How many solutions will a pair of linear equations have, if the equations are ` 4 x + 5y - 6 = 0` and `16 x + 20 y + 20 = 0` ?

Options

A. 0
B. 1
C. 2
D. Infinite

Answer

To determine how many solutions the given pair of linear equations has, we will analyze the equations step by step.

### Given Equations:
1. \( 4x + 5y - 6 = 0 \)
2. \( 16x + 20y + 20 = 0 \)

### Step 1: Rewrite the equations in standard form
We can rewrite both equations in the form \( Ax + By + C = 0 \).

1. The first equation is already in the standard form:
\[ 4x + 5y - 6 = 0 \]

2. For the second equation, we can rearrange it:
\[ 16x + 20y + 20 = 0 \implies 16x + 20y = -20 \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation \( 4x + 5y - 6 = 0 \):
- \( a_1 = 4 \)
- \( b_1 = 5 \)
- \( c_1 = -6 \)

- For the second equation \( 16x + 20y + 20 = 0 \):
- \( a_2 = 16 \)
- \( b_2 = 20 \)
- \( c_2 = -20 \)

### Step 3: Calculate the ratios
Now we will calculate the ratios \( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \), and \( \frac{c_1}{c_2} \).

1. Calculate \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{4}{16} = \frac{1}{4} \]

2. Calculate \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{5}{20} = \frac{1}{4} \]

3. Calculate \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{-6}{-20} = \frac{6}{20} = \frac{3}{10} \]

### Step 4: Analyze the ratios
Now we compare the ratios:
- \( \frac{a_1}{a_2} = \frac{1}{4} \)
- \( \frac{b_1}{b_2} = \frac{1}{4} \)
- \( \frac{c_1}{c_2} = \frac{3}{10} \)

### Conclusion
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) but \( \frac{c_1}{c_2} \) is not equal to these ratios, it indicates that the two lines represented by the equations are parallel and do not intersect. Therefore, there are no solutions to the system of equations.

### Final Answer:
**The pair of linear equations has no solutions.**

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