https://file.aiquickdraw.com/mathbot/file/1726902896582ockn87ag.jpg

Let's solve the three rational equations step by step.

### Problem 1:
\[
\frac{x}{x+3} = 12
\]

**Solution:**
1. Multiply both sides by \(x + 3\) to eliminate the denominator:
   \[
   x = 12(x + 3)
   \]
2. Expand the right-hand side:
   \[
   x = 12x + 36
   \]
3. Bring all terms involving \(x\) to one side:
   \[
   x - 12x = 36
   \]
4. Simplify:
   \[
   -11x = 36
   \]
5. Divide by \(-11\):
   \[
   x = \frac{36}{-11} = -\frac{36}{11}
   \]

### Problem 2:
\[
\frac{x - 2}{x + 2} = \frac{x}{6}
\]

**Solution:**
1. Cross-multiply to eliminate the fractions:
   \[
   6(x - 2) = x(x + 2)
   \]
2. Expand both sides:
   \[
   6x - 12 = x^2 + 2x
   \]
3. Bring all terms to one side to set the equation to zero:
   \[
   x^2 + 2x - 6x + 12 = 0
   \]
   \[
   x^2 - 4x + 12 = 0
   \]
4. Solve this quadratic equation using the quadratic formula:
   \[
   x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(12)}}{2(1)}
   \]
   \[
   x = \frac{4 \pm \sqrt{16 - 48}}{2}
   \]
   \[
   x = \frac{4 \pm \sqrt{-32}}{2}
   \]
   \[
   x = \frac{4 \pm 4i\sqrt{2}}{2}
   \]
   \[
   x = 2 \pm 2i\sqrt{2}
   \]

Thus, the solutions are complex numbers:
\[
x = 2 + 2i\sqrt{2}, \quad x = 2 - 2i\sqrt{2}
\]

### Problem 3:
\[
\frac{x}{x+2} = \frac{8}{x^2 - 4}
\]

**Solution:**
1. Notice that \(x^2 - 4\) is a difference of squares, so:
   \[
   x^2 - 4 = (x - 2)(x + 2)
   \]
   The equation becomes:
   \[
   \frac{x}{x+2} = \frac{8}{(x - 2)(x + 2)}
   \]
2. Multiply both sides by \((x - 2)(x + 2)\) to eliminate the denominators:
   \[
   x(x - 2) = 8
   \]
3. Expand the left-hand side:
   \[
   x^2 - 2x = 8
   \]
4. Bring all terms to one side:
   \[
   x^2 - 2x - 8 = 0
   \]
5. Solve the quadratic equation using the quadratic formula:
   \[
   x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}
   \]
   \[
   x = \frac{2 \pm \sqrt{4 + 32}}{2}
   \]
   \[
   x = \frac{2 \pm \sqrt{36}}{2}
   \]
   \[
   x = \frac{2 \pm 6}{2}
   \]
6. Thus, the solutions are:
   \[
   x = \frac{2 + 6}{2} = 4, \quad x = \frac{2 - 6}{2} = -2
   \]

However, \(x = -2\) is not valid because it would make the original denominator zero. So the only solution is:
\[
x = 4
\]

---

Let me know if you'd like any more details! Here are some related questions:

1. What are the conditions for the existence of a solution in rational equations?
2. How do you handle complex numbers when solving quadratic equations?
3. Can a rational equation have no solution? If so, why?
4. How does cross-multiplying work in rational equations?
5. Why must denominators in rational equations not equal zero?

**Tip:** Always check for extraneous solutions, especially when dealing with denominators, as some solutions might make the denominator zero and invalidate them.

Solve the following rational equation and check your solution: 1) x/(x+3) = 12, 2) (x-2)/(x+2) = x/6, 3) x/(x+2) = 8/(x^2 - 4).

This article provides step-by-step solutions for solving three rational equations, covering key algebraic concepts such as rational expressions, quadratic equations, and complex numbers. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation