How many solutions does the pair of equations `x + 2y-3=0 and 1/2 x + y - 3/2=0` have ?

Answer

To determine how many solutions the pair of equations \( x + 2y - 3 = 0 \) and \( \frac{1}{2}x + y - \frac{3}{2} = 0 \) has, we will follow these steps:

### Step 1: Identify the coefficients
We need to identify the coefficients of \( x \), \( y \), and the constant term from both equations.

1. For the first equation \( x + 2y - 3 = 0 \):
- Coefficient of \( x \) (denoted as \( a_1 \)) = 1
- Coefficient of \( y \) (denoted as \( b_1 \)) = 2
- Constant term (denoted as \( c_1 \)) = -3

2. For the second equation \( \frac{1}{2}x + y - \frac{3}{2} = 0 \):
- Coefficient of \( x \) (denoted as \( a_2 \)) = \( \frac{1}{2} \)
- Coefficient of \( y \) (denoted as \( b_2 \)) = 1
- Constant term (denoted as \( c_2 \)) = \( -\frac{3}{2} \)

### Step 2: Calculate the ratios
Next, we calculate the ratios \( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \), and \( \frac{c_1}{c_2} \).

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

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

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

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

Since all three ratios are equal, we have:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = 2 \]

### Step 4: Conclusion
According to the conditions for the existence of solutions for a pair of linear equations:
- If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), then the equations have infinitely many solutions.

Thus, the pair of equations \( x + 2y - 3 = 0 \) and \( \frac{1}{2}x + y - \frac{3}{2} = 0 \) has **infinitely many solutions**.

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