Solving Absolute Value Systems: |x + 3| = |2x + 6| with |x| = 3

Q: Why does x = 3 satisfy |x| = 3 but not the first equation?

When $ x = 3 $, we get |3| = 3 ✓, but |3+3| = 6 while |2(3)+6| = 12. Since 6 ≠ 12, this candidate fails the first equation!

Q: How do I know which equation to solve first?

Always start with the simpler absolute value equation! $ |x| = 3 $ is much easier than $ |x+3| = |2x+6| $, so solve it first to get candidate values.

Q: What if both candidates worked in the first equation?

Then you'd have two solutions! Systems can have multiple solutions. Always test each candidate thoroughly - don't assume only one will work.

Q: Why does |0| = 0 when x = -3?

When $ x = -3 $: $ |x+3| = |-3+3| = |0| = 0 $$ |2x+6| = |2(-3)+6| = |-6+6| = |0| = 0 $Both expressions inside the absolute values equal zero, so 0 = 0 ✓

Q: Can a system of absolute value equations have no solution?

Yes! If none of the candidates from the simpler equation satisfy the more complex equation, then the system has no solution. Always test all candidates!

Absolute Value Systems with Candidate Testing

$\begin{cases} |x+3|=|2x+6| \\ \lvert x\rvert=3 \end{cases}$

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Understand the problem

$\begin{cases} |x+3|=|2x+6| \\ \lvert x\rvert=3 \end{cases}$

Step-by-step solution

To solve this system of equations, we will follow these steps:

Step 1: Solve the second equation $|x| = 3$ .
Step 2: Check both potential solutions from Step 1 in the first equation $|x+3| = |2x+6|$ .

Now, let's work through each step:
Step 1: Solve $|x| = 3$ . This gives us two potential solutions: $x = 3$ and $x = -3$ .

Step 2: Check these solutions in $|x+3| = |2x+6|$ :
- For $x = 3$ :
$|x+3| = |3+3| = |6| = 6$ and $|2x+6| = |2(3)+6| = |6+6| = |12| = 12$ .
Since $6 \neq 12$ , $x = 3$ does not satisfy the first equation.

- For $x = -3$ :
$|x+3| = |-3+3| = |0| = 0$ and $|2x+6| = |2(-3)+6| = |-6+6| = |0| = 0$ .
Both sides equal 0, so $x = -3$ satisfies the first equation.

Therefore, the solution to the system of equations is $x = -3$ .

Final Answer

$x=-3$

Key Points to Remember

Essential concepts to master this topic

Rule: Solve simpler absolute value equation first for candidates
Technique: Test $x = 3$ : |6| ≠ |12| and $x = -3$ : |0| = |0|
Check: Both equations must be satisfied simultaneously for valid solution ✓

Common Mistakes

Avoid these frequent errors

Solving only one equation and ignoring the other
Don't solve just $|x| = 3$ and assume both x = 3 and x = -3 work! This gives incomplete answers since both equations must be satisfied. Always test each candidate from the simpler equation in the more complex equation.

Practice Quiz

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$\left|x\right|=5$

FAQ

Everything you need to know about this question

Why does x = 3 satisfy |x| = 3 but not the first equation?

When $x = 3$ , we get |3| = 3 ✓, but |3+3| = 6 while |2(3)+6| = 12. Since 6 ≠ 12, this candidate fails the first equation!

How do I know which equation to solve first?

Always start with the simpler absolute value equation! $|x| = 3$ is much easier than $|x+3| = |2x+6|$ , so solve it first to get candidate values.

What if both candidates worked in the first equation?

Then you'd have two solutions! Systems can have multiple solutions. Always test each candidate thoroughly - don't assume only one will work.

Why does |0| = 0 when x = -3?

When $x = -3$ :

$|x+3| = |-3+3| = |0| = 0$
$|2x+6| = |2(-3)+6| = |-6+6| = |0| = 0$

Both expressions inside the absolute values equal zero, so 0 = 0 ✓

Can a system of absolute value equations have no solution?

Yes! If none of the candidates from the simpler equation satisfy the more complex equation, then the system has no solution. Always test all candidates!

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