- Start with the given equation \(\frac{2}{3}y + 15 = 9\).
- Apply the subtraction property of equality by subtracting 15 from both sides to get \(\frac{2}{3}y = -6\).
- Use the multiplication property of equality to multiply both sides by \(\frac{3}{2}\), the reciprocal of \(\frac{2}{3}\).
- Conclude that the solution is \(\boxed{y = -9}\).
Explanation
- Problem Analysis
The goal is to solve the linear equation \(\frac{2}{3}y + 15 = 9\) for the variable \(y\). We are provided with a sequence of logical steps and justifications that outline the algebraic process to isolate the variable. Our analysis confirms that these steps are mathematically sound and result in a complete solution. We will break down each step to provide a clear explanation of how the equation is simplified to arrive at the final value of \(y\).
- Isolating the Variable Term
The equation starts as \(\frac{2}{3}y + 15 = 9\). The first objective is to isolate the term containing \(y\). By applying the subtraction property of equality, we subtract \(15\) from both sides of the equation:
\[\frac{2}{3}y + 15 - 15 = 9 - 15\]
This simplifies to:
\[\frac{2}{3}y = -6\]
This step effectively removes the constant term from the side with the variable, leaving us with a simpler equation where \(y\) is multiplied by a fraction.
- Solving for y
Now that we have \(\frac{2}{3}y = -6\), we need to isolate \(y\). To do this, we use the multiplication property of equality. Multiplying both sides by the reciprocal of the coefficient \(\frac{2}{3}\), which is \(\frac{3}{2}\), will isolate \(y\):
\[\frac{2}{3}y \cdot \frac{3}{2} = -6 \cdot \frac{3}{2}\]
Performing the multiplication on the left side, \(\frac{2}{3} \cdot \frac{3}{2} = 1\), so we are left with \(y\). On the right side, \(-6 \cdot \frac{3}{2} = -3 \cdot 3 = -9\). Thus, the solution is:
\[y = -9\]
This step confirms the value of \(y\) that satisfies the original equation. The steps presented in the table are indeed complete and logically consistent.
Examples
Solving linear equations like \(\frac{2}{3}y + 15 = 9\) is a foundational skill used in various practical situations. For example, if a company is tracking its budget, where \(\frac{2}{3}\) of a specific fund (\(y\)) plus a base cost of \(\$15,000\) must equal a target outcome of \(\$9,000\), they can use this algebraic method to determine the necessary initial amount of the fund. Understanding how to isolate variables allows for precise financial planning, resource allocation, and problem-solving in everyday business and personal tasks.