# Thread: Absolute Value equation with three terms

1. ## 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.

2. ## 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 $x \ge \dfrac{3}{2}$.

3. ## Re: Absolute Value equation with three terms

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

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

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

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?

4. ## Re: Absolute Value equation with three terms

Originally Posted by slothmab
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 $x$).

5. ## Re: Absolute Value equation with three terms

Originally Posted by slothmab
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 ...