Round #641

Find x: 5 times the quantity x minus 3 equals 20

To solve the equation \(5(x - 3) = 20\), we need to isolate \(x\). Here are the steps: 1. **Divide both sides by 5**: \[ \frac{5(x - 3)}{5} = \frac{20}{5} \] Simplifying this, we get: \[ x - 3 = 4 \] 2. **Add 3 to both sides**: \[ x - 3 + 3 = 4 + 3 \] Simplifying this, we get: \[ x = 7 \] So, the solution is \(x = 7\).

To solve the equation \(5x - 3 = 20\), we need to isolate \(x\). Here's how you can do it step-by-step: 1. Add 3 to both sides of the equation: \[ 5x - 3 + 3 = 20 + 3 \] Simplifying this, we get: \[ 5x = 23 \] 2. Divide both sides by 5: \[ x = \frac{23}{5} \] So, the solution to the equation \(5x - 3 = 20\) is \(x = \frac{23}{5}\). Here's how you can implement this in C#: ```csharp using System; class Program { static void Main() { try { double x = SolveEquation(); Console.WriteLine($"The solution to the equation 5x - 3 = 20 is x = {x}"); } catch (Exception ex) { Console.WriteLine($"An error occurred: {ex.Message}"); } } static double SolveEquation() { // Coefficients from the equation 5x - 3 = 20 double coefficientA = 5; double constantB = -3; double resultC = 20; // Solving for x: (resultC - constantB) / coefficientA return (resultC - constantB) / coefficientA; } } ``` ### Explanation: 1. **Main Method**: This is the entry point of the program where we call the `SolveEquation` method and print the result. 2. **SolveEquation Method**: This method calculates the value of \(x\) by isolating it using the steps derived from the equation. This implementation is straightforward and follows best practices in C#, including proper error handling with a try-catch block to catch any exceptions that might occur during execution.