Problem 39 - College Algebra

Question

Consider the relationship \(3 r+2 t=18\). a. Write the relationship as a function \(r=f(t)\). b. Evaluate \(f(-3)\). c. Solve \(f(t)=2\).

Step-by-Step Solution

Answer

a. \( f(t) = \frac{18 - 2t}{3} \); b. \( f(-3) = 8 \); c. \( t = 6 \).

Step 1: Isolate r

To express the relationship as a function with respect to \(t\), we need to solve the equation \(3r + 2t = 18\) for \(r\). Start by isolating \(3r\) on one side:\[ 3r = 18 - 2t \]

Step 2: Divide by 3

Next, divide both sides by \(3\) to completely isolate \(r\):\[ r = \frac{18 - 2t}{3} \]Therefore, the function can be expressed as:\[ f(t) = \frac{18 - 2t}{3} \]

Step 3: Substitute and Evaluate f(-3)

To find \(f(-3)\), substitute \(-3\) in place of \(t\) in the function:\[ f(-3) = \frac{18 - 2(-3)}{3} \]Simplify the expression:\[ f(-3) = \frac{18 + 6}{3} = \frac{24}{3} = 8 \]

Step 4: Set the function equal to 2

For part (c), solve the equation \(f(t) = 2\). Start by setting the function equal to 2:\[ \frac{18 - 2t}{3} = 2 \]

Step 5: Multiply both sides by 3

Clear the fraction by multiplying both sides by 3:\[ 18 - 2t = 6 \]

Step 6: Solve for t

Isolate \(t\) by first subtracting 18 from both sides:\[ -2t = 6 - 18 \]Simplify:\[ -2t = -12 \]Now divide both sides by -2:\[ t = 6 \]

Key Concepts

Linear Equations

Linear equations are the building blocks of algebra and are essential for understanding the relationships between variables. In a linear equation, the highest power of the variable is 1, and the equation forms a straight line when graphed.

The general form of a linear equation can be expressed as \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. The equation \(3r + 2t = 18\) is a perfect example of a linear equation. Here, both \(r\) and \(t\) are variables. The goal is often to solve for one variable in terms of the other, making the equation easier to work with or understand.

To transform such an equation into a function form, isolate one variable on one side of the equation, as seen in the solution where \(r\) is expressed as a function of \(t\) (\(r = f(t)\)).

Function Evaluation

Function evaluation involves finding the value of a function for a given input. Once a function is defined, such as \(f(t) = \frac{18 - 2t}{3}\), you can find specific values of the output by substituting different values of the input into this equation.

For example, to evaluate the function \(f\) at \(t = -3\), substitute \(-3\) in place of \(t\):

• Replace \(t\) with \(-3\): \(f(-3) = \frac{18 - 2(-3)}{3}\).
• Simplify the expression step-by-step: First, multiply \(-2\) by \(-3\), getting \(6\).
• Add this result to \(18\): \(18 + 6 = 24\).
• Finally, divide by \(3\): \(\frac{24}{3} = 8\).

Thus, \(f(-3) = 8\). Evaluation helps understand how changes in \(t\) affect the result \(r\).

Solving Equations

Solving equations is a core part of algebra and involves finding the value of the variable that makes the equation true. Once you have the function \(f(t) = \frac{18 - 2t}{3}\), setting this equal to a number allows you to determine the input that produces that output.

For instance, to solve \(f(t) = 2\), set the function equal to 2:

• Start with \(\frac{18 - 2t}{3} = 2\).
• Multiply both sides by 3 to eliminate the fraction: \(18 - 2t = 6\).
• Isolate \(t\) by subtracting 18 from both sides: \(-2t = -12\).
• Divide by \(-2\) to find \(t\): \(t = 6\).

In each step, perform inverse operations to both sides, ensuring you maintain the equation's balance. This approach helps uncover the 'mystery' value that satisfies the original relationship.