If 2a=b, the pair of equations `ax+by=2a^(2)-3b^(2),x+2y=2a-6b` possess

Options

A. no solution
B. only one solution
C. only two solutions
D. an infinite number of solutions

Answer

To solve the problem, we need to analyze the given pair of equations under the condition that \( 2a = b \). The equations are:

1. \( ax + by = 2a^2 - 3b^2 \)
2. \( x + 2y = 2a - 6b \)

### Step 1: Substitute \( b \) in terms of \( a \)

Given that \( b = 2a \), we can substitute \( b \) in both equations.

**Substituting in the first equation:**
\[ ax + (2a)y = 2a^2 - 3(2a)^2 \]
This simplifies to:
\[ ax + 2ay = 2a^2 - 3(4a^2) = 2a^2 - 12a^2 = -10a^2 \]

**Substituting in the second equation:**
\[ x + 2(2a)y = 2a - 6(2a) \]
This simplifies to:
\[ x + 4ay = 2a - 12a = -10a \]

### Step 2: Write the modified equations

Now we have the modified equations:
1. \( ax + 2ay = -10a^2 \) (Equation 1)
2. \( x + 4ay = -10a \) (Equation 2)

### Step 3: Analyze the system of equations

We can express these equations in the standard form \( A_1x + B_1y = C_1 \) and \( A_2x + B_2y = C_2 \).

From Equation 1:
- \( A_1 = a \)
- \( B_1 = 2a \)
- \( C_1 = -10a^2 \)

From Equation 2:
- \( A_2 = 1 \)
- \( B_2 = 4a \)
- \( C_2 = -10a \)

### Step 4: Check for consistency

To determine if the equations have a unique solution, no solution, or infinitely many solutions, we can check the ratios:
\[ \frac{A_1}{A_2} = \frac{a}{1}, \quad \frac{B_1}{B_2} = \frac{2a}{4a} = \frac{1}{2}, \quad \frac{C_1}{C_2} = \frac{-10a^2}{-10a} = a \]

For the equations to have infinitely many solutions, the ratios must be equal:
\[ \frac{a}{1} = \frac{1}{2} = a \]

This implies that \( a = 0 \) or \( a = \frac{1}{2} \).

### Step 5: Conclusion

Since the equations are consistent and represent the same line when \( a \) is not equal to zero, we conclude that:

- If \( a \neq 0 \), the equations represent the same line, thus there are infinitely many solutions.
- If \( a = 0 \), both equations become \( 0 = 0 \), which is also consistent and has infinitely many solutions.

Thus, the final answer is that the pair of equations possesses **infinitely many solutions**.

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