Solve the Absolute Equation: |2x + 4 + |x - 1|| = 10

Q: Why do I need to split the equation into different cases?

Because absolute value expressions behave differently depending on whether what's inside is positive or negative. When $ x - 1 \geq 0 $, we get $ |x - 1| = x - 1 $, but when $ x - 1 , we get \( |x - 1| = -(x - 1) $.

Q: How do I know where to split the cases?

Find the critical points where expressions inside absolute values equal zero. For $ |x - 1| $, set $ x - 1 = 0 $ to get $ x = 1 $. This gives you the boundary between cases.

Q: What if I get multiple solutions from one case?

That's normal! Each case can produce 0, 1, or 2 solutions. Just remember to check each solution against the case condition. If $ x = 5 $ comes from case \( x , it's invalid!

Q: Why does the equation become |3x + 3| = 10 in the first case?

When $ x \geq 1 $, we have $ |x - 1| = x - 1 $. Substituting into the original equation: $ |2x + 4 + (x - 1)| = |3x + 3| = 10 $.

Q: How do I verify my final answers are correct?

Substitute each solution back into the original equation $ |2x + 4 + |x - 1|| = 10 $. For $ x = \frac{7}{3} $: $ |2(\frac{7}{3}) + 4 + |\frac{7}{3} - 1|| = |\frac{14}{3} + 4 + \frac{4}{3}| = |10| = 10 $ ✓

Q: Can I solve this without cases by just removing absolute value signs?

No! You cannot simply remove absolute value signs. Each absolute value creates a piecewise function, and you must consider all possible combinations of positive/negative values inside each absolute value expression.

Step-by-step video solution

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Step-by-step written solution

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1

Understand the problem

$|2x+4+\left|x-1\right||=10$

2

Step-by-step solution

To solve the equation $|2x + 4 + |x - 1|| = 10$ , we need to consider multiple cases based on the behaviour of the absolute values.

Case 1: $x - 1 \geq 0 \Rightarrow x \geq 1$

In this case, $|x - 1| = x - 1$ .
Thus, the equation becomes $|2x + 4 + x - 1| = 10$ or $|3x + 3| = 10$ .
Solving $|3x + 3| = 10$ , we have two sub-cases:
$3x + 3 = 10$ leads to $3x = 7$ or $x = \frac{7}{3} \approx 2.33$ .
$3x + 3 = -10$ leads to $3x = -13$ or $x = -\frac{13}{3} \approx -4.33$ (which does not satisfy $x \geq 1$ ).

Case 2: $x - 1 < 0 \Rightarrow x < 1$

In this case, $|x - 1| = -(x - 1) = -x + 1$ .
Thus, the equation becomes $|2x + 4 - x + 1| = 10$ or $|x + 5| = 10$ .
Solving $|x + 5| = 10$ , we have two sub-cases:
$x + 5 = 10$ leads to $x = 5$ (but $x < 1$ is required).
$x + 5 = -10$ leads to $x = -15$ (this satisfies $x < 1$ ).

Thus, considering the regions and restrictions, we find the solutions to be $x = 2.3$ and $x = -15$ .

Therefore, the solution to the problem is $x = 2.3$ , and $x = -15$ .

3

Final Answer

$x=2.3$ , $x=-15$

Key Points to Remember

Essential concepts to master this topic

Case Analysis: Split at critical points where expressions inside absolute values equal zero
Technique: For |x - 1|, split at x = 1 to handle x ≥ 1 and x < 1 separately
Check: Verify solutions satisfy original equation and region constraints ✓

Common Mistakes

Avoid these frequent errors

Ignoring domain restrictions when solving sub-cases
Don't accept x = 5 from case x < 1 = contradicts the region requirement! This creates false solutions that don't work in the original equation. Always check that each solution satisfies both the equation and the case condition.

Practice Quiz

Test your knowledge with interactive questions

$\left|x\right|=3$

FAQ

Everything you need to know about this question

Why do I need to split the equation into different cases?

+

Because absolute value expressions behave differently depending on whether what's inside is positive or negative. When $x - 1 \geq 0$ , we get $|x - 1| = x - 1$ , but when $x - 1 < 0$ , we get $|x - 1| = -(x - 1)$ .

How do I know where to split the cases?

+

Find the critical points where expressions inside absolute values equal zero. For $|x - 1|$ , set $x - 1 = 0$ to get $x = 1$ . This gives you the boundary between cases.

What if I get multiple solutions from one case?

+

That's normal! Each case can produce 0, 1, or 2 solutions. Just remember to check each solution against the case condition. If $x = 5$ comes from case $x < 1$ , it's invalid!

Why does the equation become |3x + 3| = 10 in the first case?

+

When $x \geq 1$ , we have $|x - 1| = x - 1$ . Substituting into the original equation: $|2x + 4 + (x - 1)| = |3x + 3| = 10$ .

How do I verify my final answers are correct?

+

Substitute each solution back into the original equation $|2x + 4 + |x - 1|| = 10$ . For $x = \frac{7}{3}$ : $|2(\frac{7}{3}) + 4 + |\frac{7}{3} - 1|| = |\frac{14}{3} + 4 + \frac{4}{3}| = |10| = 10$ ✓

Can I solve this without cases by just removing absolute value signs?

+

No! You cannot simply remove absolute value signs. Each absolute value creates a piecewise function, and you must consider all possible combinations of positive/negative values inside each absolute value expression.

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Solve the Absolute Equation: |2x + 4 + |x - 1|| = 10

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