`cx-5y=6`
`2x-3y=8`
In the system of equations above, c is a constant and x and y are variables. For what values of c will the system have no solutions?

Options

A. `-(10)/(3)`
B. `(-13)/(11)`
C. `(13)/(11)`
D. `(10)/(3)`

Answer

To determine the values of \( c \) for which the system of equations has no solutions, we can follow these steps:

### Step 1: Write the equations in standard form
The given equations are:
1. \( cx - 5y = 6 \)
2. \( 2x - 3y = 8 \)

We can rewrite these equations in the form \( ax + by + c = 0 \):
1. \( cx - 5y - 6 = 0 \) (Let's call this Equation 1)
2. \( 2x - 3y - 8 = 0 \) (Let's call this Equation 2)

### Step 2: Identify coefficients
From the equations, we identify the coefficients:
- For Equation 1: \( a_1 = c \), \( b_1 = -5 \), \( c_1 = -6 \)
- For Equation 2: \( a_2 = 2 \), \( b_2 = -3 \), \( c_2 = -8 \)

### Step 3: Use the condition for no solutions
For the system of equations to have no solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]

### Step 4: Set up the equality
Using the coefficients:
\[ \frac{c}{2} = \frac{-5}{-3} \]
This simplifies to:
\[ \frac{c}{2} = \frac{5}{3} \]

### Step 5: Solve for \( c \)
To solve for \( c \), we cross-multiply:
\[ c \cdot 3 = 2 \cdot 5 \]
\[ 3c = 10 \]
\[ c = \frac{10}{3} \]

### Step 6: Verify the condition for no solutions
Now we need to check that:
\[ \frac{c}{2} \neq \frac{-6}{-8} \]
Calculating \( \frac{-6}{-8} \):
\[ \frac{-6}{-8} = \frac{3}{4} \]
Now substituting \( c = \frac{10}{3} \):
\[ \frac{10/3}{2} = \frac{10}{6} = \frac{5}{3} \]
We need to check if \( \frac{5}{3} \neq \frac{3}{4} \). Since \( \frac{5}{3} \) is not equal to \( \frac{3}{4} \), the condition is satisfied.

### Conclusion
Thus, the value of \( c \) for which the system of equations has no solutions is:
\[ \boxed{\frac{10}{3}} \]