Problem 29 - Algebra and Trigonometry

Question

Solve the equation both algebraically and graphically. $$ \frac{2}{x}+\frac{1}{2 x}=7 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{5}{14} \). This can also be verified by graphical plotting, where the curves intersect at this x-value.

Step 1: Find a Common Denominator

The equation is \( \frac{2}{x} + \frac{1}{2x} = 7 \). To combine the fractions on the left side, we need a common denominator. Here it is \(2x\), because \(x\) can be written as \( \frac{2x}{2} \), and multiplication by 2 achieves the same denominator. Re-write the first fraction by multiplying both the numerator and the denominator by 2: \( \frac{2 \cdot 2}{x \cdot 2} = \frac{4}{2x} \). So, the equation becomes: \( \frac{4}{2x} + \frac{1}{2x} = 7 \).

Step 2: Simplify the Left Side

Since both fractions are over the common denominator \(2x\), we can combine them: \( \frac{4 + 1}{2x} = \frac{5}{2x} \). The equation is now \( \frac{5}{2x} = 7 \).

Step 3: Clear the Fraction

To eliminate the fraction, multiply both sides of the equation by \(2x\): \(5 = 7 \times 2x\). Simplifying gives us: \(5 = 14x\).

Step 4: Solve for x

Now solve for \(x\) by dividing both sides by 14: \(x = \frac{5}{14}\).

Step 5: Graphical Solution

To solve graphically, plot the equations \( y = \frac{2}{x} + \frac{1}{2x} \) and \( y = 7 \) on the same set of axes. The solution for \(x\) would be the x-coordinate where the two graphs intersect. For \( y = \frac{2}{x} + \frac{1}{2x} \), as \(x\) approaches \(\frac{5}{14}\), the graph should intersect the horizontal line \(y = 7\). If plotted correctly, the graphs intersect at approximately \(x = \frac{5}{14}\).

Key Concepts

Common Denominator

When dealing with algebraic equations involving fractions, finding a common denominator is crucial. A common denominator allows us to combine fractions, making them easier to work with. In our equation, we have two terms: \( \frac{2}{x} \) and \( \frac{1}{2x} \). To combine these, we need to identify a common denominator. The least common denominator in this case is \( 2x \), because it can encompass both denominators ('x' and '2x').
To adjust the fractions to this common denominator, we multiply the numerator and denominator of the first fraction by 2: \( \frac{2 \cdot 2}{x \cdot 2} = \frac{4}{2x} \). Now our equation becomes \( \frac{4}{2x} + \frac{1}{2x} = 7 \).
This simplification step is essential as it allows us to directly compare and combine the fractions. When fractions have the same denominator, we can add or subtract their numerators directly, simplifying the algebraic equation process.

Graphical Solution

A graphical solution involves plotting the equations on a graph to visualize where they intersect. This method helps confirm the solution you derived algebraically, and can sometimes reveal additional solutions or insights.
In our case, the equations are \( y = \frac{2}{x} + \frac{1}{2x} \) and \( y = 7 \). By plotting these, we observe where the curve \( y = \frac{2}{x} + \frac{1}{2x} \) intersects with the horizontal line \( y = 7 \). This intersection represents the solution for \( x \).
As you plot, note that the curve approaches \( x = \frac{5}{14} \) where it intersects the line. This graphically confirms the algebraic solution. Graphical methods are particularly useful if the equation does not resolve neatly using algebra alone.

Solving Fractions

Solving equations with fractions requires careful manipulation to remove the fractions from the equation, making the problem easier to solve. Essentially, you want to "clear" fractions to simplify the equation.
Take our simplified equation \( \frac{5}{2x} = 7 \). To solve for \( x \), multiply both sides by \( 2x \) to eliminate the fraction. This gives you: \( 5 = 14x \). Now it's a simple linear equation, easily solved by isolating \( x \).
Divide both sides by 14 to solve: \( x = \frac{5}{14} \).
This method of dealing with fractions—finding common denominators, simplifying equations, and eliminating fractions—is a powerful strategy. It transforms complex algebraic terms into simpler linear forms, making solutions more straightforward.