Just as there are simultaneous algebraic equations (where a pair of numbers have to satisfy a pair of equations) there are systems of differential equations, (where a pair of functions have to satisfy a pair of differential equations).
Indicate which pairs of functions satisfy this system. It will take some time to make all of the calculations.
y_1' = y_1 -2 y_2 \qquad y_2' = 3y_1 - 4 y_2

A. y_1 = \sin(x) +\cos(x) \qquad y_2 = \cos(x) - \sin(x)
B. y_1 = \sin(x) \qquad y_2 = \cos(x)
C. y_1 = \cos(x) \qquad y_2 = -\sin(x)
D. y_1 = e^{-x} \qquad y_2=e^{-x}
E. y_1 = e^x \qquad y_2=e^x
F. y_1 = e^{4x} \qquad y_2 = e^{4x}
G. y_1 = 2e^{-2x} \qquad y_2 = 3e^{-2x}

As you can see, finding all of the solutions, particularly of a system of equations, can be complicated and time consuming. It helps greatly if we study the structure of the family of solutions to the equations. Then if we find a few solutions we will be able to predict the rest of the solutions using the structure of the family of solutions.

For options D and G we will show that both differential equations are satisfied. For the other options we will show the pairs don't solve one of the equations.

A. [tex] y_1 '= \cos(x)-\sin x[/tex] and [tex] y_1-2y_2= \sin x+\cos x -2(\cosx -\sin x )=3\sin x- \cos x \neq \cos x-\sin x [/tex] (when x=0 the left side is -1 and the right side is 1) so the equation [tex] y_1'=y_1 - 2y_2 [/tex] is not satisfied.

B. [tex] y_2 '= -\sin x[/tex] and [tex] 3y_1-4y_2= 3\sin x-4\cos x \neq -\sin x[/tex] so the equation [tex] y_2'=3y_1-4y_2 [/tex] is not satisfied.

C. [tex] y_1 '= -\sin(x)[/tex] and [tex] y_1-2y_2= \cos x -2\sin x \neq -\sin x[/tex] so these pairs don't solve the equation [tex] y_1'=y_1-2y_2 [/tex].

D. Since [tex]y_1=y_2=e^{-x}[/tex] then [tex]y_1'=y_2'=-e^{-x}[/tex]. The first equation is satisfied, because [tex] y_1-2y_2=e^{-x}-2e^{-x}=-e^{-x}=y_1'[/tex]. The second equation is also satisfied: [tex] 3y_1-4y_2=3e^{-x}-4e^{-x}=-e^{-x}=y_2'[/tex].

E. [tex] y_2'=e^x[/tex] and [tex] 3y_1-4y_2= 3e^x-4e^x=-e^x\neq -e^x[/tex] so they don't satisfy the equation [tex] y_2'=3y_1-4y_2 [/tex].

F. [tex] y_1 '= 4e^{4x}[/tex] and [tex] y_1-2y_2= e^{4x}-2e^{4x}=-e^{4x} \neq 4e^{4x}[/tex], then the equation [tex] y_1'=y_1-2y_2 [/tex] is not satisfied.

G. In this case, [tex]y_1=2e^{-2x}[/tex] and [tex]y_2=3e^{-2x}[/tex]. Computing derivatives, [tex]y_1'=-4e^{-2x}[/tex] and [tex]y_2'=-6e^{-2x}[/tex]. The first equation is satisfied, because [tex] y_1-2y_2=2e^{-2x}-6e^{-2x} =-4e^{-2x}=y_1'[/tex]. The second equation is also satisfied: [tex] 3y_1-4y_2= 6e^{-2x}-12e^{-2x}=-6e^{-2x}=y_2'[/tex].