Problem 56 - Basic Technical Mathematics with Calculus

Question

Solve the given equations involving fractions. $$\frac{x}{2}+\frac{1}{x-3}=3$$

Step-by-Step Solution

Answer

The solutions for \( x \) are 4 and 5.

Step 1: Identify the Denominators

The equation is \( \frac{x}{2} + \frac{1}{x-3} = 3 \). The denominators are 2 and \( x-3 \). To eliminate them, we'll find a common denominator.

Step 2: Find the Common Denominator

The common denominator for 2 and \( x-3 \) is \( 2(x-3) \). We'll use this to clear the fractions.

Step 3: Multiply to Clear Fractions

Multiply every term in the equation by \( 2(x-3) \) to eliminate the fractions: \[ 2(x-3)\left(\frac{x}{2}\right) + 2(x-3)\left(\frac{1}{x-3}\right) = 3 \cdot 2(x-3) \].

Step 4: Simplify Each Term

Simplify each term: \( x(x-3) + 2 = 6(x-3) \). Distribute where needed to simplify this further.

Step 5: Expand the Terms

Expand the expression: \( x^2 - 3x + 2 = 6x - 18 \).

Step 6: Rearrange the Equation

Move all terms to one side to form a quadratic equation: \( x^2 - 3x + 2 - 6x + 18 = 0 \). Simplify to \( x^2 - 9x + 20 = 0 \).

Step 7: Factor the Quadratic Equation

Factor \( x^2 - 9x + 20 = 0 \) to find: \( (x-5)(x-4) = 0 \).

Step 8: Solve for \( x \)

Set each factor to zero and solve for \( x \): \( x - 5 = 0 \) gives \( x = 5 \), and \( x - 4 = 0 \) gives \( x = 4 \).

Step 9: Verify Solution Within Original Constraints

Check that neither solution makes a denominator zero in the original equation \( \frac{x}{2} + \frac{1}{x-3} = 3 \). Here, \( x = 5 \) and \( x = 4 \) do not cause a denominator to be zero, so both are valid.

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra, typically written in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). This form is crucial because it represents a parabola when graphed in a two-dimensional plane. The solutions or "roots" of a quadratic equation can be found using various methods such as factoring, completing the square, or the quadratic formula.

• Factoring: This involves expressing the quadratic as a product of two binomials.
• Completing the Square: We turn the equation into a perfect square trinomial.
• Quadratic Formula: Provides a direct way to find the roots: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

To solve the quadratic equation from the exercise \( x^2 - 9x + 20 = 0 \), we can use factoring because it simplifies our work. Factoring reveals the roots of the equation by setting each factor equal to zero and solving for \( x \). This is illustrated in the step by step solution.

Common Denominator

When dealing with equations containing fractions, finding a common denominator allows you to combine or eliminate fractions, simplifying the equation. A common denominator is a shared multiple of the denominators in a given problem.In the original exercise \( \frac{x}{2} + \frac{1}{x-3} = 3 \), the denominators are 2 and \( x-3 \). To eliminate these fractions, we identify the common denominator, which is \( 2(x-3) \). By multiplying the entire equation by this common denominator, you clear the fractions easily:

• This turns the equation into a polynomial form.
• Ensures each term can be expressed without fractions.

Finding a common denominator is a foundational technique in solving equations with fractions, making it easier to manipulate and solve them.

Factoring Polynomials

Factoring polynomials is a key technique in algebra to simplify expressions and solve equations. It involves writing a polynomial as a product of its factors, which are simpler polynomials.For example, in the equation \( x^2 - 9x + 20 = 0 \), factoring transforms it into its roots \( (x-5)(x-4) = 0 \). This transformation follows a process:

• Identify: Look for two numbers that multiply to the constant term and add to the middle coefficient.
• Rewrite: Express the quadratic as two binomials set to zero.
• Solve: Set each binomial equal to zero, and solve for \( x \).

Factoring is particularly useful because it provides immediate solutions, or "roots," for quadratic equations, allowing you to solve them efficiently. It’s often the first strategy used when equations can be easily decomposed into simpler linear factors.