`{:(5x + 2y = 16),(3x + (6)/(5)y = 2):}`

Answer

To solve the given system of equations and determine whether it has a solution, we will follow these steps:

### Step 1: Write down the equations
We have the following two equations:
1. \( 5x + 2y = 16 \) (Equation 1)
2. \( 3x + \frac{6}{5}y = 2 \) (Equation 2)

### Step 2: Rewrite Equation 2 in standard form
To compare the coefficients easily, we can rewrite Equation 2 in the standard form \( Ax + By + C = 0 \).
Starting from:
\[ 3x + \frac{6}{5}y = 2 \]
We can rearrange it as:
\[ 3x + \frac{6}{5}y - 2 = 0 \]
Multiplying through by 5 to eliminate the fraction gives:
\[ 15x + 6y - 10 = 0 \]
Thus, we can express it as:
\[ 15x + 6y + (-10) = 0 \] (Equation 2 in standard form)

### Step 3: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1: \( a_1 = 5, b_1 = 2, c_1 = -16 \)
- For Equation 2: \( a_2 = 15, b_2 = 6, c_2 = -10 \)

### Step 4: Calculate the ratios
Now we calculate the ratios:
1. \( \frac{a_1}{a_2} = \frac{5}{15} = \frac{1}{3} \)
2. \( \frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3} \)
3. \( \frac{c_1}{c_2} = \frac{-16}{-10} = \frac{16}{10} = \frac{8}{5} \)

### Step 5: Compare the ratios
Now we compare the ratios:
- \( \frac{a_1}{a_2} = \frac{1}{3} \)
- \( \frac{b_1}{b_2} = \frac{1}{3} \)
- \( \frac{c_1}{c_2} = \frac{8}{5} \)

### Step 6: Determine the type of solution
According to the conditions:
- If \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), there is a unique solution.
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), there are infinitely many solutions.
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \neq \frac{c_1}{c_2} \), there is no solution.

Here, we have:
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) (both equal to \( \frac{1}{3} \))
- \( \frac{c_1}{c_2} \neq \frac{8}{5} \)

Thus, the equations have **no solution**.

### Final Conclusion
The given set of equations does not have a solution.

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