Question

Solve for the indicated variable. Discount Solve for \(L\) in \(S=L-R L\).

Short answer

\(L=\frac{S}{1-R}\)

Step-by-step solution

Analyze the equation

We start with the equation \(S=L-RL\). This equation includes \(L\) in two terms, so our objective is to factor out \(L\) to isolate it on one side.

Factor out \(L\)

By factoring out \(L\) from the right side, we get \(S=L(1-R)\).

Solve for \(L\)

To isolate \(L\), divide both sides of the equation by \(1-R\). We find that \(L=\frac{S}{1-R}\).

Key concepts

Algebraic Manipulation

The journey to solving equations begins with a fundamental skill known as algebraic manipulation. This is the ability to rearrange, simplify, or change the form of an algebraic expression according to the rules of algebra. Mastering algebraic manipulation is crucial, as it sets the groundwork for solving complex equations.

When we look at an equation, our goal is often to rearrange it in such a way that we get a clearer view of the relationship between the variables involved. In the case of the provided exercise, we are dealing with a scenario in the form of a discount, where we need to solve for the original price before a reduction. To reach the solution, one should first understand how to properly expand, combine like terms, and factor expressions. This process may involve distributing multiplication over addition or subtraction, combining terms, and using properties of equality to maintain balance in the equation.

Furthermore, strategies such as 'reverse operations' come into play. Reverse operations exploit the idea that something done to one side of the equation must be done to the other to keep it balanced. For instance, if a quantity is subtracted on one side, we can reverse this by adding the same quantity to both sides. This concept is essential when trying to isolate a variable, as seen in our exercise when solving for the original price, labeled as variable 'L'.

Factoring Expressions

Moving onto the concept of factoring expressions, this method is a powerful tool in algebra that allows us to break down complex expressions into simpler parts. Factoring can be thought of as a way to simplify equations or to find what's common between different terms. It's particularly important when expressions contain common factors or when we aim to isolate certain variables.

In our discount problem, factoring was required because the variable 'L' appeared in more than one term. By identifying 'L' as a common factor in the expression 'L - RL', we can rewrite it in the factored form as 'L(1-R)'. This is achieved by applying the distributive property in reverse. Factoring expressions helps not only in simplifying and solving equations but also in understanding the underlying relationships between variables. It is a pivotal step that lays the groundwork for the following operations to isolate the variable, which in our original problem was the final goal.

Isolate Variable

The climax of solving equations is often when we isolate the variable of interest, meaning we manipulate the equation to get the variable by itself on one side of the equality sign. It's like finding the character of a story who holds the key to the entire plot.

In the context of our exercise on solving for 'L' in the equation 'S=L - RL', isolating 'L' requires us to eliminate all other terms on the same side as 'L'. After factoring, we obtained 'S = L(1-R)'. To get 'L' on its own, we perform the reverse operation of multiplication, which is division. Therefore, we divide both sides by '(1-R)' leading us to our isolated variable, 'L = S/(1-R)'.

This step is essential in obtaining the solution to the equation. Isolating the variable allows us to see the direct relationship between 'S' (the sale price, in this case) and the variable 'L' (the original price), thus giving us the clear answer we seek. By mastering the skill of isolating a variable, students can tackle a wide range of algebra problems, as it is a fundamental aspect of solving equations in mathematics.