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We are given a triangle \(VWX\) with sides \(VW = 3y + 3\), \(VX = 4y\), and \(WX = y + 1\). The perimeter of the triangle is 44 units, and we are asked to find the length of side \(VW\).

The perimeter of the triangle is the sum of the lengths of all its sides. Therefore, the equation for the perimeter is:

\[
VW + VX + WX = 44
\]

Substituting the expressions for the sides:

\[
(3y + 3) + 4y + (y + 1) = 44
\]

Now let's solve this equation step by step.

### 1. Combine like terms:

\[
3y + 3 + 4y + y + 1 = 44
\]
\[
8y + 4 = 44
\]

### 2. Subtract 4 from both sides:

\[
8y = 40
\]

### 3. Solve for \(y\):

\[
y = \frac{40}{8} = 5
\]

Now that we know \(y = 5\), substitute this value back into the expression for \(VW = 3y + 3\):

\[
VW = 3(5) + 3 = 15 + 3 = 18
\]

Thus, the length of side \(VW\) is \( \boxed{18} \).

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Do you want more details on this? Feel free to ask!

Here are some related questions:
1. What would be the length of side \(VX\) after solving for \(y\)?
2. Can you calculate the length of side \(WX\) after solving for \(y\)?
3. How does the formula for the perimeter change for different triangles?
4. What are some other methods to solve for the perimeter of a triangle?
5. How can we verify that our solution for \(y\) is correct?

**Tip**: Always double-check your work by substituting the value you found back into the original equation.

The perimeter of the triangle is 44 units. Find the length of side VW. The sides are VW = 3y + 3, VX = 4y, and WX = y + 1.

Learn how to solve for the side length VW of a triangle with a perimeter of 44 units. The triangle's sides are represented by algebraic expressions, including VW = 3y + 3, VX = 4y, and WX = y + 1. This solution uses basic algebra and perimeter calculations. Suitable for Grades 6-8.

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