Problem 54 - Elementary Algebra

Question

For the following problems, solve the literal equations for the indicated variable. When directed, find the value of that variable for the given values of the other variables. Solve \(P=R-C\) for \(R\). Find the value of \(R\) when \(P=2\) 7and \(C=85\).

Step-by-Step Solution

Answer

Question: Solve the equation \(P=R-C\) for R, and find the value of R when \(P=2\) and \(C=85\). Answer: When \(P=2\) and \(C=85\), the value of \(R\) is \(87\).

Step 1: Solve the literal equation for R

In order to solve the equation P=R-C for R, we have to isolate the R term. To do this, we can add C to both sides of the equation: \(P + C = R - C + C\)

Step 2: Simplify the equation

By adding C to both sides, we cancel out the -C term on the right side of the equation, resulting in: \(P + C = R\) Now we have the equation solved for R.

Step 3: Find the value of R when P=2 and C=85

Now that the equation is solved for R, we can plug the given values of P and C into the equation to find R: \(R = P + C = 2 + 85\)

Step 4: Calculate the value of R

Add the values of P and C together to get the value of R: \(R = 2 + 85 = 87\) So, when \(P=2\) and \(C=85\), the value of \(R\) is \(87\).

Key Concepts

Algebraic Manipulation

Algebraic manipulation is a fundamental skill in solving mathematical equations, particularly when we aim to find the value of a specific variable. It involves performing operations that will rearrange the equation in a way that makes a variable easy to identify.

In our original exercise, the equation at hand was \(P = R - C\). To manipulate this equation algebraically with the goal to solve for \(R\), operations such as addition, subtraction, multiplication, and division are used strategically. Here, one algebraic manipulation involved adding \(C\) to both sides of the equation to move it from one side to the other. This process keeps the equation balanced, as whatever is done to one side must be done to the other according to the properties of equality.

This manipulation eliminated the \(C\) from the right side, leaving \(R\) isolated which brings us to the next important concept – isolating the variable.

Isolating Variables

Isolating the variable, in the context of our exercise, means restructuring the equation so that \(R\), the variable of interest, stands alone on one side of the equals sign. This is achieved after performing algebraic manipulation. Isolating the variable is key in solving literal equations—which are equations with multiple variables—for a particular variable.

Once \(C\) was added to both sides of the original equation, it isolated \(R\) on the right side in the form of \(P + C = R\). This explicit expression allows us to directly determine the value of \(R\) given any values for \(P\) and \(C\), and simplifies the substitution of these values into the equation. Through proper isolation of variables, we can clearly see the relationship between variables and how they influence each other.

Substituting Values

Substituting values into an equation is a step that comes after isolating the desired variable. This step involves replacing the variables with actual numbers to calculate a specific outcome. In the case of our exercise, after isolating \(R\), the next task was to substitute the given values of \(P\) and \(C\) into the equation.

Substitution would look like this: \(R = P + C = 2 + 85\). Execute this step with care, since errors in substitution can lead to incorrect answers. After substitution, we performed a basic arithmetic operation to find that \(R = 87\). This step concluded the process by providing a concrete value for \(R\) based on the specific values of \(P\) and \(C\) given in the problem. It illustrated the practical use of solving literal equations in exercises or real-world scenarios where such values may represent measures or quantities in various contexts.