Problem 29 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation analytically. Check it analytically, and then support the solution graphically. $$\frac{1}{2}(x-3)=\frac{5}{12}+\frac{2}{3}(2 x-5)$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{17}{10} \).

Step 1: Simplify Both Sides

Let's start by simplifying the given equation. The equation is: \[ \frac{1}{2}(x-3) = \frac{5}{12} + \frac{2}{3}(2x-5) \]First, simplify the left-hand side: \[ \frac{1}{2}(x-3) = \frac{x}{2} - \frac{3}{2} \]Next, simplify the right-hand side: \[ \frac{2}{3}(2x-5) = \frac{4x}{3} - \frac{10}{3} \]So, the equation becomes: \[ \frac{x}{2} - \frac{3}{2} = \frac{5}{12} + \frac{4x}{3} - \frac{10}{3} \]

Step 2: Equalize the Expression

We need a common denominator to handle the fractional terms effectively. The common denominator for 2, 12, and 3 is 12.Rewrite each term with this common denominator:- Left-hand side: \[ \frac{x}{2} = \frac{6x}{12},\quad \frac{3}{2} = \frac{18}{12} \] So, \( \frac{x}{2} - \frac{3}{2} = \frac{6x}{12} - \frac{18}{12} \)- Right-hand side (combine fractions initially): \[ \frac{5}{12} + \frac{4x}{3} - \frac{10}{3} \]Convert \( \frac{4x}{3} \) and \( \frac{10}{3} \) to have a denominator of 12:- \[ \frac{4x}{3} = \frac{16x}{12},\quad \frac{10}{3} = \frac{40}{12} \]Combine these: \[ \frac{5}{12} + \frac{16x}{12} - \frac{40}{12} = \frac{16x - 35}{12} \]

Step 3: Clear the Fractions

Now combine both sides. Since both sides have a denominator of 12, multiply through by 12 to eliminate the fractions:\[12 \left(\frac{6x - 18}{12}\right) = 12 \left(\frac{16x - 35}{12}\right)\]This simplifies to:\[6x - 18 = 16x - 35\]

Step 4: Solve for x

Now, solve the linear equation without fractions:Subtract \(6x\) from both sides:\[ -18 = 10x - 35 \] Add 35 to both sides:\[ 35 - 18 = 10x \]Simplify:\[ 17 = 10x \]Divide both sides by 10:\[ x = \frac{17}{10} \]

Step 5: Verify the Solution

Substitute \( x = \frac{17}{10} \) back into the original equation to verify:- Left side: \[ \frac{1}{2}\left(\frac{17}{10}-3\right) = \frac{1}{2}\left(\frac{17}{10}-\frac{30}{10}\right) = \frac{1}{2}\left(-\frac{13}{10}\right) = -\frac{13}{20}\]- Right side: \[\frac{5}{12} + \frac{2}{3}\left(2\cdot\frac{17}{10}-5\right) = \frac{5}{12} + \frac{2}{3}\left(\frac{34}{10}-\frac{50}{10}\right)= \frac{5}{12} + \frac{2}{3}(\frac{-16}{10}) = \frac{5}{12} - \frac{16}{15} \]Convert to a common denominator and verify they are equal:\[\frac{5}{12} = \frac{25}{60},\quad -\frac{16}{15} = -\frac{64}{60}\]Finally:\[-\frac{13}{20} = -\frac{39}{60}\] \( -\frac{39}{60} = -\frac{13}{20} \) showing they are equivalent.

Step 6: Graphical Representation

Graph both sides of the original equation as two separate functions and observe the intersection point:- Let \( f(x) = \frac{1}{2}(x-3) \) and \( g(x) = \frac{5}{12} + \frac{2}{3}(2x-5) \).Plot both functions on the same graph. The point where they intersect confirms the solution.Upon plotting, you should find that both functions intersect at \( x = \frac{17}{10}\) or approximately \( x = 1.7 \), confirming the analytical solution graphically.

Key Concepts

Graphical Representation

Visualizing mathematical problems can greatly enhance our understanding, especially in the case of solving equations. When we talk about graphical representation, we're referring to plotting equations on a graph to observe their behavior and interactions.

For the given algebraic equation, we represent each side as separate functions:

• f(x) = \frac{1}{2}(x-3)\
• g(x) = \frac{5}{12} + \frac{2}{3}(2x-5)\

By graphing these functions, we can determine their intersection point. The intersection is crucial because it's where both sides of the equation are equal, reflecting the solution to the equation. In this case, plotting the functions reveals they intersect at \( x = \frac{17}{10} \) or approximately \( x = 1.7 \). This confirms the solution obtained through algebraic manipulation, offering another layer of verification.

Utilizing graphs not only aids in verifying solutions but also provides a tangible way to visualize how varying inputs can affect the outputs. It's a powerful tool that complements analytical methods.

Fraction Operations

Performing operations with fractions is a fundamental skill necessary for solving equations that contain these elements. When working with fractions, it's crucial to follow basic guidelines to ensure clarity and accuracy:

• Identify a common denominator when adding or subtracting fractions, which simplifies the expression. In the given equation, the common denominator was 12. This made all the terms compatible for straightforward addition or subtraction.
• Simplify fractions wherever possible to make calculations easier. This involves canceling out common factors in the numerator and the denominator.
• Always double-check fraction arithmetic to avoid calculation errors, especially when denominators differ.

Handling fractions aptly in this problem involved converting terms like \( \frac{x}{2} \) to \( \frac{6x}{12} \) for easier manipulation. Mastering these operations ensures that no fractional stone is left unturned in problem-solving.

Solving Linear Equations

Understanding how to solve linear equations is a crucial aspect of algebra. A linear equation is an equation between two variables that produces a straight line when graphed. Solving such equations typically involves the following steps:

• Simplify each side separately. This involves distributing multiplication over addition or subtraction, and combining like terms. For instance, the original equation required simplification of both sides for a clearer view.
• Eliminate fractions by dedicating efforts to reach a common denominator if necessary. In this exercise, rewriting fractions with a denominator of 12 simplified the process.
• Isolate the variable by performing operations that inverse the equation components. For example, removing \( 6x \) from one side and adding a constant to both sides helped isolate \( x \) here.
• Verify your solution by substituting it back into the original equation. A correct solution should satisfy the original coefficients and constants.

Linear equations are foundational not only in pure math but also in fields like physics and economics, where relationships between quantities need clear expression and resolution. Thus, mastering solving linear equations forms a core component of a solid math education.