Problem 48 - Intermediate Algebra

Question

Solve each formula for the specified variable. \(S=\frac{a-\ell r}{1-r}\) for \(r\) (from mathematics)

Step-by-Step Solution

Answer

The solution for \(r\) is \(r = \frac{S-a}{S-\ell}\).

Step 1: Multiply Both Sides by the Denominator

To eliminate the fraction, multiply both sides of the equation by \((1-r)\). This gives us: \[ S(1-r) = a - \ell r \]

Step 2: Distribute 'S' on the Left Side

Distribute \(S\) through the parentheses on the left side of the equation: \[ S - Sr = a - \ell r \]

Step 3: Rearrange the Equation

Move all terms involving \(r\) to one side and constant terms to the other side. Add \(Sr\) to both sides and subtract \(a\) from both sides: \[ S - a = Sr - \ell r \]

Step 4: Factor Out \(r\)

Notice that both terms on the right side contain \(r\). Factor \(r\) out: \[ S - a = r(S - \ell) \]

Step 5: Solve for \(r\)

Divide each side by \((S - \ell)\) to solve for \(r\): \[ r = \frac{S-a}{S-\ell} \]

Key Concepts

Solving Equations

When faced with an equation, our primary goal is to find the value of the unknown variable that makes the equation true. In the given exercise, we are solving the equation \(S = \frac{a - \ell r}{1-r}\) for the variable \(r\). The key is to isolate \(r\) on one side of the equation, making it the subject of the formula. Each step brings us closer to expressing \(r\) in terms of other variables. We began by eliminating the fraction through multiplication. This is a standard procedure when dealing with fractions in equations. By multiplying both sides by \((1-r)\), we ensured that \(r\) was part of a simpler expression. This simplification allows us to focus on rearranging other parts of the equation to isolate \(r\).

Rearranging Formulas

Rearranging formulas is an essential algebraic skill that involves changing the format of an equation without changing its meaning. In our example, the process involved several key steps to isolate \(r\).

• Firstly, we distributed \(S\) across the terms inside the parenthesis \((1-r)\), resulting in \(S - Sr\).
• Next, we moved all terms containing \(r\) to one side of the equation. This often involves using addition or subtraction to rearrange terms.
• By simplifying and combining like terms, it becomes easier to see how the formula can be interpreted differently while retaining the same relationships.

This systematic rearrangement is crucial because it lays the groundwork for isolating the variable of interest.

Factorization

Factorization is a powerful tool in algebra, used to simplify expressions and solve equations. In the context of our problem, factorization helps us to further isolate \(r\).

• Once we've arranged the terms so that all \(r\)-related terms are on one side, we notice that both these terms contain a common factor \(r\). This allows us to factor \(r\) out, condensing them into a single product \(r(S - \ell)\).
• Factoring simplifies the equation because it highlights the relationship between terms, making algebraic manipulation more straightforward.

Finally, by dividing both sides by the remaining coefficient \((S - \ell)\), we solve for \(r\). Factorization gave us a clear path to find the solution.