Given a linear equation in two variables: 5x + 7y - 8 = 0, which of the following linear equations, along with the given equation, forms a system of linear equations having no solution?

First linear equation: 5x + 7y - 8 = 0

Two lines are parallel and have no solution if:

​$$\frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{but} \quad \frac{c_1}{c_2} \neq \frac{a_1}{a_2} = \frac{b_1}{b_2}$$​​

Option A:

Second equation: 5x + 7y - 16 = 0

Coefficients of x and y are the same as the given equation (5 and 7).

The constant term is different: −16 instead of −8.

Since the slopes are equal but the constant terms are different, the lines are parallel and have no solution.

Option B:

Second equation: 7x + 5y - 8 = 0

Coefficients of x and y are 7 and 5, respectively, which are not proportional to 5x + 7y.

Since the lines are not parallel, they have a solution.

Option C:

Second equation: 5x - 7y - 8 = 0

Coefficient of x is 5, but the coefficient of y is −7, which is not proportional to 5x + 7y.

Therefore, the lines are not parallel and have a solution.

Option D:

Second equation: 10x + 14y - 16 = 0

Coefficients of x and y are proportional to the first equation (10/5=2, 14/7=2), and the constant term is also proportional (−16/−8 = 2).

This means the two equations represent the same line, and thus the system has infinite solutions.

Thus, only option (a) 5x + 7y - 16 = 0 is correct.