https://file.aiquickdraw.com/mathbot/file/1725934736857chsj741e.jpg

Let's solve the equation from the image:

\[
\frac{3a}{2} + 15 + \frac{2a}{3} = 25
\]

### Step 1: Combine like terms
First, isolate the terms involving \( a \) on one side of the equation. Start by subtracting 15 from both sides:

\[
\frac{3a}{2} + \frac{2a}{3} = 25 - 15
\]

This simplifies to:

\[
\frac{3a}{2} + \frac{2a}{3} = 10
\]

### Step 2: Find a common denominator
To add the fractions \( \frac{3a}{2} \) and \( \frac{2a}{3} \), we need to find a common denominator. The least common denominator (LCD) of 2 and 3 is 6.

Rewrite both fractions with a denominator of 6:

\[
\frac{3a}{2} = \frac{9a}{6}, \quad \frac{2a}{3} = \frac{4a}{6}
\]

Now, combine the fractions:

\[
\frac{9a}{6} + \frac{4a}{6} = \frac{13a}{6}
\]

So the equation becomes:

\[
\frac{13a}{6} = 10
\]

### Step 3: Solve for \( a \)
To get rid of the fraction, multiply both sides of the equation by 6:

\[
13a = 10 \times 6
\]

\[
13a = 60
\]

Now, divide both sides by 13:

\[
a = \frac{60}{13}
\]

### Final Answer:
The value of \( a \) is \( \frac{60}{13} \), which corresponds to option **b**.

Would you like more details or have any questions?

### Related Questions:
1. How would you solve the equation if the fractions had different variables?
2. What is the least common denominator and how is it useful in solving equations with fractions?
3. How can you verify the solution by substituting the value of \( a \) back into the original equation?
4. How would the process differ if there was a negative term in the equation?
5. How can we solve equations with three or more fractions?

### Tip:
Always check your work by substituting the solution back into the original equation to verify its accuracy.

Solve for a: \(\frac{3a}{2} + 15 + \frac{2a}{3} = 25\). Select the correct option from the following: a. \(\frac{6}{13}\), b. \(\frac{60}{13}\), c. \(\frac{13}{60}\), d. \(\frac{10}{13}\).

Learn how to solve a linear equation with fractions, step-by-step. This problem involves combining like terms and finding the least common denominator (LCD) to simplify and solve. Ideal for students in Grades 7-9.

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