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Rezolvati exercitiul prin metoda reducerii sau substitutiei:
3(x-3y)+7=11(y-3)-27
6(3y-x)+24=2(x-3y)-8


Răspuns :

3x-9y+7=11y-33-27 <=>  3x-20y=-67   
18y-6x+24=2x-6y-8 <=> -8x+24y=-32  l:(-8) =>x-3y=4

3x-20y=-67
  x-3y=4  l·(-3)

3x-20y=-67
-3x+9y=-12
___________
  /  -11y=-79 =>y=79/11

3x-20y=-67 l·3
  x-3y=4      l·(-20)

9x-60y=-201
-20x+60y=-80
___________
-11x  /   =-281 =>x=281/11