Problem 59
Question
Solve the formula for the specified variable. \(E K+L=D-T K\) for \(K\)
Step-by-Step Solution
Verified Answer
\(K = \frac{D - L}{E + T}\).
1Step 1: Move terms involving K to one side
To solve for \(K\), first, we need to bring all the terms containing \(K\) to one side of the equation. Start by adding \(TK\) to both sides to remove it from the right side: \[ EK + TK + L = D \]
2Step 2: Isolate the K terms
Next, we need to isolate all terms that contain \(K\). Here, we already have the terms with \(K\) grouped together: \[ (E + T)K + L = D \] Now, subtract \(L\) from both sides to isolate the \(K\) terms: \[ (E + T)K = D - L \]
3Step 3: Solve for K
Now that all \(K\) terms are grouped together, divide both sides of the equation by \(E + T\) to solve for \(K\): \[ K = \frac{D - L}{E + T} \]
Key Concepts
Variable IsolationAlgebraic ManipulationEquation Rearrangement
Variable Isolation
To find the value of a specific variable, we often aim to isolate it on one side of the equation. This means separating the variable from other terms. In our exercise, we aim to solve for the variable \( K \).
The first step is to gather all terms involving \( K \) on one side. This requires moving the terms using opposite operations. If you encounter subtraction, you add the same value on both sides; if faced with addition, you subtract it instead.
In our example, the term \( TK \) is on the right side of the equation. We need it on the left side, so we add \( TK \) to both sides. This action leaves us with:
The first step is to gather all terms involving \( K \) on one side. This requires moving the terms using opposite operations. If you encounter subtraction, you add the same value on both sides; if faced with addition, you subtract it instead.
In our example, the term \( TK \) is on the right side of the equation. We need it on the left side, so we add \( TK \) to both sides. This action leaves us with:
- \( EK + TK + L = D \)
- The key is balancing the equation after each operation.
Algebraic Manipulation
Once you've gathered all the terms involving the variable you're interested in, focus on simplifying the equation as much as possible. This process is called algebraic manipulation, and it's essential for making the equation easier to work with.
Looking at our equation:
For this problem, we subtract \( L \) from both sides to isolate \( K \) terms, transforming the equation into:
Looking at our equation:
- \( (E + T)K + L = D \)
For this problem, we subtract \( L \) from both sides to isolate \( K \) terms, transforming the equation into:
- \( (E + T)K = D - L \)
Equation Rearrangement
Finding the solution often involves rearranging the equation into a form that directly gives us the variable we need. In our problem, the ultimate goal is to solve for \( K \), which involves dividing by the coefficient of \( K \).
In the equation from our previous step:
In the equation from our previous step:
- \( (E + T)K = D - L \)
- \( K = \frac{D - L}{E + T} \)
- Check your manipulation steps to ensure each preserves the equation's balance.
- This final step is crucial as it simplifies the answers into their simplest form.
Other exercises in this chapter
Problem 58
Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ 2|-11-7 x|-2>10 $$
View solution Problem 58
Exer. 57-62: Verify the property. $$ \overline{z-w}=\bar{z}-\bar{w} $$
View solution Problem 59
The power \(P\) (in watts) generated by a windmill that has efficiency \(E\) is given by the formula \(P=0.31 E D^{2} V^{3}\), where \(D\) is the diameter (in f
View solution Problem 59
Exer. 21-70: Solve the inequality, and express the solutions in terms of intervals whenever possible. $$ |7 x+2|>-2 $$
View solution