$$ \frac{x+3}{4}=2 x-1 $$ and $$ 3 x+2 y=15 $$

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Let's solve the system of equations step by step:

$$
\begin{cases}
\frac{x + 3}{4} = 2x - 1 \
3x + 2y = 15
\end{cases}
$$

### Step 1: Solve the first equation for $$ x $$

Starting with the first equation:

$$
\frac{x + 3}{4} = 2x - 1
$$

**Multiply both sides by 4** to eliminate the fraction:

$$
x + 3 = 4(2x - 1)
$$

$$
x + 3 = 8x - 4
$$

**Rearrange terms to isolate $$ x $$:**

$$
x - 8x = -4 - 3
$$

$$
-7x = -7
$$

**Divide both sides by -7:**

$$
x = 1
$$

### Step 2: Substitute $$ x = 1 $$ into the second equation to find $$ y $$

Now, use the second equation:

$$
3x + 2y = 15
$$

Substitute $$ x = 1 $$:

$$
3(1) + 2y = 15
$$

$$
3 + 2y = 15
$$

**Subtract 3 from both sides:**

$$
2y = 12
$$

**Divide both sides by 2:**

$$
y = 6
$$

### Solution

The solution to the system is:

$$
x = 1 \quad \text{and} \quad y = 6
$$

You can verify the solution by plugging these values back into both original equations:

1. First equation:
   $$
   \frac{1 + 3}{4} = \frac{4}{4} = 1 \quad \text{and} \quad 2(1) - 1 = 2 - 1 = 1
   $$
   
2. Second equation:
   $$
   3(1) + 2(6) = 3 + 12 = 15
   $$

Both equations are satisfied, confirming that $$ x = 1 $$ and $$ y = 6 $$ are the correct solutions.
The solution is $$ x = 1 $$ and $$ y = 6 $$.

\( \frac{x+3}{4}=2 x-1 \) and \( 3 x+2 y=15 \)

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