Do the following pair of linear equations have no solution? Justify your answer.
`2x+4y= 3`
`12y + 6x= 6`

Answer

To determine whether the given pair of linear equations has no solution, we will analyze the equations step by step.

**Given equations:**
1. \( 2x + 4y = 3 \) (Equation 1)
2. \( 12y + 6x = 6 \) (Equation 2)

**Step 1: Rewrite the second equation in standard form.**
We can rearrange Equation 2 to match the standard form \( ax + by = c \).

Starting with Equation 2:
\[ 12y + 6x = 6 \]
Rearranging gives:
\[ 6x + 12y = 6 \]

**Step 2: Identify coefficients.**
From the equations, we can identify the coefficients:
- For Equation 1: \( a_1 = 2, b_1 = 4, c_1 = 3 \)
- For Equation 2: \( a_2 = 6, b_2 = 12, c_2 = 6 \)

**Step 3: Calculate the ratios of the coefficients.**
We will calculate the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2}, \) and \( \frac{c_1}{c_2} \).

1. Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{2}{6} = \frac{1}{3} \]

2. Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{4}{12} = \frac{1}{3} \]

3. Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{3}{6} = \frac{1}{2} \]

**Step 4: Compare the ratios.**
Now we compare the calculated ratios:
- \( \frac{a_1}{a_2} = \frac{1}{3} \)
- \( \frac{b_1}{b_2} = \frac{1}{3} \)
- \( \frac{c_1}{c_2} = \frac{1}{2} \)

**Step 5: Determine if the system has no solution.**
According to the condition for a system of linear equations to have no solution:
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \) is not equal to them, then the system has no solution.

In our case:
- \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{1}{3} \)
- \( \frac{c_1}{c_2} = \frac{1}{2} \)

Since \( \frac{c_1}{c_2} \) is not equal to \( \frac{a_1}{a_2} \) and \( \frac{b_1}{b_2} \), we conclude that the system of equations has no solution.

**Final Conclusion:**
The given pair of linear equations has no solution.

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