Problem 1 - Elementary Algebra

Question

\(f=32 a\). Determine the value of \(f\) if \(a=6\)

Step-by-Step Solution

Answer

Answer: When \(a = 6\), the value of the function \(f = 32a\) is \(f = 192\).

Step 1: Write down the function

We are given the function \(f = 32a\).

Step 2: Substitute the given value of \(a\)

We are given that \(a=6\). Substitute this value into the function: \(f=32(6)\).

Step 3: Simplify the expression

Now, simplify the expression: \(f=32(6)=192\).

Step 4: Write down the answer

So, when \(a=6\), the value of \(f\) is \(f=192\).

Key Concepts

Function Evaluation

When we talk about function evaluation in algebra, we are referring to the process of determining the value of a function for a specific input. Generally, functions are like machines that take an input, do some calculations, and then spit out an output. In our example with the function \(f = 32a\), \(f\) represents the output and \(a\) is the input variable.
To evaluate the function, we need to know the input value. In this case, we are given that \(a = 6\). By inserting this value into the function, we calculate the output which is \(f\) in this case. This is what is meant by evaluating a function: entering a specific value for your input and calculating the corresponding output.

• Identify the input variable in the function.
• Substitute this variable with the given number or value.
• Calculate the function's output based on this substitution.

Function evaluation is an essential skill in algebra since it helps you find outputs for varying inputs, confirming the versatility and flexibility of algebraic expressions.

Substitution Method

The substitution method is a straightforward and powerful strategy used in algebra. It involves replacing a variable in an equation with a given number, value, or another expression. This process allows us to simplify and solve equations, making calculations much easier.
Let's look at our function \(f = 32a\) again. We know \(a = 6\). By substituting \(6\) for \(a\), we transform our algebraic expression from \(f = 32a\) to \(f = 32(6)\). This substitution changes the problem from an algebraic expression to a straightforward multiplication calculation, which is much simpler to solve.

• Identify the variable to substitute with a given value.
• Replace the variable with the given value carefully.
• Re-evaluate the equation or expression to find the solution.

By using substitution, we can handle more complex calculations by breaking them down into manageable parts. This method is also useful for solving systems of equations and making algebraic proofs more understandable.

Multiplication in Algebra

Multiplication in algebra involves scaling a number or an algebraic expression by another number. When we multiply, we combine multiple identical groups, which is a key operation in simplifying algebraic expressions and solving equations.
In our example \(f = 32(6)\), multiplication takes the coefficient, which is \(32\), and the variable term, which is \(6\), and combines them to produce a single number. The product in this case is \(f = 192\), which tells you how many units of \(32\) are needed to reach \(192\).

• To multiply, align the coefficients and variables.
• Multiply the numbers directly (e.g., \(32 \times 6 = 192\)).
• Verify that the operations are correctly ordered, especially when dealing with complex expressions.

Understanding multiplication in algebra is crucial, as it helps simplify expressions, solve equations, and understand mathematical patterns and relationships effectively.