Whow many of solution of these equations `5x- y= 7 and x-y +1 = 0 ` will be ?

Options

A. 1
B. 0
C. 3
D. infinite

Answer

To determine how many solutions exist for the equations \(5x - y = 7\) and \(x - y + 1 = 0\), we can follow these steps:

### Step 1: Write the equations in standard form
The standard form of a linear equation is typically expressed as \(Ax + By + C = 0\).

1. For the first equation \(5x - y = 7\):
- Rearranging gives us:
\[ 5x - y - 7 = 0 \]
Thus, \(A_1 = 5\), \(B_1 = -1\), and \(C_1 = -7\).

2. For the second equation \(x - y + 1 = 0\):
- This is already in standard form:
\[ x - y + 1 = 0 \]
Thus, \(A_2 = 1\), \(B_2 = -1\), and \(C_2 = 1\).

### Step 2: Compare the coefficients
To determine the number of solutions, we will compare the ratios of the coefficients:

\[ \frac{A_1}{A_2} = \frac{5}{1} = 5 \]
\[ \frac{B_1}{B_2} = \frac{-1}{-1} = 1 \]

### Step 3: Analyze the ratios
Since \(\frac{A_1}{A_2} \neq \frac{B_1}{B_2}\) (5 is not equal to 1), this indicates that the lines represented by these equations are not parallel.

### Step 4: Conclusion on the number of solutions
When two lines are not parallel, they will intersect at exactly one point. Therefore, there is exactly one solution to the system of equations.

### Final Answer
The number of solutions to the equations \(5x - y = 7\) and \(x - y + 1 = 0\) is **1**.

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