Problem 40 - Algebra for College Students

Question

Solve each equation. \(2 x+5+\frac{1}{2}(6 x-1)=-\frac{1}{2}\)

Step-by-Step Solution

Answer

The solution is \( x = -1 \).

Step 1: Distribute the Fraction

Distribute the \( \frac{1}{2} \) across the terms inside the parentheses in the equation \( 2x + 5 + \frac{1}{2}(6x - 1) = -\frac{1}{2} \). This results in \( \frac{1}{2} \times 6x = 3x \) and \( \frac{1}{2} \times -1 = -\frac{1}{2} \). Thus, the equation becomes \( 2x + 5 + 3x - \frac{1}{2} = -\frac{1}{2} \).

Step 2: Simplify the Equation

Combine like terms on the left side of the equation. Add \( 2x \) and \( 3x \) to get \( 5x \). This simplifies the equation to \( 5x + 5 - \frac{1}{2} = -\frac{1}{2} \).

Step 3: Eliminate the Fraction

To eliminate the \( \frac{1}{2} \) from the equation, add \( \frac{1}{2} \) to both sides of the equation. This simplifies to \( 5x + 5 = 0 \).

Step 4: Solve for x

Subtract 5 from both sides of the equation to isolate the term with \( x \), resulting in \( 5x = -5 \).

Step 5: Divide to Find x

Divide both sides of the equation by 5 to solve for \( x \). This gives \( x = -1 \).

Key Concepts

Distributive Property

The distributive property is a fundamental principle in algebra that allows you to multiply a single factor across terms within parentheses. In our exercise, we use this concept to handle the fraction \(\frac{1}{2}\). When you see an expression like \(\frac{1}{2}(6x - 1)\), it means that you have to distribute \(\frac{1}{2}\) to each term inside the parenthesis:

• \(\frac{1}{2} \times 6x = 3x\)
• \(\frac{1}{2} \times -1 = -\frac{1}{2}\)

This step is crucial as it allows you to eliminate the parentheses and work with a simplified equation. Remember, the distributive property applies not only to numbers but also to variables.

Combining Like Terms

Like terms in an equation are terms that contain the same variable raised to the same power. In solving equations, combining like terms is an essential step to simplify the equation. For our problem, combining like terms boils down to adding coefficients of terms with the same variable:

• The terms \(2x\) and \(3x\) are like terms because both have the variable \(x\).
• Add them to get \(5x\).

This process reduces clutter, making equations simpler and easier to solve. It also aligns various mathematical expressions into a more manageable form, setting the stage for solving the equation effectively.

Solving Linear Equations

Solving linear equations involves isolating the variable to find its value. After simplifying our equation, the next step was to get the variable \(x\) alone on one side of the equation. Here's how:

• First, eliminate constants from the side with the variable. In our example, subtract 5 from both sides, which gives \(5x = -5\).
• Now isolate \(x\) by dividing each side by the coefficient of \(x\), which is 5.
• This step results in \(x = -1\).

Persistence and organization in each step ensure accuracy when solving equations. Always perform the same operation on both sides to maintain the equation's balance.

Fraction Operations

Fractions often seem tricky in algebraic equations, but breaking them into simpler operations can demystify them. In our exercise, we encountered fractions in the equation that required handling with care:

• Distribute fractions correctly as shown with \(\frac{1}{2}(6x - 1)\).
• Operations needed us to both add and subtract fractions. To get rid of \(\frac{1}{2}\) in \(5x + 5 - \frac{1}{2} = -\frac{1}{2}\), one method is to add \(\frac{1}{2}\) to both sides of the equation for easier simplification.

Whenever fractions are involved, ensure you adjust operations to fit the context, be it addition, subtraction, or distribution. Mastering these operations makes navigating through complex equations simpler.