \leq 3x-4 \leq x

    3x-4 \ge -x \text{ and }3x-4 \le x

    4x \ge 4 \text{ and } 2x \le 4

    x \ge 1 \text { and } x \le 2

    Solution set: \{x|1 \le x \le 2\}

    Thank you! This was helpful as it was a reiteration of the first response, but in a different form

    PEMDAS has been engrained into my brain to the point where i began to use it in an unhelpful way that led me in circles. with this problem i was shuffling constants back and forth resulting in x always being in terms of x. how silly i feel. -3x+4=x yields x=1. so simple!
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