Problem 19
Question
Solve the formula for the given variable. $$A=S w+w ; w \quad \text { (Physics) }$$
Step-by-Step Solution
Verified Answer
The solution of the given equation for \(w\) is \(w = \frac{A}{S+1}\)
1Step 1: Rewrite the equation
Rewrite the given equation \(A=Sw + w\) to \(A = w(S + 1)\). This is done by factoring out the common term \(w\) from the right side of the equation.
2Step 2: Solve for \(w\)
To isolate \(w\) on one side of the equation, perform the inverse operation of multiplication on both sides by dividing by \((S+1)\). This results in the equation \(w = \frac{A}{S+1}\)
Key Concepts
Algebraic ManipulationFactoring in AlgebraInverse Operations
Algebraic Manipulation
Mastering algebraic manipulation is fundamental when solving for a variable in any equation. Let's demystify this process using the given physics exercise as our example. First things first, what is algebraic manipulation? In essence, it's the process of rearranging and simplifying equations using a variety of mathematical operations.
Our starting equation is \(A=Sw + w\). At a glance, the variable we want to isolate, \(w\), appears to be on equal terms with \(A\) and \(S\). However, by using our algebraic tools, we can rewrite the equation. Here, we look for terms that share a common factor—in this case, \(w\). By factoring out \(w\), we effectively gather all instances of \(w\) together: \(A = w(S + 1)\). This illuminates the relationship between \(A\) and \(w\), showing how we can manipulate one to isolate the other.
Using algebraic manipulation, we tease out the underlying structure of equations, preparing them for further operations that will solve for the desired variable. As we saw here, it involves discerning patterns like common factors and using them to our advantage.
Our starting equation is \(A=Sw + w\). At a glance, the variable we want to isolate, \(w\), appears to be on equal terms with \(A\) and \(S\). However, by using our algebraic tools, we can rewrite the equation. Here, we look for terms that share a common factor—in this case, \(w\). By factoring out \(w\), we effectively gather all instances of \(w\) together: \(A = w(S + 1)\). This illuminates the relationship between \(A\) and \(w\), showing how we can manipulate one to isolate the other.
Using algebraic manipulation, we tease out the underlying structure of equations, preparing them for further operations that will solve for the desired variable. As we saw here, it involves discerning patterns like common factors and using them to our advantage.
Factoring in Algebra
When we talk about factoring in algebra, we refer to the process of breaking down complicated expressions into simpler pieces. It's akin to splitting a whole into its parts, or finding what multiplies together to give the original expression.
In the equation \(A=Sw + w\), our task is to factor out the term \(w\). Why? Because factoring provides clarity and simplicity for further manipulation. Observe how \(Sw\) and \(w\) share the same variable \(w\). By factoring out \(w\), we combine them into a single term, \(w(S + 1)\), unfolding the equation to a clearer form. It's a pivotal step, essential for isolating \(w\).
Useful tip: When factoring, look for the greatest common factor. Here, it's \(w\), but in other scenarios, it may be a number or a more complex expression. By making factoring second nature, solving equations becomes much more intuitive.
In the equation \(A=Sw + w\), our task is to factor out the term \(w\). Why? Because factoring provides clarity and simplicity for further manipulation. Observe how \(Sw\) and \(w\) share the same variable \(w\). By factoring out \(w\), we combine them into a single term, \(w(S + 1)\), unfolding the equation to a clearer form. It's a pivotal step, essential for isolating \(w\).
Useful tip: When factoring, look for the greatest common factor. Here, it's \(w\), but in other scenarios, it may be a number or a more complex expression. By making factoring second nature, solving equations becomes much more intuitive.
Inverse Operations
Finding the inverse operation is essential in identifying the steps needed to solve an equation. They're like a dance partner to the original operation, allowing you to reverse the steps and isolate your desired variable. In our example, we're faced with the equation \(A = w(S + 1)\) after factoring, and we want to find \(w\).
Because \(w\) is multiplied by the bracket \((S+1)\), the inverse operation is division. By dividing both sides of the equation by \((S + 1)\), we reverse the multiplication and move towards our goal: \(w = \frac{A}{S+1}\). Et voilà, we've isolated \(w\) using its inverse operation, clearing the stage for it to shine alone on one side of the equal sign.
Remember: Inverse operations are not just about undoing multiplication with division; they're also about replacing addition with subtraction, and so forth. Recognizing and applying inverse operations is vital to solving any algebraic equation.
Because \(w\) is multiplied by the bracket \((S+1)\), the inverse operation is division. By dividing both sides of the equation by \((S + 1)\), we reverse the multiplication and move towards our goal: \(w = \frac{A}{S+1}\). Et voilà, we've isolated \(w\) using its inverse operation, clearing the stage for it to shine alone on one side of the equal sign.
Remember: Inverse operations are not just about undoing multiplication with division; they're also about replacing addition with subtraction, and so forth. Recognizing and applying inverse operations is vital to solving any algebraic equation.
Other exercises in this chapter
Problem 19
Simplify. $$\frac{2 x+3}{x^{2}-x-30}-\frac{x-2}{x^{2}-x-30}$$
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Solve. $$3+\frac{8}{n}=5$$
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Simplify. $$\frac{1-\frac{6}{x}+\frac{8}{x^{2}}}{\frac{4}{x^{2}}+\frac{3}{x}-1}$$
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Find the LCM of the polynomials. $$\begin{aligned} &x^{2}-x-6\\\ &x^{2}+x-12 \end{aligned}$$
View solution