How to solve this multiple-absolute-value equation using three-region number line?
I can solve it with combination of giving each absolute value a negative sign and leaving it as it is. There are four combinations. The method using regions in number line only requires three combinations instead of four. But it fails. So please help me to solve it using region-in-a number line method. (The answer should be {-6, -2/3})
On
Here's how to solve this using a quadratic formula approach: $$\vert2x+4\vert-\vert3-x\vert=-1$$ $$\implies\vert2x+4\vert-\vert3-x\vert+1=-1+1$$ $$\implies\vert2x+4\vert-\vert3-x\vert+1=0$$ $$\implies\vert2x+4\vert-\vert3-x\vert+\vert3-x\vert+1=\vert3-x\vert$$ $$\implies\vert2x+4\vert+1=\vert3-x\vert$$ $$\therefore\implies\vert2x+4\vert^2+1^2=\vert3-x\vert^2\iff4x^2+16x+16+1=x^2-6x+9$$ $$\implies4x^2-x^2+16x+6x+17-9=x^2-x^2-6x+6x+9-9\iff3x^2+22x+8=0$$ $$\therefore\implies\frac{-22\pm\sqrt{22^2-4(3)(8)}}{6}\iff\frac{-22\pm\sqrt{484-96}}{6}\iff\frac{-22\pm\sqrt{388}}{6}$$ $$\iff\frac{-22\pm2\sqrt{97}}{6}$$ $$\iff-\frac{11}{3}\pm\frac{\sqrt{97}}{3}=x$$ $$x\approx-6\,\text{(although the actual value achieved is closer to}\,-7),\,-\frac{2}{3}\,\text{(although the actual value achieved is closer to}\,-\frac{7}{6})$$ Using the quadratic formula method gives us a result that has a percent error on the left hand side of$\,\approx19.2\%$, and on the right hand side a percent error of$\,\approx72.7\%$, both of which you can see are invalid values if you plug them in ($-6\neq-1\,$and$\,-\frac{1}{6}\neq-1$).
Therefore, the following would be a faster (and more accurate) method:
From Mostafa Ayaz's answer:
Case 1: Let $x\lt-2$ therefore $$\vert2x+4\vert-\vert3-x\vert=-2x-4-(3-x)=-1$$ which yields to valid answer $x=-6$ Case 2: Let $-2\leq x\leq3$ therefore $$\vert2x+4\vert-\vert3-x\vert=2x+4-(3-x)=-1$$ which yields to valid answer $x=-\frac{2}{3}$
$$\vert2x+4\vert-\vert3-x\vert=-1$$ $$\iff\vert2x+4\vert-\vert-x+3\vert=-1$$ $$\implies\vert2x+4\vert-\vert-(x+3)\vert=-1$$ $$\iff\vert2x+4\vert-\vert x-3\vert=-1$$ As you can see, when you plug in $-6$ and $-\frac{2}{3}$ for $x$, we get this: $$\text{1) Plugging in 6 for x}$$ $$\vert2(-6)+4\vert-\vert(-6)-3\vert=-1$$ $$\iff\vert-12+4\vert-\vert-9\vert=-1$$ $$\iff\vert-8\vert-9=-1$$ $$\iff8-9=-1$$ $$\iff\,-1\overset{\checkmark}{=}-1$$ $$\text{2) Now, plugging in}\,-\frac{2}{3}\,\text{for x:}$$ $$\left\vert2\left(-\frac{2}{3}\right)+4\right\vert-\left\vert\left(-\frac{2}{3}\right)-3\right\vert=-1$$ $$\iff\left\vert-\frac{4}{3}+4\right\vert-\left\vert\frac{-2-9}{3}\right\vert=-1$$ $$\iff\frac{8}{3}-\frac{11}{3}=-1$$ $$\iff-\frac{3}{3}=-1$$ $$\iff-1\overset{\checkmark}{=}-1$$ Here's a Desmos graph I made that shows the solutions to your question on a number line from $-8$ to $8$. Hope this helps!
