Problem 36 - Algebra and Trigonometry

Question

\(9- 46\) The given equation is either linear or equivalent to a linear equation. Solve the equation. $$ \frac{12 x-5}{6 x+3}=2-\frac{5}{x} $$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{15}{19}.\)

Step 1: Clear the Denominators

To eliminate the fractions, find a common denominator. The common denominator for the terms in the equation is \[(6x + 3)x.\]Multiply every term by this common denominator to clear the fractions:\[\frac{12x - 5}{6x + 3} \cdot (6x + 3)x = 2 \cdot (6x + 3)x - \frac{5}{x} \cdot (6x + 3)x.\]Simplifying gives us:\[(12x - 5)x = 2(6x + 3)x - 5(6x + 3).\]

Step 2: Expand and Simplify

Expand both sides of the equation from Step 1:\[12x^2 - 5x = (12x^2 + 6x) - (30x + 15).\]Simplify the right side:\[12x^2 + 6x - 30x - 15.\]Combine like terms:\[12x^2 - 24x - 15.\]

Step 3: Equate and Rearrange Terms

Set the equation from Step 2 equal:\[12x^2 - 5x = 12x^2 - 24x - 15.\]Subtract \(12x^2\) from both sides:\[-5x = -24x - 15.\]

Step 4: Solve for x

Add \(24x\) to both sides:\[19x = -15.\]Divide both sides by 19 to solve for \(x\):\[x = -\frac{15}{19}.\]

Key Concepts

Clearing Denominators

Clearing denominators is a crucial first step when solving equations that involve fractions. In the case of the equation \( \frac{12x - 5}{6x + 3} = 2 - \frac{5}{x} \), the presence of fractions makes solving directly quite challenging. To handle this:

• Identify a common denominator for all the fractional terms.
• The goal is to eliminate these denominators by multiplying each term in the equation by this common denominator.
• Here, the common denominator is \((6x + 3)x\).

By multiplying every term in the equation by this common denominator, you effectively clear the fractions. This transformation simplifies the process significantly because you won't have to deal with additions and subtractions involving fractions further. It creates a new equation that’s easier to manipulate since it primarily involves polynomials.

Expanding Expressions

Once denominators are cleared, the next step is expanding expressions. This process is essential when the equation includes terms that need to be distributed across others, as simplified equations are easier to work with.

• For instance, in our cleared equation, \((12x - 5)x = 2(6x + 3)x - 5(6x + 3)\), each side of the equation needs expansion.
• When you expand, you apply the distributive property, multiplying each term within parentheses by other terms.

Breaking down, you’ll distribute each component:

• On the left-hand side: \((12x - 5)x\), expand it to \(12x^2 - 5x\).
• On the right-hand side: Expand \(2(6x + 3)x\) to get \(12x^2 + 6x\), and \(-5(6x + 3)\) to obtain \(-30x - 15\).

This step, expanding expressions, helps in further simplifying the equation into its most tractable form.

Combining Like Terms

After expanding, the next target is to combine like terms. This step consolidates the equation, making it easier to work with and solve.

• "Like terms" refer to terms that contain exactly the same variables raised to the same power.
• For the equation \(12x^2 + 6x - 30x - 15\), combine the \(x\) terms by adding them (or subtracting, as required).

In this case:

• Combine \(6x\) and \(-30x\) to get \(-24x\).

With combined like terms, the equation becomes: \(12x^2 - 24x - 15\). Reducing the complexity of the equation by consolidating similar parts allows us to focus on fewer components, thereby simplifying the path to the solution.

Algebraic Simplification

The final part is algebraic simplification, which is all about solving for the variable in terms that make the equation straightforward.

• Once like terms are condensed, you set both sides of the simplified equation equal.
• Adjust the equation using algebraic operations to isolate the variable of interest.

For example, when you have:

• Set the equation \(12x^2 - 5x = 12x^2 - 24x - 15\).
• First, subtract \(12x^2\) from each side to eliminate these terms.
• Then, solve for \(x\) by isolating the variable.

This process involved strategically adding or subtracting terms to both sides until \(x\) stood alone. For instance:

• From \(-5x = -24x - 15\), add \(24x\) to both sides to gather all \(x\) terms together.
• Dividing by the coefficient of \(x\), you conclude \(x = -\frac{15}{19}\).

Algebraic simplification ensures that the solution is presented clearly and accurately, making it reachable with just a few more straightforward steps.