2x-y=6 y-x=3+x The system of equations above has how many solutions?

In any question on equations, your first job is to analyse whether the equations given are dependent or independent, consistent or inconsistent. Dependent equations have infinite solutions, inconsistent equations have NO solution and consistent independent equations have a UNIQUE solution/set.

The standard form of a linear equation is ax + by = c. If we have two instances of this equation, say, \( a_1x + b_1y = c_1\) and \(a_2x + b_2y = c_2\), then,

If \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\), the equations are consistent, dependent and have infinite solutions.

If \(\frac{a_1}{a_2}\) ≠ \(\frac{b_1}{b_2}\), the equations are consistent, independent and have a UNIQUE solution.

If \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\), the equations are inconsistent and have NO solution.

With this information, let us look at the equations given. We have,

2x – y = 6 and y-x = 3+x. Re-organising the second equation, we have 2x – y = -3. Comparing with the standard form, we have, \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{1}{1}\).

But, \(\frac{c_1}{c_2}\) = \(\frac{6}{-3}\) = -2. Therefore,

\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) ≠ \(\frac{c_1}{c_2}\)

We have a set of inconsistent equations and hence these equations will not have any solution. The correct answer option is A.

Hope that helps!