Problem 26 - Algebra for College Students

Question

Find the value of the indicated variable. Round approximate answers to three decimal places. $$\text { Find } p \text { if } f=2.3, q=1.7, \text { and } \frac{1}{p}+\frac{1}{q}=\frac{1}{f}$$

Step-by-Step Solution

Answer

p \approx -6.536

Step 1: Write the given equation

Start with the given equation: \[ \frac{1}{p} + \frac{1}{q} = \frac{1}{f} \] Substitute the given values \( f = 2.3 \) and \( q = 1.7 \).

Step 2: Substitute known values

Substitute \( f = 2.3 \) and \( q = 1.7 \) into the equation: \[ \frac{1}{p} + \frac{1}{1.7} = \frac{1}{2.3} \]

Step 3: Simplify the equation

Isolate \( \frac{1}{p} \) by moving \( \frac{1}{1.7} \) to the other side of the equation: \[ \frac{1}{p} = \frac{1}{2.3} - \frac{1}{1.7} \]

Step 4: Calculate right-hand side

Compute \( \frac{1}{2.3} \) and \( \frac{1}{1.7} \): \[ \frac{1}{2.3} \approx 0.435 \] \[ \frac{1}{1.7} \approx 0.588 \] Then, substitute these values into the equation: \[ \frac{1}{p} = 0.435 - 0.588 \]

Step 5: Perform the subtraction

Subtract the values: \[ \frac{1}{p} = 0.435 - 0.588 = -0.153 \]

Step 6: Solve for p

Invert \( -0.153 \): \[ p = \frac{1}{-0.153} \approx -6.536 \] Round to three decimal places.

Key Concepts

algebraic manipulation

Algebraic manipulation is essential when solving rational equations. This involves rearranging terms and simplifying expressions to isolate the variable you need to solve for.
In our exercise, we begin with the equation: \[ \frac{1}{p} + \frac{1}{q} = \frac{1}{f} \] First, we substitute the given values for \( f \) and \( q \): \[ \frac{1}{p} + \frac{1}{1.7} = \frac{1}{2.3} \] Our goal is to isolate \( \frac{1}{p} \).
Thus, we move \( \frac{1}{1.7} \) to the other side by subtracting it: \[ \frac{1}{p} = \frac{1}{2.3} - \frac{1}{1.7} \]Understanding how to manipulate and rearrange these terms is crucial for solving equations in algebra.

substitution

In algebra, substitution involves replacing variables with their known values to simplify the equation.
In this problem, we are given that \( f = 2.3 \) and \( q = 1.7 \).
Starting with the initial equation: \[ \frac{1}{p} + \frac{1}{q} = \frac{1}{f} \] We substitute the known values: \[ \frac{1}{p} + \frac{1}{1.7} = \frac{1}{2.3} \] This step is essential because it transforms an abstract equation into one with specific numerical values.
This makes the subsequent steps of solving the equation much more straightforward.

fractions and decimals

Understanding fractions and decimals is key when working with rational equations.
In our example, we need to calculate \( \frac{1}{2.3} \) and \( \frac{1}{1.7} \).
These values are: \[ \frac{1}{2.3} \approx 0.435 \] \[ \frac{1}{1.7} \approx 0.588 \] These calculations convert fractions to decimals, making the next steps easier. We then substitute these back into the equation: \[ \frac{1}{p} = 0.435 - 0.588 \] It's important to be comfortable switching between fractions and decimals, as it helps in performing the required operations.

inverse operations

Inverse operations allow you to undo actions performed on a variable. In this problem, we perform subtraction followed by inversion to solve for \( p \).
First, we subtract 0.588 from 0.435: \[ \frac{1}{p} = 0.435 - 0.588 \] This results in: \[ \frac{1}{p} \approx -0.153 \] To solve for \( p \), we perform the inverse of taking a reciprocal by inverting \( -0.153 \): \[ p = \frac{1}{-0.153} \approx -6.536 \] Thus, \( p \) is approximately -6.536.
Understanding and using inverse operations, such as subtraction and inversion, are crucial for solving equations correctly.