3x+3y=12 2x-4y=14

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3x+3y=12 ------ eq'1' 2x-4y=14 -------eq'2' multiplying eq'1' by 4 & eq'2' by 3 we get 12x+12y=48 ------- eq'3' 6x-12y=42  --------eq'4' adding eq'3' & eq'4' (eliminating 'y') we get 18x=90 x=90/18 x=5 substituting value of 'x' in eq '1' 3(5)+3y=12 15+3y=12 3y=12-15 3y=-3 y=-3/3 y=-1 hence, solution set is {5,-1}