Problem 327
Question
In the following exercises, simplify. $$ 12\left(\frac{5}{6} p\right) $$
Step-by-Step Solution
Verified Answer
10p
1Step 1: Distribute the 12
Multiply the constant 12 by the fraction inside the parenthesis: \[ 12 \times \frac{5}{6} p \]
2Step 2: Simplify inside the parenthesis
Simplify the expression by multiplying 12 and the fraction \( \frac{5}{6} \). \[ 12 \times \frac{5}{6} = \frac{12 \times 5}{6} \]
3Step 3: Perform the multiplication
Calculate the multiplication: \[ \frac{12 \times 5}{6} = \frac{60}{6} \]
4Step 4: Simplify the fraction
Simplify the fraction \( \frac{60}{6} \): \[ \frac{60}{6} = 10 \]
5Step 5: Multiply by the variable
Attach the variable \( p \) to the simplified expression: \[ 10p \]
Key Concepts
Distributive PropertyMultiplying FractionsSimplifying FractionsAlgebraic Multiplication
Distributive Property
The distributive property is a fundamental concept in algebra. It states that for any numbers or variables, multiplying a sum by a number gives the same result as multiplying each addend individually by the number and then adding the products.
In the given example, we distribute the 12 over the \(\frac{5}{6} p\) inside the parentheses:
\[ 12 \times \frac{5}{6} p \]
This means we are multiplying 12 by \(\frac{5}{6}\) first, then by the variable \(p\). Understanding this property helps break down more complex expressions and simplifies problem solving.
In the given example, we distribute the 12 over the \(\frac{5}{6} p\) inside the parentheses:
\[ 12 \times \frac{5}{6} p \]
This means we are multiplying 12 by \(\frac{5}{6}\) first, then by the variable \(p\). Understanding this property helps break down more complex expressions and simplifies problem solving.
Multiplying Fractions
When multiplying fractions, multiply the numerators together and the denominators together. It's often easier to simplify the fractions before performing the multiplication to make calculations easier.
In our case:
\[ 12 \times \frac{5}{6} \]
Here, 12 can be written as \( \frac{12}{1} \). So it becomes:
\[ \frac{12}{1} \times \frac{5}{6} = \frac{12 \times 5}{1 \times 6} = \frac{60}{6} \]
This illustrates that multiplying fractions involves straightforward arithmetic but benefits greatly from simplification wherever possible.
In our case:
\[ 12 \times \frac{5}{6} \]
Here, 12 can be written as \( \frac{12}{1} \). So it becomes:
\[ \frac{12}{1} \times \frac{5}{6} = \frac{12 \times 5}{1 \times 6} = \frac{60}{6} \]
This illustrates that multiplying fractions involves straightforward arithmetic but benefits greatly from simplification wherever possible.
Simplifying Fractions
Simplifying fractions is about reducing them to their smallest form. This is done by dividing the numerator and the denominator by their greatest common divisor.
In our example:
\[ \frac{60}{6} \]
Here, both 60 and 6 are divisible by 6:
\[ \frac{60 \div 6}{6 \div 6} = \frac{10}{1} = 10 \]
This step is crucial, as it makes the fraction easier to work with and aligns with the goal of simplifying algebraic expressions.
In our example:
\[ \frac{60}{6} \]
Here, both 60 and 6 are divisible by 6:
\[ \frac{60 \div 6}{6 \div 6} = \frac{10}{1} = 10 \]
This step is crucial, as it makes the fraction easier to work with and aligns with the goal of simplifying algebraic expressions.
Algebraic Multiplication
Algebraic multiplication involves multiplying coefficients and variables in expressions. After simplification of numerical factors, the variable is included in the final product.
For our step-by-step problem, the final step involves multiplying the simplified number by the variable \(p\):
\[ 10 \times p = 10p \]
This final product shows the result of multiplying the constant term by the given variable. Such practices are common in algebra and help bridge simpler arithmetic to more complex algebraic techniques.
For our step-by-step problem, the final step involves multiplying the simplified number by the variable \(p\):
\[ 10 \times p = 10p \]
This final product shows the result of multiplying the constant term by the given variable. Such practices are common in algebra and help bridge simpler arithmetic to more complex algebraic techniques.
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