Real-Life Situation:
Maria is throwing a birthday party for her son, and she needs to buy balloons. She plans to buy a total of 36 balloons. She knows that each pack of balloons contains 6 balloons. She will buy \( p \) packs of balloons. The equation representing the situation can be written as:
\[ 6p = 36 \]
where \( p \) is the number of packs Maria needs to buy.
Method 1: Solving by Division
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Start with the equation: \[ 6p = 36 \]
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To isolate \( p \), divide both sides by 6: \[ p = \frac{36}{6} \]
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Calculate the division: \[ p = 6 \]
So, Maria needs to buy 6 packs of balloons.
Method 2: Solving by Subtraction and Checking
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Start with the equation: \[ 6p = 36 \]
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First, let’s express the number of balloons per pack. To check how many balloons each pack contributes:
- We know that \( 6p \) represents the total balloons.
- We can subtract the contribution of one pack at a time until we reach zero.
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Subtracting the equivalent of one pack at a time: \[ 6p - 6 = 36 - 6 \Rightarrow 6p = 30 \] \[ 6p - 6 = 30 - 6 \Rightarrow 6p = 24 \] \[ 6p - 6 = 24 - 6 \Rightarrow 6p = 18 \] \[ 6p - 6 = 18 - 6 \Rightarrow 6p = 12 \] \[ 6p - 6 = 12 - 6 \Rightarrow 6p = 6 \] \[ 6p - 6 = 6 - 6 \Rightarrow 6p = 0 \]
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This process indicates it took 6 subtractions to get down to zero, leading us again to: \[ p = 6 \]
Comparison of Solution Methods:
Between the two methods, I prefer Method 1: Solving by Division.
Reason:
- It is straightforward and uses fewer steps, making it quicker and more efficient.
- The division method allows you to directly isolate the variable without the need for adjusting and checking each step.
- It is also easier to communicate and understand, especially in practical situations like calculating costs or quantities.
While Method 2 provides a visual understanding of how many packs you would have to subtract to reach zero, it can be unnecessarily complex for simple arithmetic problems like this one. The division method is more systematic and preferred in algebraic contexts.