solve 3 times the absolute value of (w-5) equal to the absolute value of (2w+10)

We are given the equation:

\[
3 \cdot |w - 5| = |2w + 10|
\]

To solve this, we need to consider the properties of absolute values. The equation can be split into different cases depending on the sign of the expressions inside the absolute values. Let's solve it step by step.

### Case 1: \( w - 5 \geq 0 \) and \( 2w + 10 \geq 0 \)

If both expressions are non-negative, the equation becomes:

\[
3(w - 5) = 2w + 10
\]

Simplify and solve for \( w \):

\[
3w - 15 = 2w + 10
\]

Subtract \( 2w \) from both sides:

\[
w - 15 = 10
\]

Add 15 to both sides:

\[
w = 25
\]

For \( w = 25 \), we check if it satisfies both conditions \( w - 5 \geq 0 \) and \( 2w + 10 \geq 0 \):
- \( 25 - 5 = 20 \geq 0 \) (True)
- \( 2(25) + 10 = 60 \geq 0 \) (True)

So, \( w = 25 \) is a valid solution.

### Case 2: \( w - 5 \geq 0 \) and \( 2w + 10 < 0 \)

If \( w - 5 \geq 0 \) but \( 2w + 10 \) is negative, the equation becomes:

\[
3(w - 5) = -(2w + 10)
\]

Simplify and solve for \( w \):

\[
3w - 15 = -2w - 10
\]

Add \( 2w \) to both sides:

\[
5w - 15 = -10
\]

Add 15 to both sides:

\[
5w = 5
\]

Divide by 5:

\[
w = 1
\]

For \( w = 1 \), check if the conditions hold:
- \( 1 - 5 = -4 \geq 0 \) (False)

Since the first condition \( w - 5 \geq 0 \) is not satisfied, \( w = 1 \) is not a valid solution.

### Case 3: \( w - 5 < 0 \) and \( 2w + 10 \geq 0 \)

If \( w - 5 < 0 \) but \( 2w + 10 \) is non-negative, the equation becomes:

\[
-3(w - 5) = 2w + 10
\]

Simplify and solve for \( w \):

\[
-3w + 15 = 2w + 10
\]

Add \( 3w \) to both sides:

\[
15 = 5w + 10
\]

Subtract 10 from both sides:

\[
5 = 5w
\]

Divide by 5:

\[
w = 1
\]

For \( w = 1 \), check if the conditions hold:
- \( 1 - 5 < 0 \) (True)
- \( 2(1) + 10 = 12 \geq 0 \) (True)

So, \( w = 1 \) is a valid solution.

### Case 4: \( w - 5 < 0 \) and \( 2w + 10 < 0 \)

If both expressions are negative, the equation becomes:

\[
-3(w - 5) = -(2w + 10)
\]

Simplify and solve for \( w \):

\[
-3w + 15 = -2w - 10
\]

Add \( 3w \) to both sides:

\[
15 = w - 10
\]

Add 10 to both sides:

\[
w = 25
\]

For \( w = 25 \), check if the conditions hold:
- \( 25 - 5 < 0 \) (False)

Since the first condition \( w - 5 < 0 \) is not satisfied, \( w = 25 \) is not a valid solution for this case.

### Final Answer
The valid solutions are \( w = 25 \) and \( w = 1 \).

Would you like more details, or do you have any questions?

Here are five related questions to expand your understanding:

1. How do you solve absolute value equations involving inequalities?
2. Can absolute value equations have no solution? If so, under what conditions?
3. What are some real-life applications of absolute value functions?
4. How does the concept of absolute value relate to distance on a number line?
5. How do you graph the solution to an absolute value equation on a coordinate plane?

**Tip:** Always check the conditions of the case you are solving to ensure that your solution is valid.

Learn how to solve the equation 3 times the absolute value of (w-5) equals the absolute value of (2w+10) with a step-by-step solution. Explore different cases and conditions involving absolute values and inequalities. Suitable for high school students and anyone interested in solving absolute value equations.

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