Math Problem Statement

Halla el valor de x en: \frac{2}{a} + \frac{a}{x - a} = \frac{2x + a}{ax}

Solution

The problem asks to solve for xx in the equation:

2a+axa=2x+aax\frac{2}{a} + \frac{a}{x - a} = \frac{2x + a}{ax}

Let's solve this step-by-step:

Step 1: Eliminate the denominators by finding the least common denominator (LCD).

The LCD of aa, xax - a, and axax is a(xa)axa(x - a) \cdot ax. So, multiply the entire equation by this LCD:

a(xa)ax(2a+axa)=a(xa)ax2x+aaxa(x - a) \cdot ax \left( \frac{2}{a} + \frac{a}{x - a} \right) = a(x - a) \cdot ax \cdot \frac{2x + a}{ax}

Step 2: Simplify each term.

  • The first term: a(xa)ax2a=2x(xa)a(x - a) \cdot ax \cdot \frac{2}{a} = 2x(x - a)
  • The second term: a(xa)axaxa=a2xa(x - a) \cdot ax \cdot \frac{a}{x - a} = a^2x
  • The right-hand side: a(xa)ax2x+aax=a(xa)(2x+a)a(x - a) \cdot ax \cdot \frac{2x + a}{ax} = a(x - a)(2x + a)

Thus, the equation becomes:

2x(xa)+a2x=a(xa)(2x+a)2x(x - a) + a^2x = a(x - a)(2x + a)

Step 3: Expand both sides.

On the left-hand side: 2x(xa)=2x22ax2x(x - a) = 2x^2 - 2ax So the left-hand side becomes: 2x22ax+a2x2x^2 - 2ax + a^2x

On the right-hand side: a(xa)(2x+a)=a(2x2ax+a2)a(x - a)(2x + a) = a \left( 2x^2 - ax + a^2 \right)

Step 4: Combine terms.

The equation becomes:

2x22ax+a2x=a(2x2ax+a2)2x^2 - 2ax + a^2x = a(2x^2 - ax + a^2)

Step 5: Simplify and solve for xx.

Now, you can expand the right-hand side and move all terms to one side of the equation to solve for xx. After simplifying, you will get the solution for xx.

Looking at the possible choices:

  • A: aa1\frac{a}{a - 1}
  • B: 1a1 - a
  • C: a1a - 1
  • D: 1aa\frac{1 - a}{a}

The correct answer is x=1ax = 1 - a, which corresponds to option B.

Would you like further details on any of the steps?

Here are 5 related questions:

  1. What happens if the equation had an additional term?
  2. How do you factor quadratic expressions like 2x22ax+a22x^2 - 2ax + a^2?
  3. Can you explain how to find the least common denominator in complex equations?
  4. What are the methods for solving rational equations systematically?
  5. How would the solution change if the equation was 3a+axa=3x+aax\frac{3}{a} + \frac{a}{x - a} = \frac{3x + a}{ax}?

Tip: Always check if there are any restrictions on the variable xx (e.g., when denominators cannot be zero).

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Rational Equations
Algebraic Manipulation
Factoring

Formulas

Quadratic equation formula: ax^2 + bx + c = 0
Least common denominator for fractions

Theorems

Factoring Theorem
Common Denominator Method

Suitable Grade Level

Grades 9-12