Problem 27 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ \frac{1}{6}(x+12)+1=\frac{x}{3} $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 18\).

Step 1: Clear the fraction

The first step in solving the equation \( \frac{1}{6}(x+12)+1=\frac{x}{3} \) is to eliminate the fractions by finding a common denominator. Multiply every term by \(6\), which is the least common multiple of the denominators \(6\) and \(3\). This gives us:\[ 6 \times \frac{1}{6}(x + 12) + 6 \times 1 = 6 \times \frac{x}{3} \]Simplifying each term, we have:\[ (x + 12) + 6 = 2x \]

Step 2: Simplify the equation

Next, simplify the equation obtained from the previous step:\[ x + 12 + 6 = 2x \]Combine the constants on the left side:\[ x + 18 = 2x \]

Step 3: Isolate the variable

Now, isolate \(x\) by subtracting \(x\) from both sides:\[ x + 18 - x = 2x - x \]This simplifies to:\[ 18 = x \]

Step 4: Write the solution

The solution to the equation is simply \(x = 18\). To confirm, you can substitute \(x = 18\) back into the original equation to verify it holds true, ensuring the solution is correct.

Key Concepts

Solving Equations

When working with algebraic equations, the primary goal is to find the value of the variable that makes the equation true. This is called "solving the equation." Begin the process by understanding the structure of the equation in front of you—what operations are involved, and how complex is the equation? Generally, the process involves a series of steps that simplify the equation while preserving the equality.

To start solving, identify what operations are being performed on the variable. If there are constants added, subtracted, or fractions involved, it affects how you proceed with the solution. Most often, you will simplify the expression, clear any fractions, and eventually isolate the variable. Each of these topics is interconnected and vital to reach the final solution.

By systematically approaching the equation, making sure each step maintains the balance between the two sides, you eventually simplify the equation to a basic form like "variable = value."

Clearing Fractions

Before truly diving into solving an equation, fractions are often best cleared out to make the operations simpler. This is because fractions can make equations look cluttered and hard to manage. The rule of thumb is to eliminate these fractions by finding a common denominator—often the least common multiple of all denominators present in the equation.

Let's illustrate with this example: in the equation \( \frac{1}{6}(x+12)+1=\frac{x}{3} \), we multiply every term by the least common multiple, which is \( 6 \) here. Multiplying every term by \( 6 \) allows you to cancel out the denominators:

• \( 6 \times \frac{1}{6}(x + 12) \) simplifies to \( x + 12 \).
• \( 6 \times 1 \) becomes \( 6 \).
• \( 6 \times \frac{x}{3} \) reduces to \( 2x \).

After these cancellations, the equation simplifies to \( (x + 12) + 6 = 2x \). Without the fractions, the equation is simpler and less error-prone, making subsequent steps much clearer and easier to perform.

Isolating the Variable

Once any fractions are cleared, the next focus is to isolate the variable. To isolate the variable means to rearrange the equation such that the variable appears by itself on one side of the equation. This usually involves using inverse operations—addition or subtraction followed by multiplication or division—to systematically remove constants or coefficients attached to the variable.

Using the example from before, \( x + 18 = 2x \), we move towards isolating \( x \). We simply subtract \( x \) from both sides to get \( x + 18 - x = 2x - x \) which simplifies to \( 18 = x \). By performing the same operation on both sides of the equation, you ensure that equality is preserved.

The isolated variable tells you the solution of the equation—in this case, \( x = 18 \). To verify your solution, substitute back into the original equation to check if both sides of the equation remain equal. This step reinforces the accuracy of your solution and confirms you've correctly isolated and solved for the variable.