Sprawdzimy po kolei wstawiając dane a, który z układów nie ma rozwiązania

A. dla a = 3 mamy

$\{\begin{array}{l}3x-y=5\mid +y\\ 4y-3=4\cdot 3x\end{array}$  

$\{\begin{array}{l}3x=5+y\\ 4y-3=4\cdot 3x\end{array}$  

$\{\begin{array}{l}3x=5+y\\ 4y-3=4\cdot \left(5+y\right)\end{array}$  

$\{\begin{array}{l}3x=5+y\\ 4y-3=20+4y\mid -4y\end{array}$  

$\{\begin{array}{l}3x=5+y\\ -3\ne 20\end{array}$  

sprzeczność

B. dla a = 1 mamy

$\{\begin{array}{l}3x=5+y\mid -5\\ 4y-3=4x\end{array}$  

$\{\begin{array}{l}3x-5=y\\ 4\cdot \left(3x-5\right)-3=4x\end{array}$  

$\{\begin{array}{l}y=3x-5\\ 12x-20-3=4x\end{array}$  

$\{\begin{array}{l}y=3x-5\\ 12x-23=4x\mid -4x\end{array}$  

$\{\begin{array}{l}y=3x-5\\ 8x-23=0\mid +23\end{array}$  

$\{\begin{array}{l}y=3x-5\\ 8x=23\mid :8\end{array}$  

$\{\begin{array}{l}y=3x-5\\ x=\frac{23}{8}\end{array}$  

$\{\begin{array}{l}y=3\cdot \frac{23}{8}-5\\ x=\frac{23}{8}\end{array}$  

$\{\begin{array}{l}y=\frac{69}{8}-\frac{40}{8}\\ x=\frac{23}{8}\end{array}$  

$\{\begin{array}{l}y=\frac{29}{8}\\ x=\frac{23}{8}\end{array}$  

C. dla a = 5 mamy

$\{\begin{array}{l}3x-y=5\mid +y\\ 4y-3=4\cdot 5x\end{array}$  

$\{\begin{array}{l}3x=5+y\mid -5\\ 4y-3=20x\end{array}$  

$\{\begin{array}{l}3x-5=y\\ 4\cdot \left(3x-5\right)-3=20x\end{array}$  

$\{\begin{array}{l}y=3x-5\\ 12x-20-3=20x\end{array}$  

$\{\begin{array}{l}y=3x-5\\ 12x-23=20x\mid -12x\end{array}$  

$\{\begin{array}{l}y=3x-5\\ -23=8x\mid :8\end{array}$  

$\{\begin{array}{l}y=3x-5\\ x=-\frac{23}{8}\end{array}$  

$\{\begin{array}{l}y=3\cdot \frac{-23}{8}-5\\ x=-\frac{23}{8}\end{array}$  

$\{\begin{array}{l}y=\frac{-69}{8}-\frac{40}{8}\\ x=-\frac{23}{8}\end{array}$  

$\{\begin{array}{l}y=-\frac{109}{8}\\ x=-\frac{23}{8}\end{array}$  

D. dla a = 0 mamy

$\{\begin{array}{l}3x-y=5\mid +y\\ 4y-3=0\mid +3\end{array}$  

$\{\begin{array}{l}3x=5+y\\ 4y=3\mid :4\end{array}$  

$\{\begin{array}{l}3x=5+\frac{3}{4}\\ y=\frac{3}{4}\end{array}$  

$\{\begin{array}{l}3x=\frac{23}{4}\mid :3\\ y=\frac{3}{4}\end{array}$  

$\{\begin{array}{l}x=\frac{23}{12}\\ y=\frac{3}{4}\end{array}$  

Układ w podpunkcie A. jest sprzeczny, więc nie ma rozwiązania.

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Poprawna odpowiedź to A.