How many solution does the pair of equations x + y = 1 and x + y= -5 have?

Options

A. Unique
B. No Solution
C. Infinitely many
D. Can't decide

Answer

To determine how many solutions the pair of equations \(x + y = 1\) and \(x + y = -5\) has, we will follow these steps:

### Step 1: Write the equations in standard form
We will convert both equations into the standard form \(Ax + By + C = 0\).

1. For the first equation \(x + y = 1\):
\[ x + y - 1 = 0 \]
This gives us \(A_1 = 1\), \(B_1 = 1\), and \(C_1 = -1\).

2. For the second equation \(x + y = -5\):
\[ x + y + 5 = 0 \]
This gives us \(A_2 = 1\), \(B_2 = 1\), and \(C_2 = 5\).

### Step 2: Find the ratios of the coefficients
Next, we will find the ratios of the coefficients \(A\), \(B\), and \(C\).

1. Calculate the ratio \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{1}{1} = 1 \]

2. Calculate the ratio \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{1}{1} = 1 \]

3. Calculate the ratio \( \frac{C_1}{C_2} \):
\[ \frac{C_1}{C_2} = \frac{-1}{5} = -\frac{1}{5} \]

### Step 3: Analyze the ratios
Now we compare the ratios:

- \( \frac{A_1}{A_2} = 1 \)
- \( \frac{B_1}{B_2} = 1 \)
- \( \frac{C_1}{C_2} = -\frac{1}{5} \)

### Step 4: Determine the number of solutions
According to the rules for the number of solutions of a pair of linear equations:
- If \( \frac{A_1}{A_2} = \frac{B_1}{B_2} \) but \( \frac{C_1}{C_2} \) is not equal to these ratios, then the equations are parallel and have no solutions.

Since \( \frac{A_1}{A_2} = \frac{B_1}{B_2} \) (both equal to 1) and \( \frac{C_1}{C_2} \) is not equal to 1 (it is \(-\frac{1}{5}\)), we conclude that the equations are parallel and do not intersect.

### Final Answer
Thus, the pair of equations has **no solutions**.

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