- Start with the given equation: \(\frac{2}{3} y + 15 = 9\).
- Apply the Subtraction property of equality to isolate the variable term: \(\frac{2}{3} y + 15 - 15 = 9 - 15\).
- Simplify to obtain \(\frac{2}{3} y = -6\).
- Apply the Multiplication property of equality: \(\frac{2}{3} y \cdot \frac{3}{2} = -6 \cdot \frac{3}{2}\).
- Simplify the final result to find \(\boxed{y = -9}\).
Explanation
- Analyzing the Given Equation
We begin with the linear equation \(\frac{2}{3} y + 15 = 9\). Our objective is to isolate the variable \(y\) by performing a series of algebraic operations. We will document each transformation and justify it using appropriate properties of equality or algebraic simplification.
- Subtracting 15 from Both Sides
Starting with the initial equation, we observe that \(15\) is added to the term containing \(y\). To isolate the term \(\frac{2}{3} y\), we apply the Subtraction property of equality by subtracting 15 from both sides. The expression becomes: \[\frac{2}{3} y + 15 - 15 = 9 - 15\] This step is justified by the Subtraction property of equality.
- Simplifying the Equation
Next, we simplify the expression on both sides of the equation. Calculating \(9 - 15 = -6\), the equation simplifies to: \[\frac{2}{3} y = -6\] This is justified by simple simplification.
- Multiplying by the Reciprocal
Now that we have \(\frac{2}{3} y = -6\), we need to isolate \(y\). We do this by multiplying both sides by the reciprocal of the coefficient \(\frac{2}{3}\), which is \(\frac{3}{2}\). This gives us: \[\frac{2}{3} y \cdot \frac{3}{2} = -6 \cdot \frac{3}{2}\] This step is justified by the Multiplication property of equality.
- Final Simplification and Result
Finally, we perform the arithmetic simplification on both sides: \(\frac{2}{3} \cdot \frac{3}{2} = 1\) and \(-6 \cdot \frac{3}{2} = -9\). The equation becomes \(y = -9\). This final step is justified by simplification. We can now complete the required table based on these steps and justifications.
Examples
Solving linear equations is a foundational skill used in many real-life applications. For instance, if you are planning a budget for a monthly subscription service where you pay a flat fee plus a per-usage cost, you can model your total cost with a linear equation to determine how many times you can use the service within your budget. By isolating the variable representing usage, you can find the maximum number of times you can afford to use the service, helping you make informed financial decisions.