It is true to say that the pair of equations `- 2x + y + 3 =0 and 1/3 x + 2y -1=0` has a unique solution ? Justify your answer.

Answer

To determine whether the pair of equations \(-2x + y + 3 = 0\) and \(\frac{1}{3}x + 2y - 1 = 0\) has a unique solution, we can use the condition for a unique solution of a system of linear equations.

### Step 1: Rewrite the equations in standard form
The given equations are:
1. \(-2x + y + 3 = 0\) can be rewritten as \(-2x + y = -3\).
2. \(\frac{1}{3}x + 2y - 1 = 0\) can be rewritten as \(\frac{1}{3}x + 2y = 1\).

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \(-2x + 1y = -3\):
- \(a_1 = -2\)
- \(b_1 = 1\)
- \(c_1 = -3\)

- For the second equation \(\frac{1}{3}x + 2y = 1\):
- \(a_2 = \frac{1}{3}\)
- \(b_2 = 2\)
- \(c_2 = 1\)

### Step 3: Calculate the ratios
To check for a unique solution, we need to calculate the ratios \(\frac{a_1}{a_2}\) and \(\frac{b_1}{b_2}\):
- Calculate \(\frac{a_1}{a_2}\):
\[ \frac{a_1}{a_2} = \frac{-2}{\frac{1}{3}} = -2 \times 3 = -6 \]

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

### Step 4: Compare the ratios
Now we compare the two ratios:
- \(\frac{a_1}{a_2} = -6\)
- \(\frac{b_1}{b_2} = \frac{1}{2}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) (i.e., \(-6 \neq \frac{1}{2}\)), the condition for a unique solution is satisfied.

### Conclusion
Thus, the pair of equations \(-2x + y + 3 = 0\) and \(\frac{1}{3}x + 2y - 1 = 0\) has a unique solution.

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