Problem 32 - Algebra and Trigonometry

Question

\(31-42\) . Solve the equation both algebraically and graphically. $$ \frac{1}{2} x-3=6+2 x $$

Step-by-Step Solution

Answer

The solution is \( x = -6 \).

Step 1: Simplify the Equation

First, we need to simplify both sides of the equation \( \frac{1}{2}x - 3 = 6 + 2x \). For simplicity, let's eliminate the fraction by multiplying both sides by 2, resulting in \( x - 6 = 12 + 4x \).

Step 2: Rearrange Terms

Next, rearrange the equation to bring all terms involving \( x \) to one side and the constant terms to the other side. Subtract \( 4x \) from both sides to get \( x - 4x = 12 + 6 \). This simplifies to \( -3x = 18 \).

Step 3: Solve for x

Now, solve for \( x \) by dividing both sides by -3. This gives us \( x = \frac{18}{-3} \), which simplifies to \( x = -6 \).

Step 4: Graphical Solution

Graph each side of the equation as a function. The left side becomes \( y = \frac{1}{2}x - 3 \) and the right side \( y = 6 + 2x \). Plot both lines on the same graph to find the intersection point. Upon graphing, the lines intersect at \( x = -6 \), confirming the algebraic solution.

Key Concepts

Algebraic Methods

Algebraic methods are very useful when solving linear equations such as \( \frac{1}{2}x - 3 = 6 + 2x \). The main goal is to find the value of \( x \) that makes both sides of the equation equal. To do this algebraically, follow these general steps:

• Eliminate Fractions: If any fractions are present, multiplying through by a common denominator is helpful, making calculations easier.
• Rearrange the equation: Move all terms containing \( x \) to one side and constants to the other side.
• Solve for the variable: Isolate \( x \) by performing arithmetic operations, like addition, subtraction, or division, to find the solution.

In this exercise, simplifying by multiplying both sides by 2 removed the fraction, leading us to \( x - 6 = 12 + 4x \), aiding in a clearer path to solve for \( x = -6 \).

Graphical Methods

Graphical methods provide a visual approach to solving equations by plotting each side of the equation as a separate function. This can be particularly helpful to understand the solution's structure or cross-check algebraic solutions.
To solve the equation \( \frac{1}{2}x - 3 = 6 + 2x \) graphically, consider these steps:

• Define the functions: Write the equation in terms of two functions: \( y = \frac{1}{2}x - 3 \) and \( y = 6 + 2x \).
• Plot the functions: Create a graph with a suitable scale, and sketch both lines.
• Identify the intersection: The point where the two graphs intersect gives the solution to the equation.

In this exercise, the intersection point, \( x = -6 \), verifies the algebraic solution. Graphical methods offer a way to see the relationship between expressions in a tangible way.

Equation Simplification

Equation simplification is key to making complex problems easier to solve. It involves reducing equations to their simplest form by removing fractions, combining like terms, and simplifying expressions. In the provided example:

• Eliminate Fractions: Multiplying through by the least common denominator (2 in this case) simplifies \( \frac{1}{2}x - 3 \) into \( x - 6 \).
• Combine Like Terms: Move all terms involving \( x \) to one side of the equation, and constants to the other.
• Solve the Simplified Equation: After rearranging, solving \( -3x = 18 \) is straightforward, leading to \( x = -6 \).

Simplifying equations reduces computation errors and highlights the path to finding solutions quickly and accurately.

Intersection of Lines

Understanding the intersection of lines is a cornerstone of solving linear equations graphically. When two lines intersect, it means they meet at a particular point in the coordinate plane. This point represents the solution to the corresponding linear equation.
In our exercise:

• The two lines represented by \( y = \frac{1}{2}x - 3 \) and \( y = 6 + 2x \) intersect at \( x = -6 \).
• The \( x \)-coordinate of the intersection point is the solution to the equation.

Check visual consistency by ensuring that the discovered intersect point matches the algebraic solution. Graphing tools and graph paper are helpful to sketch and confirm that the intersection lies exactly where calculations predict. This visual endorsement provides confidence in the correct solution of the equation.