(x-342)/15 + (x-323)/17 + (x-300)/19 + (x-273)/21 =10
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Let's solve the equation step by step:

The equation given is:

\[
\frac{x - 342}{15} + \frac{x - 323}{17} + \frac{x - 300}{19} + \frac{x - 273}{21} = 10
\]

### Step 1: Clear the Fractions
To eliminate the fractions, find the least common multiple (LCM) of the denominators: \(15\), \(17\), \(19\), and \(21\).

- LCM of \(15\) and \(21\): \(15 = 3 \times 5\), \(21 = 3 \times 7\) → LCM = \(3 \times 5 \times 7 = 105\).
- LCM of \(105\) and \(17\) (since \(17\) is prime) = \(105 \times 17 = 1785\).
- LCM of \(1785\) and \(19\) (since \(19\) is prime) = \(1785 \times 19 = 33915\).

The LCM of the denominators is \(33915\).

### Step 2: Multiply the entire equation by \(33915\) to clear the fractions:
\[
33915 \times \left(\frac{x - 342}{15}\right) + 33915 \times \left(\frac{x - 323}{17}\right) + 33915 \times \left(\frac{x - 300}{19}\right) + 33915 \times \left(\frac{x - 273}{21}\right) = 33915 \times 10
\]

Simplifying each term:

\[
\frac{33915}{15} = 2261, \quad \frac{33915}{17} = 1995, \quad \frac{33915}{19} = 1785, \quad \frac{33915}{21} = 1615
\]

So, the equation becomes:

\[
2261(x - 342) + 1995(x - 323) + 1785(x - 300) + 1615(x - 273) = 339150
\]

### Step 3: Expand and simplify
Expanding each term:

\[
2261x - 2261 \times 342 + 1995x - 1995 \times 323 + 1785x - 1785 \times 300 + 1615x - 1615 \times 273 = 339150
\]

Calculate the constants:

\[
2261 \times 342 = 773862, \quad 1995 \times 323 = 644235, \quad 1785 \times 300 = 535500, \quad 1615 \times 273 = 441495
\]

So the equation becomes:

\[
2261x + 1995x + 1785x + 1615x - (773862 + 644235 + 535500 + 441495) = 339150
\]

Combine the terms:

\[
7656x - 2395092 = 339150
\]

### Step 4: Solve for \(x\)
Add \(2395092\) to both sides:

\[
7656x = 2734242
\]

Finally, divide by \(7656\):

\[
x = \frac{2734242}{7656} \approx 357
\]

### Solution
The solution to the equation is approximately:

\[
x \approx 357
\]

Would you like more details or have any questions?

### Related Questions:
1. How do you find the LCM of more than two numbers?
2. Why is clearing fractions a useful step in solving equations?
3. Can this equation be solved using substitution or elimination methods?
4. What if the denominators were not integers? How would the approach change?
5. How does the accuracy of the solution change if we approximate the LCM?
6. How would you check if the solution \(x = 357\) is correct?
7. Could we have solved this using a numerical method like Newton's method?
8. What happens if we vary the coefficients in the original equation?

**Tip:** When solving equations with fractions, clearing the fractions early in the process can simplify your work and reduce errors.

Learn how to solve the linear equation (x-342)/15 + (x-323)/17 + (x-300)/19 + (x-273)/21 = 10 step-by-step. This problem involves clearing fractions and solving linear equations. Suitable for high school students.

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