Problem 55 - Algebra A Combined Function

Question

Solve each equation. See Section \(2.3 .\) \(4\left(\frac{y+1}{2}\right)+3 y=0\)

Step-by-Step Solution

Answer

The solution is \(y = -\frac{2}{5}\).

Step 1: Distribute the 4 inside the bracket

Begin by distributing the 4 across the expression inside the parenthesis on the left-hand side of the equation. This simplifies the equation:\[ 4 \left( \frac{y+1}{2} \right) + 3y = 0 \]Distribute the 4:\[ 4 \times \frac{y}{2} + 4 \times \frac{1}{2} \]This simplifies to:\[ 2y + 2 + 3y = 0 \]

Step 2: Combine like terms

Now combine the like terms on the left-hand side of the equation:\[ 2y + 3y + 2 = 0 \]This simplifies to:\[ 5y + 2 = 0 \]

Step 3: Isolate the variable term

Subtract 2 from both sides to isolate the term with the variable:\[ 5y + 2 - 2 = 0 - 2 \]This simplifies to:\[ 5y = -2 \]

Step 4: Solve for y

Now divide both sides by 5 to solve for \(y\):\[ \frac{5y}{5} = \frac{-2}{5} \]This simplifies to:\[ y = -\frac{2}{5} \]

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that allows us to simplify expressions and equations. It's often used when dealing with parentheses. Essentially, it means multiplying each term inside the parentheses by the term outside. This helps spread or 'distribute' the multiplication across each term within the parentheses.

For example, in our original equation:

• We have the expression \(4\left(\frac{y+1}{2}\right)\).
• By applying the distributive property, we multiply 4 by each term inside the bracket: \(4 \times \frac{y}{2}\) and \(4 \times \frac{1}{2}\).

This simplifies to \(2y + 2\) after performing the multiplication.

Understanding the distributive property is crucial because it forms the basis for more complex algebraic operations. It's a powerful tool that helps in breaking down larger problems into manageable parts.

Combining Like Terms

Combining like terms is a necessary step in simplifying expressions. Like terms have the same variable raised to the same power, although they might have different coefficients.

For the equation we're working on, after distributing the 4, we get:

• \(2y + 2 + 3y = 0\)

The next step is to combine the terms that have the variable \(y\).
Like terms in this context are \(2y\) and \(3y\), which both contain the variable \(y\).

When combined:\(2y + 3y\) results in \(5y\). The constant term, 2, remains unchanged for now.
So, the expression is simplified to \(5y + 2 = 0\).

This process reduces clutter in equations and brings clarity, making it easier to solve them. Remembering to match up terms carefully is key while combining.

Isolating Variables

Solving equations often involves isolating the variable you're looking for. This means getting the variable by itself on one side of the equation.

In our equation, once simplified to \(5y + 2 = 0\), we need to clear away other terms around \(y\).

• First, subtract 2 from both sides: \(5y + 2 - 2 = 0 - 2\), which reduces to \(5y = -2\).
• Then, to isolate \(y\), divide both sides by 5, giving us: \(\frac{5y}{5} = \frac{-2}{5}\).

The solution is \(y = -\frac{2}{5}\).

By carefully isolating the variable step-by-step, you can solve for it accurately. This method is used frequently in algebra to solve all types of linear equations.