Problem 75
Question
The formula occurs in the indicated application. Solve for the specified variable. \(\frac{1}{f}=\frac{1}{p}+\frac{1}{q}\) for \(q\)
Step-by-Step Solution
Verified Answer
The solution for \( q \) is \( q = \frac{fp}{p-f} \).
1Step 1: Set Up the Equation for q
We start with the given equation \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \). Our goal is to solve for the variable \( q \). To isolate \( \frac{1}{q} \) on one side of the equation, we need to subtract \( \frac{1}{p} \) from both sides: \[ \frac{1}{q} = \frac{1}{f} - \frac{1}{p} \].
2Step 2: Simplify the Right Side
To find a common denominator for the fractions on the right side, we use \( fp \) (the product of \( f \) and \( p \)): \[ \frac{1}{q} = \frac{p}{fp} - \frac{f}{fp} = \frac{p-f}{fp} \].
3Step 3: Reciprocal to Solve for q
Now that we have \( \frac{1}{q} = \frac{p-f}{fp} \), we can solve for \( q \) by taking the reciprocal of both sides: \[ q = \frac{fp}{p-f} \].
Key Concepts
Solving for a variableReciprocalsFractions in equations
Solving for a variable
In algebra, solving for a variable means finding the value of the unknown that satisfies the equation. It's like unwrapping a present to see what's inside. Let's look at this operation step by step. Here we have the equation \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \) and we need to isolate \( q \). This means rearranging the equation so that \( q \) is on one side by itself.
To do this, algebra allows us to perform the same operation on both sides of the equation. It becomes like a balance, ensuring no side gets heavier. In this case, subtract \( \frac{1}{p} \) from both sides, and you will have the expression \( \frac{1}{q} = \frac{1}{f} - \frac{1}{p} \).
Remember, solving for a variable often requires putting it by itself, and balancing the equation by using operations like addition, subtraction, multiplication, or division in both sides. Keep practicing and it becomes as simple as second nature!
To do this, algebra allows us to perform the same operation on both sides of the equation. It becomes like a balance, ensuring no side gets heavier. In this case, subtract \( \frac{1}{p} \) from both sides, and you will have the expression \( \frac{1}{q} = \frac{1}{f} - \frac{1}{p} \).
Remember, solving for a variable often requires putting it by itself, and balancing the equation by using operations like addition, subtraction, multiplication, or division in both sides. Keep practicing and it becomes as simple as second nature!
Reciprocals
Reciprocals can seem tricky at first, but they're actually quite intuitive. The reciprocal of a number is what you multiply with that number to get 1. For instance, the reciprocal of \( 2 \) is \( \frac{1}{2} \).
In our problem, we ended up with \( \frac{1}{q} = \frac{p-f}{fp} \). To finish solving for \( q \), we need to "flip" both sides, finding the reciprocal of both the left and the right sides. It's like looking at yourself in the mirror but only flipping one way.
This action will deliver \( q = \frac{fp}{p-f} \). Taking reciprocals is a fundamental skill in solving rational equations, especially when you want to flip the fraction upside-down to solve for an unknown variable. Never hesitate to "flip" when you have everything right in a fraction form!
In our problem, we ended up with \( \frac{1}{q} = \frac{p-f}{fp} \). To finish solving for \( q \), we need to "flip" both sides, finding the reciprocal of both the left and the right sides. It's like looking at yourself in the mirror but only flipping one way.
This action will deliver \( q = \frac{fp}{p-f} \). Taking reciprocals is a fundamental skill in solving rational equations, especially when you want to flip the fraction upside-down to solve for an unknown variable. Never hesitate to "flip" when you have everything right in a fraction form!
Fractions in equations
Fractions are parts of a whole and can make solving equations a bit more challenging.
They require careful handling, but don't worry; there's a clear path to follow. When dealing with fractions in equations like \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \), remember the importance of finding a common denominator.
In our exercise, to subtract \( \frac{1}{p} \) from \( \frac{1}{f} \), we needed to align them under a shared denominator, which comes naturally as \( fp \) (the product of \( f \) and \( p \)). Once both fractions are under the same denominator, they become easier to subtract or add.
Remember these helpful tips:
They require careful handling, but don't worry; there's a clear path to follow. When dealing with fractions in equations like \( \frac{1}{f} = \frac{1}{p} + \frac{1}{q} \), remember the importance of finding a common denominator.
In our exercise, to subtract \( \frac{1}{p} \) from \( \frac{1}{f} \), we needed to align them under a shared denominator, which comes naturally as \( fp \) (the product of \( f \) and \( p \)). Once both fractions are under the same denominator, they become easier to subtract or add.
Remember these helpful tips:
- Identify a common denominator for fractions when solving equations.
- Treat all fractions carefully - each operation must be done with precision.
- Practicing with different sets of fractions can boost your confidence!
Other exercises in this chapter
Problem 74
Exer. 73-76: Express the statement in terms of an inequality involving an absolute value. The radius \(r\) of a ball bearing must be within \(0.01\) centimeter
View solution Problem 74
When a rock is dropped from a cliff into an ocean, it travels approximately \(16 t^{2}\) feet in \(t\) seconds. If the splash is heard 4 seconds later and the s
View solution Problem 75
Exer. 73-76: Express the statement in terms of an inequality involving an absolute value. The difference of two temperatures \(T_{1}\) and \(T_{2}\) within a ch
View solution Problem 75
A company sells running shoes to dealers for $$\$ 40$$ per pair if less than 50 pairs are ordered. If 50 or more pairs are ordered (up to 600 ), the price per p
View solution