- Analyze the equation: \(-0.3a = -2.5\).
- Recognize the operation needed: divide both sides by \(-0.3\) to isolate \(a\).
- Identify the property: The division property of equality states that dividing both sides of an equation by the same non-zero number yields an equivalent equation.
- Final answer: \(\boxed{\text{the division property of equality}}\).
Explanation
- Analysis of the Equation Solving Process
To determine the justification for Step 3, let's examine the transition from Step 2 to Step 3 in the provided equation-solving process.
The equation in Step 2 is \(-0.3a = -2.5\).
The equation in Step 3 is \(a = 8.\overline{3}\).
To transition from Step 2 to Step 3, we need to isolate the variable \(a\). This is achieved by dividing both sides of the equation by the coefficient of \(a\), which is \(-0.3\).
Performing this calculation:
\[\frac{-0.3a}{-0.3} = \frac{-2.5}{-0.3}\]
\[a = \frac{25}{3} = 8.\overline{3}\]
Since we divided both sides by the same non-zero constant, the property used is the division property of equality.
- Selecting the Correct Justification
Comparing our operation with the given options:
- Combining like terms: This was used in Step 1 (\(0.8a - 0.1a = 0.7a\)).
- The addition property of equality: This involves adding the same value to both sides, which is not what we did here.
- The subtraction property of equality: This was used in Step 2 to move \(a\) to the left (\(0.7a - a = -0.3a\)).
- The division property of equality: This is the operation we used to isolate \(a\) by dividing by \(-0.3\).
Therefore, the correct justification for Step 3 is the division property of equality. The final answer is \(\boxed{\text{the division property of equality}}\).
Examples
In real-life scenarios, the division property of equality is used constantly, such as when calculating unit prices. If you buy 5 identical notebooks for a total of R\$ 50, the equation for the unit price \(p\) is \(5p = 50\). To find the cost of one notebook, you divide both sides by 5: \(\frac{5p}{5} = \frac{50}{5}\), resulting in \(p = 10\). This ensures that you can always determine the individual value of an item from a total group cost.