Equations with Absolute Values: System of equations

To solve the given problem, let's break it down into manageable steps:

The given equation is $||x-11|-1|=5$ and the condition is $x > 10$.

Step 1: Solve the outer absolute value: $||y|-1|=5$, where $y = x - 11$.

This implies two scenarios:

• $|y|-1 = 5 \implies |x-11| = 6$
• $|y|-1 = -5 \implies |x-11| = -4$ (which is not possible since absolute values are non-negative)

So, we only consider $|x-11| = 6$.

Step 2: Solve the equation $|x-11| = 6$, giving two possibilities:

• $x - 11 = 6 \implies x = 17$
• $x - 11 = -6 \implies x = 5$

Step 3: Apply the condition $x > 10$.

Out of the two possible solutions, $x = 5$ does not satisfy $x > 10$.

Therefore, the only valid solution that satisfies both conditions is $x = 17$.

The final solution is $x = 17$.

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

• Step 1: Solve the polynomial equation $x^4 = 0$.
• Step 2: Verify the solution against the absolute value equation $|x+2| = |x-2|$.
• Step 3: Ensure the solution matches a provided choice, if necessary.

Now, let's work through each step:

Step 1: Solving $x^4 = 0$ yields $x = 0$, as any non-zero values to the power of 4 would not equal zero.

Step 2: Substituting $x = 0$ into the absolute value equation:

$|0+2| = |0-2|$

$|2| = |-2|$

$2 = 2$, thus the condition is satisfied.

Step 3: Compare with provided answer choices. Since the correct choice must satisfy both equations, $x = 0$ is consistent with option (1).

Therefore, the solution to the problem is $x = 0$.

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 $x = -3$.

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

• Step 1: Solve the equation $|x+2| = 0$
  This implies $x+2 = 0$, hence $x = -2$.
• Step 2: Solve the equation $|x+2| = |x-2|$.
  This leads to two cases:
  - Case 1: $x+2 = x-2$ results in $2 = -2$, which is false.
  - Case 2: $x+2 = -(x-2)$ simplifies to $2x = 0$, hence $x = 0$.
• Step 3: Check for intersection of solutions.
  We have solutions $x = -2$ from the first equation and $x = 0$ from the absolute value equation.
• Step 4: Verify if any common $x$ satisfies both equations.

Substituting $x = -2$ into $|x+2| = |x-2|$, we find that it does not satisfy $|x-2| = |-4| = 4$. Therefore, $x = -2$ is not a solution to the system.

Substituting $x = 0$ into both equations does not satisfy $|x+2| = 0$. Therefore, there is no overlap of solutions.

The solution to the system is No solution.

To solve the given problem, we'll work through these steps:

• Step 1: Solve the equation $|x+4| = |2x+20|$.
• Step 2: Consider the two cases derived from the absolute value equation.
• Step 3: Apply the condition $x < -10$ to determine the suitable solution.

Step 1: We start by considering the absolute value equation $|x+4| = |2x+20|$. This gives us two cases to explore:

• Case 1: $x+4 = 2x+20$
• Case 2: $x+4 = -(2x+20)$

Step 2: Solve each case individually:

Case 1:
Solving $x+4 = 2x+20$,
We first subtract $x$ from both sides to obtain:
$4 = x + 20$.
Subtracting 20 from both sides, we get:
$x = 4 - 20 \Rightarrow x = -16$.

Case 2:
Solving $x+4 = -(2x+20)$,
This simplifies to $x+4 = -2x - 20$.
Adding $2x$ to both sides, we have:
$3x + 4 = -20$.
Subtracting 4 from both sides gives:
$3x = -24$.
Dividing by 3, we find:
$x = -8$.

Step 3: Consider the inequality $x < -10$:

• Substitute $x = -16$ into the inequality: $-16 < -10$ is true.
• Substitute $x = -8$ into the inequality: $-8 < -10$ is false.

Therefore, the solution that satisfies both the equation and the inequality is $x = -16$.

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

• Step 1: Consider the equation $|x+4| = |2x+20|$.
• Step 2: Break it down into two separate cases.

We'll address each case separately below:

Case 1: $x + 4 = 2x + 20$

Simplifying gives:

$x + 4 = 2x + 20$
$4 = 2x - x + 20$
$4 = x + 20$
Subtracting 20 from both sides, we get:
$x = 4 - 20$
$x = -16$.

Since $x > -10$ is required, $x = -16$ is not valid.

Case 2: $x + 4 = -(2x + 20)$

Simplifying gives:

$x + 4 = -2x - 20$
Add $2x$ to both sides:
$x + 2x + 4 = -20$
$3x + 4 = -20$
Subtract 4 from both sides:
$3x = -24$
Divide by 3:
$x = -8$.

Check: Since $-8 > -10$, this solution is valid.

Thus, the solution to the problem is $x = -8$.

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

• Step 1: Solve the equation $|x-5|=0$.
• Step 2: Validate the result with the other equation $||x-11|-1|=5$.

Let's work through each step:

Step 1: Solve $|x-5|=0$.
The equation $|x-5| = 0$ implies that the expression inside the absolute value must be zero. Therefore, $x-5=0$ gives us:

$x = 5$

Step 2: Validate with $||x-11|-1|=5$.
Now, substitute $x = 5$ into the other equation:

$|5-11| = | -6 | = 6$

$|6-1| = |5| = 5$

This simplifies to $|5| = 5$, which is true. Therefore, the solution $x = 5$ satisfies both equations.

Thus, the system of equations has a single solution: $x = 5$.