On
Since absolute-value functions are piecewise-defined linear functions, one can also take an approach that looks at the slopes of relevant lines. As remarked by dxiv in a comment, the vertex of the "V"-shapes of these functions occur where the argument of the absolute-value function is zero, so the vertex of $ \ |2x + 4| \ $ lies at $ \ x \ = \ -2 \ $ and that of $ \ |3 - x | \ $ is at $ \ x \ = \ 3 \ \ ; \ $ these define the "boundaries" of the intervals of interest in the "regions" method.
We will work with $ \ |x - 3 | \ $ for convenience, since it is exactly the same as $ \ |3 - x| \ \ . \ $ In the three intervals, we can say that
• for $ \ x \ < \ -2 \ \ , \ \ |2x + 4| \ $ has a slope of -2 and $ \ | x - 3 | \ $ has a slope of -1 ;
• for $ \ -2 \ < \ x \ < \ 3 \ \ , \ \ |2x + 4| \ $ has a slope of +2 and $ \ | x - 3 | \ $ has a slope of -1 ; and
• for $ \ x \ > \ 3 \ \ , \ \ |2x + 4| \ $ has a slope of +2 and $ \ | x - 3 | \ $ has a slope of -1 .
It will also be convenient to re-arrange the equation $ \ |2x + 4| \ - \ |3 - x| \ = \ -1 \ \ $ as $ \ |x - 3| \ = \ 1 + |2x - 4| \ \ . \ $ We now want to find when the functions on each side of the equation are equal.
At $ \ x \ = \ -2 \ \ , \ $ we have $ \ |x - 3| \ = \ 5 \ $ and $ \ 1 + |2x + 4| \ = \ 1 \ \ , \ $ and at $ \ x \ = \ 3 \ \ , \ $ we have $ \ |x - 3| \ = \ 0 \ $ and $ \ 1 + |2x + 4| \ = \ 11 \ \ . \ $ We will now observe what happens in the different intervals.
At $ \ x \ = \ -2 \ \ , \ \ 1 + |2x + 4| \ $ has the smaller function value but the "steeper" slope "on the left", so for each unit $ \ x \ $ decreases, it becomes "closer" to the value of $ \ | x - 3 | \ $ by one unit; since the difference in the function values at $ \ x \ = \ -2 \ $ is 4 units, the functions will be equal 4 units "to the left" of $ \ x \ = \ -2 \ \ , \ $ thus at $ \ \mathbf{x \ = \ -6 }\ \ . $ When $ \ x \ $ "moves to the right" fron $ \ x \ = \ -2 \ $ instead, $ \ 1 + |2x + 4| \ $ is still 4 units smaller than $ \ |x - 3| \ \ , $ but the difference in slopes now leads $ \ 1 + |2x + 4| \ $ to get $ \ \mathbf{3} \ $ units closer for each unit $ \ x \ $ increases; the two functions will be equal $ \ \frac43 \ $ units "to the right" of $ \ x \ = \ -2 \ \ , \ $ so at $ \ \mathbf{x \ = \ -\frac23 } \ \ . $ Since we are dealing with linear functions in this discussion, there will be at most one intersection of the function curves in each interval.
Finally, at $ \ x \ = \ 3 \ \ , \ \ 1 + |2x + 4| \ \ $ has both the larger function value and the larger positive slope, so it will never equal $ \ |x - 3| \ \ $ in the interval $ \ x \ > \ 3 \ \ . \ $ Therefore, we have found the only two solutions to the given equation. (We can of course write this out in formal notation, but these functions have reasonably simple behavior and can be discussed usefully in the manner followed here.)
Case 1: Let $x<-2$ therefore $$|2x+4|-|3-x|=-2x-4-(3-x)=-1$$which yields to valid answer $x=-6$
Case 2: Let $-2\le x\le3$ therefore $$|2x+4|-|3-x|=2x+4-(3-x)=-1$$which yields to valid answer $x=-\dfrac{2}{3}$
Case 3: Let $x>3$ therefore $$|2x+4|-|3-x|=2x+4+(3-x)=-1$$which yields to invalid answer $x=-8$