Problem 110 - Intermediate Algebra

Question

Simplify each expression. $$\frac{1}{3}(6 t-9)-12[3(2 t-1)-t]$$

Step-by-Step Solution

Answer

The simplified expression is \(-58t + 33\).

Step 1: Distribute inside parentheses

First, distribute the numbers in front of the parentheses throughout the terms inside each parentheses. Start with \( \frac{1}{3}(6t - 9) \), which becomes \( \frac{1}{3} \times 6t - \frac{1}{3} \times 9 \). Calculate these values to get \( 2t - 3 \).

Step 2: Simplify the brackets

Now look at the terms within the brackets, \( 12[3(2t - 1) - t] \). First, distribute \( 3 \) across the \((2t - 1)\) to give \( 6t - 3 \). Then, subtract \( t \) from \( 6t \) to get \( 5t - 3 \).

Step 3: Distribute 12

Now distribute \( 12 \) across the simplified bracket expression \( 5t - 3 \) to result in \( 12 \times 5t - 12 \times 3 \), which equals \( 60t - 36 \).

Step 4: Combine like terms

Combine the expressions from Step 1 and Step 3: \( (2t - 3) - (60t - 36) \). Simplify to \( 2t - 3 - 60t + 36 \).

Step 5: Final Simplification

Combine like terms to simplify \( 2t - 60t + 36 - 3 \) to \( -58t + 33 \).

Key Concepts

Distributive Property

The distributive property is a handy tool in algebra, helping us to expand expressions efficiently. It involves multiplying a single term by all terms inside a parenthesis. Think of it like distributing gifts evenly to all members of a group! For example, when simplifying \( \frac{1}{3}(6t - 9) \), we distribute \( \frac{1}{3} \) to both \( 6t \) and \( -9 \). This means you multiply \( \frac{1}{3} \times 6t \), resulting in \( 2t \), and \( \frac{1}{3} \times 9 \), resulting in \( 3 \). This gives us a new, expanded expression: \( 2t - 3 \).

The distributive property helps break down complex equations into more manageable parts, which is particularly useful when dealing with multiple layers of parentheses as we see in the exercise above with the expression \( 12[3(2t - 1) - t] \). By using distributive property, you can simplify the expression inside the brackets before multiplying by \( 12 \). Applying this technique step by step ensures that no term is left out of the transformation process.

Combining Like Terms

After using the distributive property and expanding expressions, the next step is to make the expression simpler by combining like terms. Like terms are those with the same variables raised to the same powers, or constants, like \( 2t \) and \( -60t \). When simplifying an algebraic expression such as \( 2t - 60t + 36 - 3 \), focus on reducing it by grouping similar terms together.

- **Identify and Group**: Look for terms that have identical variable parts. In our example, \( 2t \) and \( -60t \) are like terms.
- **Combine**: Add or subtract the coefficients (numbers in front of the variables). So, \( 2t - 60t \) becomes \( -58t \).
- **Simplify Constants**: Similarly, process the constants by calculating \( 36 - 3 = 33 \).

Combining like terms reduces clutter in an equation, turning it into a tidy expression that is much easier to interpret and solve.

Order of Operations

The order of operations is like a roadmap that guides us in simplifying expressions in the correct sequence to avoid errors. It's commonly remembered by the acronym PEMDAS: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). In the exercise, following this order ensures every step is systematically and accurately executed.

- **Parentheses**: Begin by simplifying expressions inside parentheses and brackets using distribution or any other relevant operation.
- **Multiplication and Division**: Next, tackle any operations outside the brackets and inside when necessary, like distributing \( 12 \) in \( 12[5t - 3] \).
- **Addition and Subtraction**: Finally, proceed to simplify the expression by adding or subtracting like terms, as seen in combining \( -58t + 33 \).

By adhering to the order of operations, you ensure that each part of the expression is resolved at the right time, and nothing is out of place in your solution. This methodic approach is crucial for keeping complex algebraic simplification tasks on track.