Problem 49 - Precalculus: Functions and Graphs

Question

Solve the formula for the specifled variable. \(E K+L=D-T K\) for \(K\)

Step-by-Step Solution

Answer

\( K = \frac{D - L}{E + T} \)

Step 1: Isolate terms containing K

First, we want all terms with the variable \(K\) on one side of the equation and all other terms on the opposite side. We have the equation \(E K + L = D - T K\). To start, let's add \(T K\) to both sides to bring the \(K\) terms together: \[ E K + T K + L = D \].

Step 2: Factor out K

After isolating the \(K\) terms on one side, the equation now looks like this: \[ E K + T K + L = D \]. Next, factor \(K\) out of the two terms on the left: \[ K(E + T) + L = D \].

Step 3: Solve for K

To solve for \(K\), subtract \(L\) from both sides to get: \[ K(E + T) = D - L \]. Finally, divide both sides by \(E + T\): \[ K = \frac{D - L}{E + T} \].

Key Concepts

Equation Solving

Solving equations is like solving puzzles. The goal is to figure out what the unknowns in the equation represent. When you have an equation like \( E K + L = D - T K \), your task is to find the value of \( K \). To find a variable, you need to perform operations that maintain the balance of the equation. Think of an equation as a balance scale. Whatever you do to one side, you have to do to the other to keep it balanced.

When solving equations, you often perform a combination of addition, subtraction, multiplication, and division. These operations help to isolate the variable you are solving for. This takes us directly to the next point, isolating variables!

Isolating Variables

Isolating variables is a crucial step in algebra. It involves manipulating the equation until the variable you're interested in appears alone on one side of the equation. Using our equation example, \( E K + L = D - T K \), we want to isolate \( K \).

To start, move all terms containing \( K \) to one side of the equation:

• Add \( T K \) to both sides: \( E K + T K + L = D \). This step ensures all the \( K \) terms are together.
• Then, get rid of constants or terms without \( K \) from this side. That would leave you with only \( K \)-related terms on one side.

Isolating variables helps set up the equation for potentially solving it by factorization.

Factorization

Factorization in algebra simplifies equations by pulling out common factors. It’s very handy when you're left with a term that multiplies the variable multiple times. In the equation \( E K + T K + L = D \), notice how \( K \) appears in both terms. By factoring, we take \( K \) outside:

• You can rewrite this as \( K(E + T) + L = D \).
• It makes the problem simpler by reducing repetitive terms: you no longer have separate \( K \) terms.

This process of factorization is important because it often turns complex problems into much easier ones to address. Once factored, it's easier to manipulate the equation or make substitutions if necessary, especially when dealing with a system of equations.

Linear Equations

Linear equations are the backbone of algebra. They represent constant relations between variables. The term 'linear' indicates that the variable has no exponents other than 1. In our worked example, \( K(E + T) = D - L \) is a straightforward linear equation since it can be expressed in the form \( ax = b \).

Understanding linear equations helps you understand how changes in one variable affect another. Additionally, solving linear equations frequently includes identifying proportional relationships or directly solving for an unknown.


The linearity means there's a constant rate of change. To solve, you divide each side by the factor accompanying your variable: \( K = \frac{D - L}{E + T} \). This kind of equation solving is straightforward, and mastering it paves the way for tackling more complex algebraic problems.