Problem 16 - Algebra and Trigonometry

Question

The given equation is either linear or equivalent to a linear equation. Solve the equation. \(3+\frac{1}{3} X=5\)

Step-by-Step Solution

Answer

The solution to the equation is \( X = 6 \).

Step 1: Identify the Equation Structure

We are given the equation \( 3 + \frac{1}{3} X = 5 \), which appears to be equivalent to a linear equation. The equation includes a constant term and a term with \( X \).

Step 2: Isolate the Variable Term

Subtract 3 from both sides of the equation to isolate the variable term on one side. This gives us:\[\frac{1}{3} X = 5 - 3\] which simplifies to:\[\frac{1}{3} X = 2\]

Step 3: Solve for X

To solve for \( X \), multiply both sides of the equation by 3 to cancel the fraction:\[X = 2 \times 3\]Simplifying this gives:\[X = 6\]

Step 4: Verify the Solution

Substitute \( X = 6 \) back into the original equation to verify:\[3 + \frac{1}{3} \times 6 = 5\]Calculating the left side:\[3 + 2 = 5\], which matches the right side. Thus, the solution is correct.

Key Concepts

Isolate the Variable

When dealing with linear equations, the goal is often to find the value of the unknown variable. In the exercise provided, the variable is represented by \( X \). To solve the equation, we must first isolate this variable on one side of the equation.
The original equation is \( 3 + \frac{1}{3} X = 5 \). To begin isolating the variable, subtract the constant term (3) from both sides of the equation. This will remove the constant from the side with the variable, giving us:

• \( \frac{1}{3} X = 5 - 3 \)

After performing this subtraction, the equation simplifies to \( \frac{1}{3} X = 2 \). Now the variable term is isolated on one side of the equation. Breaking down linear equations into simpler parts makes them easier to solve. Remember, the key is to perform the same operation on both sides of the equation to maintain equality.

Solving Equations

Once the variable has been isolated, the next step is to solve the remaining equation. In our example, the isolated variable term is \( \frac{1}{3} X = 2 \). To solve for \( X \), we need to get rid of the fraction.
To eliminate the fraction, multiply both sides of the equation by 3. This operation cancels the denominator, leaving you with:

• \( X = 2 \times 3 \)

When you multiply the numbers, you get \( X = 6 \). Multiplying by the reciprocal of a fraction is a common technique in solving linear equations, as it helps to simplify the expression and solve for the variable directly. Practicing these steps will strengthen understanding and mastery of solving linear equations.

Verifying Solutions

After obtaining a solution, it is essential to verify that it satisfies the original equation. This step ensures that no mistakes were made during the solution process. For our equation, substitute \( X = 6 \) back into the original equation:

• \( 3 + \frac{1}{3} \times 6 = 5 \)

Now, calculate the expression on the left-hand side:

• \( 3 + 2 = 5 \)

Since both sides of the equation equal 5, the solution \( X = 6 \) is verified as correct.
Verification is an important skill because it helps catch any errors that might have been made in the calculations. Always make sure to plug the solution back into the original equation to confirm its accuracy.