Equations with Absolute Values: Double absolute value

To solve the equation $|\left|2x+4\right|+x-1|=10$, we will consider different cases due to the absolute value expressions.

First, let's set $|2x + 4| + x - 1 = 10$:

• Case 1: $2x + 4 \geq 0$ (so $|2x + 4| = 2x + 4$),
  Equation becomes: $2x + 4 + x - 1 = 10$,
  Combine and simplify: $3x + 3 = 10$,
  Solve: $3x = 7$ $\Rightarrow x = \frac{7}{3}$.
• Case 2: $2x + 4 < 0$ (so $|2x + 4| = -(2x + 4)$),
  Equation becomes: $-2x - 4 + x - 1 = 10$,
  Combine and simplify: $-x - 5 = 10$,
  Solve: $-x = 15$ $\Rightarrow x = -15$.

Next, consider $|2x + 4| + x - 1 = -10$:

• Case 3: $2x + 4 \geq 0$ (so $|2x + 4| = 2x + 4$),
  Equation becomes: $2x + 4 + x - 1 = -10$,
  Combine and simplify: $3x + 3 = -10$,
  Solve: $3x = -13$ $\Rightarrow x = -\frac{13}{3}$, which is not a solution since it doesn't satisfy the condition.
• Case 4: $2x + 4 < 0$ (so $|2x + 4| = -(2x + 4)$),
  Equation becomes: $-2x - 4 + x - 1 = -10$,
  Combine and simplify: $-x - 5 = -10$,
  Solve: $-x = -5$ $\Rightarrow x = 5$, which is not a solution since it doesn't satisfy the condition.

Thus, the possible solutions are $x = \frac{7}{3} \approx 2.3$ and $x = -15$.

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

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 $x = -15$.

To solve this problem, we'll follow these steps:

• Step 1: Consider the equation $|B| = 7$, where $B = |2-7x|-5$.
• Step 2: Solve the two separate scenarios: $B = 7$ and $B = -7$.

Now, let's work through each major step:

Step 1: Begin with $|2-7x|-5 = 7$.
This implies either $|2-7x|-5 = 7$ or $|2-7x|-5 = -7$.

Step 2: Solve $|2-7x|-5 = 7$:

- $|2-7x|-5 = 7$ becomes $|2-7x| = 12$.

- For $|2-7x| = 12$, solve $2-7x = 12$ or $2-7x = -12$.

Step 3: Solve the equations derived:

• Case 1:

- Solve $2-7x = 12$, leading to:

+ $(-7x = 10)$, hence $x = -\frac{10}{7}$.
• Case 2:

- Solve $2-7x = -12$, arrives at:

+ $(-7x = -14)$, so $x = 2$.

Check for the case $|2-7x|-5 = -7$ is invalidated since no real solution exists for absolute difference less than zero.

Conclusively, solving $||2-7x|-5|=7$ yields solutions $x=2$ and $x=-\frac{10}{7}$.

Therefore, the correct choice is "a and c correct".

To solve the equation $|2x + |x - 2|| = 1$, we need to consider different cases based on the expression within the absolute value signs.

Case 1: $x \geq 2$

• Here, $|x - 2| = x - 2$.
• The equation becomes $|2x + x - 2| = 1$ or $|3x - 2| = 1$.
• This gives two sub-cases to solve:
  • Sub-case 1.1: $3x - 2 = 1$ leads to $3x = 3$, so $x = 1$. However, $x = 1$ does not satisfy $x \geq 2$. Discard this solution.
  • Sub-case 1.2: $3x - 2 = -1$ leads to $3x = 1$, so $x = \frac{1}{3}$. But $x = \frac{1}{3}$ does not satisfy $x \geq 2$. Discard this solution.

Case 2: $x < 2$

• In this case, $|x - 2| = 2 - x$.
• The equation transforms into $|2x + 2 - x| = 1$, then simplifies to $|x + 2| = 1$.
• This leads to two sub-cases:
  • Sub-case 2.1: $x + 2 = 1$ leads to $x = -1$. Since $x = -1 < 2$, it is a valid solution.
  • Sub-case 2.2: $x + 2 = -1$ leads to $x = -3$. Again, $x = -3 < 2$ holds, so this is also a valid solution.

Thus, the solutions to the equation are $x = -1$ and $x = -3$.

Therefore, the correct answers are $x = -1$ and $x = -3$.

To solve the equation $||x-1| + x + 1| = 4$, we need to consider multiple cases based on the conditions affecting the absolute value expressions. For such cases, the key points to assess are where the internal expressions become zero or change signs. Let's start solving the problem step-by-step:

Step 1: Consider the critical point $x = 1$ where the expression $|x-1|$ changes from zero or negative to positive. We handle separate cases surrounding this point.

• Case 1: $x \leq 1$

• Case 2: x > 1

Case 1: $x \leq 1$
In this interval, $x - 1 \leq 0$, so $|x-1| = -(x-1) = -x + 1$.

Now we consider the absolute value $||x-1| + x + 1|$:

• $||-x + 1 + x + 1|$ simplifies to $|2| = 2$, which is not equal to 4. Thus, no solutions in this interval.

Case 2: x > 1
In this interval, x - 1 > 0 , so $|x-1| = x - 1$.

Now the expression becomes:

• $||(x - 1) + x + 1| = |2x| = 2x|$. Setting this equal to 4 gives $2x = 4$ or $x = 2$.

Since $x = 2$ falls within the interval x > 1 , it satisfies the conditions for this case.

We've determined that the solution to the equation is $x = 2$, which satisfies $||x-1| + x + 1| = 4$.

Therefore, after checking the conditions for both cases and ensuring that the solution adheres to the interval restrictions, the correct solution is $x = 2$.

The correct answer is choice 2: $x = 2$.