Chapter 7: Problem 56
Solve the given equations involving fractions. $$\frac{x}{2}+\frac{1}{x-3}=3$$
Short Answer
Expert verified
The solutions for \( x \) are 4 and 5.
Step by step solution
01
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.
02
Find the Common Denominator
The common denominator for 2 and \( x-3 \) is \( 2(x-3) \). We'll use this to clear the fractions.
03
Multiply to Clear Fractions
Multiply every term in the equation by \( 2(x-3) \) to eliminate the fractions: \[ 2(x-3)\left(\frac{x}{2}\right) + 2(x-3)\left(\frac{1}{x-3}\right) = 3 \cdot 2(x-3) \].
04
Simplify Each Term
Simplify each term: \( x(x-3) + 2 = 6(x-3) \). Distribute where needed to simplify this further.
05
Expand the Terms
Expand the expression: \( x^2 - 3x + 2 = 6x - 18 \).
06
Rearrange the Equation
Move all terms to one side to form a quadratic equation: \( x^2 - 3x + 2 - 6x + 18 = 0 \). Simplify to \( x^2 - 9x + 20 = 0 \).
07
Factor the Quadratic Equation
Factor \( x^2 - 9x + 20 = 0 \) to find: \( (x-5)(x-4) = 0 \).
08
Solve for \( x \)
Set each factor to zero and solve for \( x \): \( x - 5 = 0 \) gives \( x = 5 \), and \( x - 4 = 0 \) gives \( x = 4 \).
09
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
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.
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 \( x \).