A linear combination of the system is any equation formed by the algebraic addition of both equations, one or both multiplied by an arbitrary constant.

To prove that the given system has no solution you could multiply the first equation times 6 (to get rid of the fractions), multiply the second equation times - 1, and add the two results:

1. First equation times 6:

$6\times\frac{2}{3} x+6\times\frac{5}{2}y=6\times 15\\ \\ 4x+15y=90$

2. Second equation times - 1:

$-4x-15y=-12$

3. Add the two new equations:

$0=78$

4. Conclusion:

Since 0 = 78 is false, no matter what the value of x and y are, the conclusion is that the system of equations has not solution.

The only choice that represents that same situation is the second one, 0 = 26. That is a possible linear combination that represents that the system of equations has no solutions.

In fact, you might calculate the exact factors by which you had to multiply each one of the original equations to get 0 = 26, but it is not necessary to tell that that option represents a possible linear combination for the given system of equations.

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