Problem 59
Question
Apply the distributive property, then simplify. $$3\left(\frac{5}{6} a+\frac{4}{9}\right)$$
Step-by-Step Solution
Verified Answer
\(\frac{5}{2}a + \frac{4}{3}\).
1Step 1: Apply the distributive property
The distributive property states that for any numbers \(a\), \(b\), and \(c\), \(a(b+c) = ab + ac\). In this case, \(a = 3\), \(b = \frac{5}{6}a\), and \(c = \frac{4}{9}\). We distribute \(3\) across the terms inside the parentheses: 3\left(\frac{5}{6}a\right) + 3\left(\frac{4}{9}\right).
2Step 2: Multiply each term
First, calculate \(3\left(\frac{5}{6}a\right)\):\(3 \times \frac{5}{6}\) gives \(\frac{15}{6}a\). Then, calculate \(3 \times \frac{4}{9}\):\(3 \times \frac{4}{9}\) gives \(\frac{12}{9}\).
3Step 3: Simplify each term
Simplify the coefficient \(\frac{15}{6}\) by dividing the numerator and the denominator by their greatest common divisor, which is 3:\(\frac{15}{6}a = \frac{5}{2}a\).Similarly, simplify \(\frac{12}{9}\) by dividing by 3:\(\frac{12}{9} = \frac{4}{3}\).
4Step 4: Combine the terms
After simplifying, our expression becomes:\(\frac{5}{2}a + \frac{4}{3}\). This is the simplified form of the original expression.
Key Concepts
Simplifying FractionsMultiplication of FractionsAlgebraic Expressions
Simplifying Fractions
Simplifying fractions is an essential skill in math that makes fractions easier to understand and work with. But what does simplifying really mean? It means reducing a fraction to its simplest form where the numerator and the denominator have no common factors other than 1.
Let's break that down a little more clearly. A fraction like \(\frac{12}{9}\) can be simplified by finding the greatest common divisor (GCD) of the numerator (12) and the denominator (9). In this case, the GCD is 3.
Let's break that down a little more clearly. A fraction like \(\frac{12}{9}\) can be simplified by finding the greatest common divisor (GCD) of the numerator (12) and the denominator (9). In this case, the GCD is 3.
- Divide the numerator by 3: \(12 \div 3 = 4\)
- Divide the denominator by 3: \(9 \div 3 = 3\)
Multiplication of Fractions
Multiplying fractions is straightforward once you know the trick! Unlike addition or subtraction, you don't need a common denominator when multiplying.
Here's the straightforward process:
Multiplication is not limited to fractions alone. If you come across people talking about distributing a number across terms, it essentially involves multiplication, but it just looks a little different when you apply it.
Here's the straightforward process:
- Multiply the numerators (top numbers) together.
- Multiply the denominators (bottom numbers) together.
Multiplication is not limited to fractions alone. If you come across people talking about distributing a number across terms, it essentially involves multiplication, but it just looks a little different when you apply it.
Algebraic Expressions
Algebraic expressions represent numbers using symbols and variables, like \(a\), to communicate unknown or variable amounts. They're like recipes where you find out not only what ingredients, but how much you need.
When dealing with expressions such as \(3(\frac{5}{6}a + \frac{4}{9})\), we use the distributive property to take a number outside of a parenthesis and multiply it with each term inside the parenthesis.
Both understanding and applying operations like these help foster a deeper comprehension of algebra, making equations easier to solve and understand. This is the core power of algebraic expressions in mathematics.
When dealing with expressions such as \(3(\frac{5}{6}a + \frac{4}{9})\), we use the distributive property to take a number outside of a parenthesis and multiply it with each term inside the parenthesis.
- You multiply \(3\) with \(\frac{5}{6}a\), leading to \(\frac{15}{6}a\).
- Then, multiply \(3\) with \(\frac{4}{9}\), giving \(\frac{12}{9}\).
Both understanding and applying operations like these help foster a deeper comprehension of algebra, making equations easier to solve and understand. This is the core power of algebraic expressions in mathematics.
Other exercises in this chapter
Problem 59
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