Which of the following sytem of equations doesnot have a solution ?

Options

A. `2x = 3y, 4x = 5y`
B. `2x + y = 7, 4x + 2y = 14`
C. `3x - 4y = 8, 3x - 4y = 12`
D. `4x + y = 7 , 4y + x = 7`

Answer

To determine which system of equations does not have a solution, we will analyze each option using the condition for no solution in a system of linear equations. The condition states that for two equations of the form:

1. \(a_1x + b_1y + c_1 = 0\)
2. \(a_2x + b_2y + c_2 = 0\)

The system has no solution if:

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]

Now, let's evaluate each option step by step.

### Option 1:
Equations:
1. \(2x = 3y\) (or \(2x - 3y = 0\))
2. \(4x = 5y\) (or \(4x - 5y = 0\))

Here:
- \(a_1 = 2\), \(b_1 = -3\), \(c_1 = 0\)
- \(a_2 = 4\), \(b_2 = -5\), \(c_2 = 0\)

Calculating:
\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \]
\[ \frac{b_1}{b_2} = \frac{-3}{-5} = \frac{3}{5} \]
\[ \frac{c_1}{c_2} = \frac{0}{0} \text{ (undefined)} \]

Since \(\frac{1}{2} \neq \frac{3}{5}\) and the condition is not satisfied, this option has a solution.

### Option 2:
Equations:
1. \(2x + y = 7\) (or \(2x + y - 7 = 0\))
2. \(4x + 2y = 14\) (or \(4x + 2y - 14 = 0\))

Here:
- \(a_1 = 2\), \(b_1 = 1\), \(c_1 = -7\)
- \(a_2 = 4\), \(b_2 = 2\), \(c_2 = -14\)

Calculating:
\[ \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \]
\[ \frac{b_1}{b_2} = \frac{1}{2} \]
\[ \frac{c_1}{c_2} = \frac{-7}{-14} = \frac{1}{2} \]

Since \(\frac{1}{2} = \frac{1}{2} = \frac{1}{2}\), this option has a solution.

### Option 3:
Equations:
1. \(3x - 4y = 8\) (or \(3x - 4y - 8 = 0\))
2. \(3x - 4y = 12\) (or \(3x - 4y - 12 = 0\))

Here:
- \(a_1 = 3\), \(b_1 = -4\), \(c_1 = -8\)
- \(a_2 = 3\), \(b_2 = -4\), \(c_2 = -12\)

Calculating:
\[ \frac{a_1}{a_2} = \frac{3}{3} = 1 \]
\[ \frac{b_1}{b_2} = \frac{-4}{-4} = 1 \]
\[ \frac{c_1}{c_2} = \frac{-8}{-12} = \frac{2}{3} \]

Since \(1 = 1 \neq \frac{2}{3}\), this option does not have a solution.

### Option 4:
Equations:
1. \(4x + y = 7\) (or \(4x + y - 7 = 0\))
2. \(4y + x = 7\) (or \(x + 4y - 7 = 0\))

Here:
- \(a_1 = 4\), \(b_1 = 1\), \(c_1 = -7\)
- \(a_2 = 1\), \(b_2 = 4\), \(c_2 = -7\)

Calculating:
\[ \frac{a_1}{a_2} = \frac{4}{1} = 4 \]
\[ \frac{b_1}{b_2} = \frac{1}{4} \]
\[ \frac{c_1}{c_2} = \frac{-7}{-7} = 1 \]

Since \(4 \neq \frac{1}{4} \neq 1\), this option has a solution.

### Conclusion:
The system of equations that does not have a solution is **Option 3: \(3x - 4y = 8\) and \(3x - 4y = 12\)**.

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