Problem 47 - Fundamentals of Mathematics

Question

For problems \(47-56\), simplify each expression by combining like terms. $$ 4 a+5-2 a+1 $$

Step-by-Step Solution

Answer

The simplified expression is \(2a + 6\).

Step 1: Identify Like Terms

Before combining, we need to identify which terms are similar. Like terms are terms that have the same variable raised to the same power. In the expression \(4a + 5 - 2a + 1\), the like terms with the variable \(a\) are \(4a\) and \(-2a\). The constant terms are \(5\) and \(1\).

Step 2: Combine the Like Terms with 'a'

Now, we combine the terms with the variable \(a\). Add \(4a\) and \(-2a\) together: \[ 4a - 2a = 2a \]

Step 3: Combine the Constant Terms

Next, we combine the constant numbers. Add \(5\) and \(1\) together: \[ 5 + 1 = 6 \]

Step 4: Write the Simplified Expression

Combine the results from the previous steps to write the simplified expression. The term with \(a\) is \(2a\) and the constant is \(6\). Thus the simplified expression is: \[ 2a + 6 \]

Key Concepts

Simplifying Expressions

Simplifying expressions in algebra is like tidying up a room. The goal is to make the expression as neat and concise as possible. When you're faced with an expression such as \(4a + 5 - 2a + 1\), you're dealing with a mix of terms.**Why Simplify?**

• It makes the expression easier to understand.
• Simplified expressions are easier to use in further calculations.

Expressions often consist of multiple terms that may or may not be like terms. Let's break down these terms:- **Like Terms**: These are terms that have the same variables raised to the same power. In our example, \(4a\) and \(-2a\) are like terms.- **Constants**: These are numbers on their own, like 5 and 1 in the example, without any variables.When simplifying, you combine like terms and constants separately. Add or subtract the coefficients of like terms. Here, \(4a - 2a\) becomes \(2a\). Then, add the constants together to get \(5 + 1 = 6\). The neat and tidy room is represented by the simplified expression, \(2a + 6\).

Algebra

Algebra is a branch of mathematics that deals with symbols, usually letters, and the rules for manipulating these symbols. These symbols represent numbers and what may vary or be unknown.**Key Aspects of Algebra**

• **Variables**: These are symbols used to represent unknown values. From our example, \(a\) is a variable.
• **Constants**: Numbers that have fixed values. In the expression, 5 and 1 are constants.
• **Operations**: Includes addition, subtraction, multiplication, and division. In \(4a + 5 - 2a + 1\), you find addition and subtraction at play.

Algebra allows for the representation of real-life situations using mathematical sentences. In algebra, expressions like \(4a + 5\) can be manipulated and simplified by following specific rules. This makes algebra a powerful tool since it enables solving equations that can explain or predict real-world phenomena such as the movement of cars, personal finance management, and even predicting future trends.

Constants and Variables

In algebra, understanding constants and variables is crucial.**Constants**A constant is basically a number on its own that doesn't change its value. In the expression \(4a + 5 - 2a + 1\), both 5 and 1 are constants. They remain the same regardless of the value of other variables around them.**Variables**On the other hand, a variable is a symbol, often a letter like \(a\), representing a number whose exact value isn't yet known. Variables are placeholders that can take on different values, depending on the situation or problem.**How They Work Together**In an expression, constants and variables interact through mathematical operations:

• Variables are combined based on their coefficients when they're like terms.
• Constants are combined through basic arithmetic operations.

For the expression \(4a + 5 - 2a + 1\), the variable terms \(4a\) and \(-2a\) are combined to yield \(2a\). Meanwhile, the constant terms 5 and 1 sum up to 6. The interplay of constants and variables results in the simplified form \(2a + 6\). By understanding these two basic components, students can efficiently approach a variety of algebraic problems.