# Solving equations with multiple absolute values

I know the method of solving the equation like this $|2x+1|=|3x+9|$ but the problem is if the same equation would be like this $| 2x+1 | = x | 3x+9 |$, how can I solve this?

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Again, the point that has been raised to you in your past few questions: treat the absolute value as a function with two cases. $|2x+1|=\mathrm{something}$ can mean either of $2x+1=\mathrm{something}$ or $-(2x+1)=\mathrm{something}$. You will have four "cases" since you have two absolute values; work from that. – J. M. Aug 23 '10 at 12:28
actuall we can solve the first example as 2x+1=3x+9 or 2x+1=-(3x+1) we have only two cases to solve , now i want to discover that how we can solve the second example in this way – Zia ur Rahman Aug 23 '10 at 12:35

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The same technique still applies, Zia. – J. M. Aug 23 '10 at 12:37
– J. M. Aug 23 '10 at 12:47

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Eleven exclamation marks look like someone's shouting. What shall I picture for myself in the face of eleven interrogation marks? – Rasmus Aug 23 '10 at 12:54
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## 2 Answers

To solve $|f(x)|=g(x)|h(x)|$ in general, find the sets $P=\{x:f(x)g(x)\ge 0\}$ and $N=\{x:f(x)g(x)\lt 0\}$.

Then solve the two equations

1. $f(x)=g(x)h(x)$ - but discard solutions that are not in $P$.
2. $f(x)=-g(x)h(x)$ - but discard solutions that are not in $N$.

You will be left with the solutions of $|f(x)|=g(x)|h(x)|$.

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Both equations you mention can be solved in the same way. As others have said in the comments on your question here and the comments and answers on this question, $|ax-b|=\begin{cases} ax-b & \text{ if } x\ge \frac{b}{a} \\ b-ax & \text{ if } x< \frac{b}{a} \end{cases}$, where I'll call $\frac{b}{a}$ a "turning point" value. So, for any equation an absolute value, there is at least one turning point value of the variable for each absolute value. In both of your examples, the turning point values are $-\frac{1}{2}$ and $-3$. The $n$ (2 in your examples) turning point values divide the number line into $n+1$ (3 in your examples) regions where, within each region, each absolute value can be replaced by either the expression inside the absolute value or its opposite. Each resulting equation can then be solved, restricting solutions to the corresponding region on the number line.

Let's consider a different example: $x|4x-1|=|3x-2|$. The turning points are $\frac{1}{4}$ and $\frac{2}{3}$.

$$\begin{matrix} \leftarrow & \frac{1}{4} & \text{---} & \frac{2}{3} & \rightarrow \\ x(1-4x)=(2-3x) & | & x(4x-1)=(2-3x) & | & x(4x-1)=(3x-2) \\ x-4x^2=2-3x & | & 4x^2-x=2-3x & | & 4x^2-x=3x-2 \\ 0=4x^2-4x+2 & | & 4x^2+2x-2=0 & | & 4x^2-4x+2=0 \\ \text{no }x\in\mathbb{R} & | & x=-1\text{ or }x=\frac{1}{2} & | & \text{no }x\in\mathbb{R} \end{matrix}$$

(edit: Note that between any two adjacent regions, the only change in the equations at the tops of the regions is the one term with turning point at the boundary between the two regions.)

Now, the solutions for each region-specific equation apply only to that region, so the solutions -1 and $\frac{1}{2}$ in the middle region only apply to the interval $[\frac{1}{4},\frac{2}{3}]$, so -1 is not a solution (it's not in the interval) but $x=\frac{1}{2}$ is a solution (and it's the only solution to $x|4x-1|=|3x-2|$).

Regardless of the algebraic method of solution, graphing both sides of the equation and looking for points of intersection is a good way to check your solution.

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