Which of the following system of equation has infinitely many solutions ?

Options

A. `2x + 3y = 12, 4x + 6y = 12`
B. `x - 3y = 10, 2x - 6y = 20`
C. `x = 4, y = 3`
D. `3x - 4y = 0, 3x + 4y = 0`

Answer

To determine which system of equations has infinitely many solutions, we can use the condition that for two equations to have infinitely many solutions, the ratios of their coefficients must be equal. Specifically, for equations of the form:

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

The condition for infinitely many solutions is:

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

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

### Option 1:
Equations:
1. \( 2x + 3y = 12 \)
2. \( 4x + 6y = 12 \)

Rewriting them in standard form:
1. \( 2x + 3y - 12 = 0 \) (Here \( a_1 = 2, b_1 = 3, c_1 = -12 \))
2. \( 4x + 6y - 12 = 0 \) (Here \( a_2 = 4, b_2 = 6, c_2 = -12 \))

Calculating the ratios:
- \( \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2} \)
- \( \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2} \)
- \( \frac{c_1}{c_2} = \frac{-12}{-12} = 1 \)

Since \( \frac{c_1}{c_2} \) is not equal to the other two ratios, this option does not satisfy the condition.

### Option 2:
Equations:
1. \( x - 3y = 10 \)
2. \( 2x - 6y = 20 \)

Rewriting them in standard form:
1. \( x - 3y - 10 = 0 \) (Here \( a_1 = 1, b_1 = -3, c_1 = -10 \))
2. \( 2x - 6y - 20 = 0 \) (Here \( a_2 = 2, b_2 = -6, c_2 = -20 \))

Calculating the ratios:
- \( \frac{a_1}{a_2} = \frac{1}{2} \)
- \( \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2} \)
- \( \frac{c_1}{c_2} = \frac{-10}{-20} = \frac{1}{2} \)

All ratios are equal, so this option satisfies the condition for infinitely many solutions.

### Option 3:
Equations:
1. \( x = 4 \)
2. \( y = 3 \)

These are not in the form of a linear equation with both x and y. They represent vertical and horizontal lines, respectively, and do not satisfy the condition for infinitely many solutions.

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

Rewriting them in standard form:
1. \( 3x - 4y = 0 \) (Here \( a_1 = 3, b_1 = -4, c_1 = 0 \))
2. \( 3x + 4y = 0 \) (Here \( a_2 = 3, b_2 = 4, c_2 = 0 \))

Calculating the ratios:
- \( \frac{a_1}{a_2} = \frac{3}{3} = 1 \)
- \( \frac{b_1}{b_2} = \frac{-4}{4} = -1 \)
- \( \frac{c_1}{c_2} = \frac{0}{0} \) (undefined)

Since the ratios are not equal, this option does not satisfy the condition.

### Conclusion:
The only option that satisfies the condition for infinitely many solutions is **Option 2**.