To solve the equation:
[tex]\[
\frac{1}{x} + \frac{2}{x+10} = \frac{1}{3}
\][/tex]
### Step-by-Step Solution:
1. Identify the Least Common Denominator (LCD):
- The denominators of the fractions are [tex]\(x\)[/tex], [tex]\(x + 10\)[/tex], and 3.
- The least common denominator (LCD) must be a common multiple of all these denominators. Thus, the LCD is:
[tex]\[
3x(x + 10)
\][/tex]
2. Rewrite the Equation with the LCD:
- Using the LCD to eliminate the fractions, we multiply every term in the equation by [tex]\(3x(x + 10)\)[/tex]:
[tex]\[
3x(x + 10) \left(\frac{1}{x}\right) + 3x(x + 10) \left(\frac{2}{x + 10}\right) = 3x(x + 10) \left(\frac{1}{3}\right)
\][/tex]
3. Simplify the Equation:
- Simplifying each term:
[tex]\[
3(x + 10) + 6x = x(x + 10)
\][/tex]
- Distribute and combine like terms:
[tex]\[
3x + 30 + 6x = x^2 + 10x
\][/tex]
- Combine all terms on one side to form a quadratic equation:
[tex]\[
x^2 + 10x - 9x - 30 = 0
\][/tex]
- Simplified:
[tex]\[
x^2 + x - 30 = 0
\][/tex]
4. Solve the Quadratic Equation:
- Use the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -30\)[/tex]:
[tex]\[
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{1 + 120}}{2}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{121}}{2}
\][/tex]
[tex]\[
x = \frac{-1 \pm 11}{2}
\][/tex]
- Thus, the solutions are:
[tex]\[
x = \frac{10}{2} = 5 \quad \text{and} \quad x = \frac{-12}{2} = -6
\][/tex]
5. Determine Valid Solutions:
- We need the solution(s) to be realistic in the context of the problem. Since [tex]\(x = -6\)[/tex] would make the denominator of the original fractions negative and non-realistic for our problem context, we discard it. The only valid solution is:
[tex]\[
x = 5
\][/tex]
### Conclusion:
The least common denominator between all three fractions is [tex]\(\mathbf{3x(x+10)}\)[/tex].
The equation has [tex]\(\mathbf{1}\)[/tex] valid solution.
