Do the following pair of linear equations have no solution? Justify your answer.
`3x + y-3= 0`
`2x + (2)/(3)y= 2`

Answer

To determine if the given pair of linear equations has no solution, we can follow these steps:

### Given Equations:
1. \( 3x + y - 3 = 0 \)
2. \( 2x + \frac{2}{3}y = 2 \)

### Step 1: Rewrite the equations in standard form
We can rewrite both equations in the standard form \( Ax + By + C = 0 \).

For the first equation:
\[ 3x + y - 3 = 0 \quad \text{(already in standard form)} \]

For the second equation, we can transpose \( 2 \) to the left side:
\[ 2x + \frac{2}{3}y - 2 = 0 \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \( A_1 = 3, B_1 = 1, C_1 = -3 \)
- For the second equation \( A_2 = 2, B_2 = \frac{2}{3}, C_2 = -2 \)

### Step 3: Check the condition for no solution
For the system of equations to have no solution, the following condition must hold:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2} \]

#### Calculate \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{3}{2} \]

#### Calculate \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{1}{\frac{2}{3}} = \frac{1 \times 3}{2} = \frac{3}{2} \]

#### Calculate \( \frac{C_1}{C_2} \):
\[ \frac{C_1}{C_2} = \frac{-3}{-2} = \frac{3}{2} \]

### Step 4: Compare the ratios
Now we compare the ratios:
- \( \frac{A_1}{A_2} = \frac{3}{2} \)
- \( \frac{B_1}{B_2} = \frac{3}{2} \)
- \( \frac{C_1}{C_2} = \frac{3}{2} \)

Since:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]
This indicates that the system does not satisfy the condition for no solution.

### Conclusion
The given pair of linear equations has a solution because the ratios are equal. Therefore, the system of equations does not have no solution.

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