Chapter 13: Problem 3
If \(b \neq 1\) and if \(\frac{a b}{a-b}=1\), what is the value of \(a\) in terms of \(b ?\)
Short Answer
Expert verified
The value of \(a\) in terms of \(b\) is \frac{-b}{b-1}.\
Step by step solution
01
Write Down the Given Equation
The given equation is \(\frac{ab}{a-b}=1.\)
02
Eliminate the Fraction
Multiply both sides of the equation by \(a-b\) to get rid of the fraction: \(ab = 1(a - b).\)
03
Distribute and Simplify
Distribute on the right-hand side to get: \(ab = a - b.\)
04
Rearrange the Equation
Move all terms involving \(a\) to one side and constant terms to the other: \(ab - a = -b.\)
05
Factor Out the Common Term
Factor out \(a\) from the left-hand side: \(a(b - 1) = -b.\)
06
Solve for \(a\)
Divide both sides by \(b-1\): \(a = \frac{-b}{b-1}.\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation Simplification
Understanding equation simplification is crucial in solving algebraic problems. It involves using arithmetic operations and algebraic manipulations to transform equations into simpler forms that are easier to solve. Simplification often includes combining like terms, expanding expressions, and eliminating fractions.
In the given exercise, we start with the equation \(\frac{ab}{a-b} = 1\). To simplify, we need to eliminate the fraction. We do this by multiplying both sides by the denominator \(a - b\), giving us:\(ab = 1(a - b)\).
This step gets rid of the denominator and makes the equation easier to work with.
Finally, it's important to always maintain the equation balance by performing the same operation on both sides. This ensures that the equation remains valid throughout the simplification process.
In the given exercise, we start with the equation \(\frac{ab}{a-b} = 1\). To simplify, we need to eliminate the fraction. We do this by multiplying both sides by the denominator \(a - b\), giving us:\(ab = 1(a - b)\).
This step gets rid of the denominator and makes the equation easier to work with.
Finally, it's important to always maintain the equation balance by performing the same operation on both sides. This ensures that the equation remains valid throughout the simplification process.
Factoring in Algebra
Factoring is a technique used to break down complex expressions into simpler ones that can be easily solved. It's particularly useful in solving quadratic equations and other polynomials. In this exercise, we use factoring to isolate the variable.
After simplifying to \(ab = a - b\), we rearrange the terms to bring all the terms involving \(a\) to one side and constants to the other, resulting in \(ab - a = -b\).
Next, we factor out the common term \(a\):\(a(b - 1) = -b\).
This step is crucial because it allows us to express the equation in a form where we can easily isolate \(a\). By factoring, we simplify the equation further, making the next steps much more straightforward.
After simplifying to \(ab = a - b\), we rearrange the terms to bring all the terms involving \(a\) to one side and constants to the other, resulting in \(ab - a = -b\).
Next, we factor out the common term \(a\):\(a(b - 1) = -b\).
This step is crucial because it allows us to express the equation in a form where we can easily isolate \(a\). By factoring, we simplify the equation further, making the next steps much more straightforward.
Solving for a Variable
Solving for a variable means isolating the unknown variable on one side of the equation so we can find its value. This often involves inverse operations like addition, subtraction, multiplication, and division.
Let's solve for \(a\) in the equation \(a(b-1) = -b\). To isolate \(a\), we divide both sides by \(b-1\):\(a = \frac{-b}{b-1}\).
This step is essential because it gives us \(a\) expressed in terms of \(b\), which is what the exercise is asking for.
It's important to remember the given condition \(b eq 1\), as this ensures the denominator \(b-1\) is not zero, keeping the equation valid and well-defined.Understanding these key concepts—equation simplification, factoring, and solving for a variable—can help you tackle a wide range of algebraic problems with confidence.
Let's solve for \(a\) in the equation \(a(b-1) = -b\). To isolate \(a\), we divide both sides by \(b-1\):\(a = \frac{-b}{b-1}\).
This step is essential because it gives us \(a\) expressed in terms of \(b\), which is what the exercise is asking for.
It's important to remember the given condition \(b eq 1\), as this ensures the denominator \(b-1\) is not zero, keeping the equation valid and well-defined.Understanding these key concepts—equation simplification, factoring, and solving for a variable—can help you tackle a wide range of algebraic problems with confidence.