To identify which equation could represent a linear combination of the given system, we begin by considering the two equations in the system:
[tex]\[
(1) \quad \frac{2}{3} x + \frac{5}{2} y = 15
\][/tex]
[tex]\[
(2) \quad 4x + 15y = 12
\][/tex]
A linear combination of these equations is formed by multiplying each equation by a constant and then adding or subtracting the resulting equations from each other.
Consider the four given choices and how they might be derived from the original equations. Without loss of generality, let's analyze each choice:
1. [tex]\(\frac{4}{3} x = 42\)[/tex]
2. [tex]\(0 = -78\)[/tex]
3. [tex]\(\frac{15}{2} y = 33\)[/tex]
4. [tex]\(0 = 0\)[/tex]
We need to see which of these could be derived as a linear combination of [tex]\((1)\)[/tex] and [tex]\((2)\)[/tex].
### Choice 1: [tex]\(\frac{4}{3} x = 42\)[/tex]
- If we take (1) and (2) and attempt any combination, it is unlikely to get an equation in terms of [tex]\(\frac{4}{3} x = 42\)[/tex] directly. Additionally, given that [tex]\(\frac{4}{3} x = 42\)[/tex] needs to result from manipulating both [tex]\(x\)[/tex] and [tex]\(y\)[/tex] terms combined, it doesn't fit as neatly into a form of these linear combinations.
### Choice 2: [tex]\(0 = -78\)[/tex]
- This implies that through some combination of the given equations, we produce a statement that is always false, indicating no solutions exist. Considering no solution was specified for the system, this looks like it might be our candidate.
### Choice 3: [tex]\(\frac{15}{2} y = 33\)[/tex]
- Similar to choice 1, attempting to manipulate both equations to isolate [tex]\(\frac{15}{2} y\)[/tex], and have it equal to 33 seems unlikely. The nature of our linear combinations doesn't easily simplify to such a structure.
### Choice 4: [tex]\(0 = 0\)[/tex]
- While this is always true and hence a valid equation, it doesn't tell us anything about the system being consistent, providing no information about the potential linear combination needed to show inconsistency or the meaningful combination of the initial system.
From analyzing the structure and underlined intent of linear combinations and given the nature of no solution exists, we should verify our earlier correct choice stands.
Thus, the correct linear combination that represents the given system's no solution characteristic is:
[tex]\[
\boxed{0 = -78}
\][/tex]
