Answer:
Step-by-step explanation:
Elimination sounds the easiest for y:
[tex]I - 2II: 2x+3y -2x -2y = 15-12 \rightarrow y= 3[/tex]
Now we got one value, let's replace in one of the two and find the second
[tex]II: x+(3)=6 \rightarrow x=3[/tex]
[tex]2x + 3y = 15...(1) \\ x + y = 6...(2) \\ 2 \times (2) \\ 2x + 2y = 12...(3) \\ (1) - (3) \\ 0 + 3y - 2y = 15 - 12 \\ \color{red} \boxed{ y = 3 }\\ sub \: in \: (2) \\ x + 3 = 6 \rightarrow \: x = 6 - 3 \\ \color{green} \boxed{ x = 3 }[/tex]
