Problem 58 - Intermediate Algebra

Question

Multiply. See Example 4 $$6\left(-\frac{4}{3}+\frac{7}{6} s+\frac{16}{3} t\right)$$

Step-by-Step Solution

Answer

The final expression is \(-8 + 7s + 32t\).

Step 1: Distribute the first term

Distribute the 6 to the first term inside the parentheses, \(-\frac{4}{3}\). Multiply 6 by \(-\frac{4}{3}\): \\[\ 6\cdot\left(-\frac{4}{3}\right) = -8\]

Step 2: Distribute the second term

Distribute the 6 to the second term inside the parentheses, \(\frac{7}{6} s\). Multiply 6 by \(\frac{7}{6} s\):\[\ 6\cdot\left(\frac{7}{6} s\right) = 7s\]

Step 3: Distribute the third term

Distribute the 6 to the third term inside the parentheses, \(\frac{16}{3} t\). Multiply 6 by \(\frac{16}{3} t\):\[\ 6\cdot\left(\frac{16}{3} t\right) = 32t\]

Step 4: Combine all distributed terms

Combine the results from each distribution step to write the final expression:\( -8 + 7s + 32t \).

Key Concepts

Understanding the Distributive Property

When working with algebraic expressions, the distributive property is a valuable tool. It allows us to distribute a multiplier across terms inside parentheses. In simple words, you multiply the factor outside the parentheses by each term inside. This property is especially useful when dealing with expressions that have fractions or variables.

The basic formula for the distributive property is: \[ a(b + c) = ab + ac \].

In our exercise, the factor outside the parentheses is 6, and inside we have multiple terms: \(-\frac{4}{3}\), \(\frac{7}{6} s\), and \(\frac{16}{3} t\). By applying the distributive property, we multiply each of these terms by 6 individually. This step ensures that each term is distributed and expanded correctly. Using the distributive property properly helps simplify complex expressions and solve equations efficiently. Remember, the goal is to make the math simpler and more manageable.

How to Multiply Fractions

Multiplying fractions might seem daunting, but it's simpler than you think! It's all about multiplying the numerators and denominators respectively. If you're given two fractions like \(\frac{a}{b}\) and \(\frac{c}{d}\), you multiply them as follows: \[ \frac{a}{b} \times \frac{c}{d} = \frac{a \times c}{b \times d} \].

Often, you might deal with a whole number multiplied by a fraction. In such cases, treat the whole number as a fraction with a denominator of 1. For example, to multiply 6 by \(\frac{7}{6}\), write 6 as \(\frac{6}{1}\). Thus, \[ \frac{6}{1} \times \frac{7}{6} = \frac{6 \times 7}{1 \times 6} = \frac{42}{6} = 7 \]. This simplification is a critical step that reduces fractions to their simplest form.

• Multiply numerators together.
• Multiply denominators together.
• Simplify the resultant fraction if possible.

Don't forget to factor in any signs, as they tell you whether the resulting product will be positive or negative!

Combining Like Terms

Combining like terms is a simple yet essential operation when simplifying expressions. The process involves adding or subtracting terms that have the same variable raised to the same power.

For example, in terms like \(7s\) and \(-2s\), you can combine them because they both have the variable \(s\). Simply add or subtract their coefficients: \[ 7s - 2s = 5s \]. This yields a more compact and simplified expression.

In our exercise, after distributing the terms, we have \(-8 + 7s + 32t\). There's no need to combine further since each term is unique: \(-8\) is a constant, \(7s\) and \(32t\) are terms with distinct variables. Each stands on its own, and the expression is fully simplified.

• Identify like terms (same variable and exponent).
• Add or subtract the coefficients.
• Ensure the terms are fully simplified.

This skill helps keep large expressions in a neat and manageable form, making it easier to understand and solve algebraic problems.