Find the solution of the equation to X upon 3 + 1 is equal to 7 X upon 15 + 3

We are given the equation:

\[
\frac{X}{3} + 1 = \frac{7X}{15} + 3
\]

### Step 1: Eliminate the fractions
To eliminate the fractions, we multiply through by the least common denominator (LCD) of 3 and 15, which is 15:

\[
15 \times \left(\frac{X}{3} + 1\right) = 15 \times \left(\frac{7X}{15} + 3\right)
\]

This gives:

\[
5X + 15 = 7X + 45
\]

### Step 2: Simplify the equation
Now, let's move all the terms involving \(X\) to one side and the constants to the other side:

\[
5X - 7X = 45 - 15
\]

\[
-2X = 30
\]

### Step 3: Solve for \(X\)
Now, divide both sides by \(-2\) to solve for \(X\):

\[
X = \frac{30}{-2} = -15
\]

### Final Answer:
The solution to the equation is \(X = -15\).

Would you like a detailed breakdown of any step or have questions about this solution?

Here are some related questions for practice:
1. Solve: \(\frac{2X}{4} + 3 = \frac{5X}{6} + 7\).
2. Solve: \(\frac{X}{2} - 4 = \frac{3X}{8} + 2\).
3. Solve for \(X\): \(\frac{X}{5} + 2 = \frac{4X}{10} + 1\).
4. What is the solution to: \(\frac{3X}{7} - 2 = \frac{X}{14} + 5\)?
5. Solve for \(X\): \(\frac{X}{4} + 6 = \frac{5X}{12} - 3\).

**Tip:** When solving equations with fractions, multiplying through by the least common denominator simplifies the equation and helps eliminate the fractions.

Find the solution of the equation to X upon 3 + 1 is equal to 7 X upon 15 + 3.

Discover how to solve the linear equation X/3 + 1 = 7X/15 + 3 with a clear step-by-step explanation. This algebra problem covers fractions, eliminating denominators, and rearranging terms to find X. Ideal for students in Grades 6-8 looking to improve their algebra skills.

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