Problem 27 - College Algebra and Calculus: An Applied Approach

Question

Solve the equation and check your solution. (Some equations have no solution.) $$ \frac{5 x}{4}+\frac{1}{2}=x-\frac{1}{2} $$

Step-by-Step Solution

Answer

The equation has no solution.

Step 1: Simplify both sides of equation

To simplify both sides of the equation, one must find a common denominator. Since 4 and 2 have a common multiple 4. Multiply the right side by 2 to make denominators the same:\[ \frac{5 x}{4}+\frac{2}{4} = \frac{2x}{1} - \frac{2}{4} \].

Step 2: Combine like fractions

Combine fractions on both sides that have the same denominator:\[ \frac{5x + 2}{4} = \frac{2x - 2}{4} \].

Step 3: Simplify equation further

Multiply both sides of the equation by 4 to eliminate the denominators:\[ 5x + 2 = 2x - 2 \].

Step 4: Solve for x

Get all terms containing x on one side and the numerical values on the other. Subtract 2x from both sides to isolate x terms on left side, and add 2 to both sides to isolate the numerical values on the right side, which yields:\[ 5x - 2x = -2 + 2 \]},Then, simplify both sides:\[ 3x = 0 \].Finally, solve for x by dividing both sides by 3:\[ x = 0 \].

Step 5: Check solution

Check the solution by substituting \( x = 0 \) in the original equation:\[ \frac{5(0)}{4}+\frac{1}{2} = 0-\frac{1}{2} \].This simplifies to:\[ 0 + \frac{1}{2} = -\frac{1}{2} \].This statement is false, hence, the equation has no solution.

Key Concepts

Common Denominator

When dealing with algebraic fractions, finding a common denominator is crucial to simplify and solve equations. A common denominator is a shared multiple of the denominators of fractions within the equation. Having common denominators allows you to combine or compare fractions easily. In our example, the equation \( \frac{5x}{4} + \frac{1}{2} = x - \frac{1}{2} \) involves denominators 4 and 2.

• First, determine the Least Common Denominator (LCD), the smallest number into which both denominators can divide evenly.
• Here, 4 is the LCD since both 4 and 2 can divide into it.
• Adjust fractions related to the equation to have this common denominator.

Using this process, rewrite \( \frac{1}{2} \) as \( \frac{2}{4} \), ensuring all fractions share the denominator of 4 to facilitate further operations.

Like Fractions

Like fractions have the same denominator, making them straightforward to add or subtract. Once each fraction in the equation has a common denominator, they are regarded as like fractions. This helps in simplifying the algebraic equation swiftly.

Our equation, \( \frac{5x}{4} + \frac{2}{4} = \frac{2x}{4} - \frac{2}{4} \), transforms into like fractions by adopting a common denominator (in this case, 4).

• With denominators matched, combine the numerators for simplicity.
• Perform arithmetic operations only on the numerators when denominators are identical.

Thus, we can represent the fractions more simply as \( \frac{5x + 2}{4} = \frac{2x - 2}{4} \), setting the stage for further computation.

Solving for Variable

Solving a variable involves isolating it on one side of the equation. The ultimate goal is to solve the algebraic equation, which expresses the quantity for the variable in terms of known numbers. Once fractions are simplified, like in the equation \( 5x + 2 = 2x - 2 \), you can solve for the variable \( x \):

• Move all terms containing \( x \) to one side of the equation.
• On the opposite side, position numerical terms independently.
• Simplify both sides by arithmetic operations.

In our situation, move \( 2x \) to get \( 5x - 2x = -2 \). Then process \( 3x = 0 \) to determine \( x = 0 \). Proceed methodically with systematic arithmetic and deport numerical errors, however minuscule.

Checking Solution

Checking the solution to an equation is a crucial step, confirming the variable's value accurately satisfies the original equation. After calculation yields \( x = 0 \), substitute back into the starting equation to ensure it holds true:

• Replace \( x \) with 0 in the original equation \( \frac{5(0)}{4} + \frac{1}{2} = 0 - \frac{1}{2} \).
• This results in \( 0 + \frac{1}{2} = -\frac{1}{2} \).
• Since this equality does not hold, the solution can't satisfy the original equation.

This invalid solution demonstrates the equation has no solutions. Any false statement confirms an unsatisfactory value, requiring reevaluation or substantiating no valid solutions exist. This diligent checking practice ensures an accurate understanding and performance of algebraic equations.