Is the following of linear equations consistent ? Justify your answer.
`2ax + by =a, 4ax+ 2by-2a=0, a,b, ne0`

Answer

To determine whether the given pair of linear equations is consistent, we will analyze the equations step by step.

### Given Equations:
1. \(2ax + by = a\)
2. \(4ax + 2by - 2a = 0\)

### Step 1: Rewrite the equations in standard form
We can rewrite the second equation for clarity:
\[ 4ax + 2by = 2a \]

Now we have:
1. \(2ax + by = a\) (Equation 1)
2. \(4ax + 2by = 2a\) (Equation 2)

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1:
- \(a_1 = 2a\)
- \(b_1 = b\)
- \(c_1 = -a\)

- For Equation 2:
- \(a_2 = 4a\)
- \(b_2 = 2b\)
- \(c_2 = -2a\)

### Step 3: Check the consistency conditions
To check if the equations are consistent, we will use the following conditions:
1. If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\), the equations are inconsistent.
2. If \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\), the equations are consistent.

#### Calculate \(\frac{a_1}{a_2}\):
\[ \frac{a_1}{a_2} = \frac{2a}{4a} = \frac{1}{2} \]

#### Calculate \(\frac{b_1}{b_2}\):
\[ \frac{b_1}{b_2} = \frac{b}{2b} = \frac{1}{2} \]

#### Calculate \(\frac{c_1}{c_2}\):
\[ \frac{c_1}{c_2} = \frac{-a}{-2a} = \frac{1}{2} \]

### Step 4: Conclusion
Since we have:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2} \]
This satisfies the second condition for consistency.

Thus, the given pair of linear equations is consistent.