91Ó°ÊÓ

The distributive property is a fundamental concept in algebra. It's all about breaking down and simplifying expressions that involve parentheses. When you have a term outside the parentheses, like a number, you multiply it by each term inside the parentheses.
In our problem, we start with the equation \(4\left(\frac{y+1}{2}\right) + 3y = 0\). The 4 outside the parentheses needs to be distributed to each term inside. This is done by multiplying 4 by \(\frac{y}{2}\) and then by \(\frac{1}{2}\).
After applying the distributive property, each term is simplified individually to give:

• \(4 \times \frac{y}{2} = 2y\)
• \(4 \times \frac{1}{2} = 2\)

This simplifies the equation to \(2y + 2 + 3y = 0\), getting rid of the parentheses altogether. Understanding this property helps in organizing calculations for solving the equation efficiently.

Once the distributive property is applied, we move on to combining like terms. This step is crucial for reducing complexity in the equation. Like terms are terms that have the same variable raised to the same power. They can be added or subtracted easily.
In the equation \(2y + 2 + 3y = 0\), we identify the like terms:

• \(2y\) and \(3y\) can be combined because they both involve the variable \(y\).

Adding these like terms together, we get:
\(2y + 3y = 5y\).
Now our equation is simplified to \(5y + 2 = 0\). Combining like terms not only makes the equation simpler but also helps focus on solving for the variable more directly.

The final step involves solving the equation for \(y\). This means isolating \(y\) on one side of the equation. From the simplified equation \(5y + 2 = 0\), we aim to get \(y\) by itself.
Start by eliminating the constant term on the left side: subtract 2 from both sides of the equation:
\(5y + 2 - 2 = 0 - 2\), which simplifies to \(5y = -2\).
Next, divide both sides by 5 to solve for \(y\):
\( \frac{5y}{5} = \frac{-2}{5} \), leaving us with \(y = -\frac{2}{5}\).
This approach ensures that \(y\) is isolated and calculated correctly. Solving for a variable helps determine its value based on the given conditions of the problem.