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Thread: Absolute Value equation with three terms

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    Absolute Value equation with three terms

    Can someone help me with how to approach an absolute value equation involving three terms? The equation in my textbook is this
    |x - 1| + |2x - 3| = |3x - 4|

    The textbook gives the answer as x <= 1 and x >= 3/2 with no explanation.
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  2. #2
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    Re: Absolute Value equation with three terms

    $|x-1| = \begin{cases} x-1 & x\ge 1 \\ 1-x & x \le 1\end{cases}$

    $|2x-3| = \begin{cases} 2x-3 & x\ge \dfrac{3}{2} \\ 3-2x & x \le \dfrac{3}{2}\end{cases}$

    $|3x-4| = \begin{cases} 3x-4 & x\ge \dfrac{4}{3} \\ 4-3x & x \le \dfrac{4}{3}\end{cases}$

    So, if $x \ge \dfrac{3}{2}$ then we have $(x-1)+(2x-3) = (3x-4)$ which is true for any $x \ge \dfrac{3}{2}$.
    If $\dfrac{4}{3} \le x \le \dfrac{3}{2}$, then we have:

    $(x-1) + (3-2x) = 2-x = 3x-4$ Solving for $x$ we get $4x=6 \Longrightarrow x = \dfrac{3}{2}$

    If $1\le x \le \dfrac{4}{3}$, then we have:

    $(x-1) + (3-2x) = 4-3x$
    $x = 1$

    Finally, for $x\le 1$ we have:

    $(1-x) + (3-2x) = (4-3x)$ which is true for all $x \le 1$

    If we combine all of the information we got for each interval, we have $x \le 1$ or $x \ge \dfrac{3}{2}$.
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    Re: Absolute Value equation with three terms

    Quote Originally Posted by SlipEternal View Post
    $|x-1| = \begin{cases} x-1 & x\ge 1 \\ 1-x & x \le 1\end{cases}$

    $|2x-3| = \begin{cases} 2x-3 & x\ge \dfrac{3}{2} \\ 3-2x & x \le \dfrac{3}{2}\end{cases}$

    $|3x-4| = \begin{cases} 3x-4 & x\ge \dfrac{4}{3} \\ 4-3x & x \le \dfrac{4}{3}\end{cases}$

    So, if $x \ge \dfrac{3}{2}$ then we have $(x-1)+(2x-3) = (3x-4)$ which is true for any $x \ge \dfrac{3}{2}$.
    If $\dfrac{4}{3} \le x \le \dfrac{3}{2}$, then we have:

    $(x-1) + (3-2x) = 2-x = 3x-4$ Solving for $x$ we get $4x=6 \Longrightarrow x = \dfrac{3}{2}$

    If $1\le x \le \dfrac{4}{3}$, then we have:

    $(x-1) + (3-2x) = 4-3x$
    $x = 1$

    Finally, for $x\le 1$ we have:

    $(1-x) + (3-2x) = (4-3x)$ which is true for all $x \le 1$

    If we combine all of the information we got for each interval, we have $x \le 1$ or $x \ge \dfrac{3}{2}$.
    So if both sides are equal for a given interval then all values in the interval are part of the solution set?
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    Re: Absolute Value equation with three terms

    Quote Originally Posted by slothmab View Post
    So if both sides are equal for a given interval then all values in the interval are part of the solution set?
    The equation simplifies to 1=1, so it is always true for any value in the interval. For example:

    $(x-1)+(2x-3) = (3x-4)$
    $3x-4 = 3x-4$

    Divide both sides by $3x-4$ (since you know $x\ge \dfrac{3}{2}$, this will always be nonzero), and you get 1=1, which is a tautology (it is true no matter what value you use for $x$).
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    Re: Absolute Value equation with three terms

    Quote Originally Posted by slothmab View Post
    So if both sides are equal for a given interval then all values in the interval are part of the solution set?
    note the values on the line graph are the x-values which make each absolute value term zero

    checking the equation at each of these values and in the intervals between them yields ...
    Attached Thumbnails Attached Thumbnails Absolute Value equation with three terms-abs_inequality.jpg  
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