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

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

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 $x = -15$.