Answer:
[tex]x=3[/tex];[tex]y=0[/tex]
Step-by-step explanation:
Double the first one, and add it to the second, to get rid of y
[tex]2I + II :8x -10y +6x -10y = 24+18\\14x = 42 \rightarrow x=3\\[/tex]
For the second value, let's take 3 times the first, subtract twice the second, and solve:
[tex]3I-2II: 12x-15y-(12x+20y)=36-36\\-15y-20y=0 \rightarrow -45y=0 \rightarrow y=0[/tex]
