[tex]x+y =6 ~~~.....(i)\\\\x =3y -2~~.....(ii)\\\\\text{Substitute}~ x =3y-2~ \text{in equation (i):}\\\\\\3y -2 +y =6\\\\\implies 4y = 6+2\\\\\implies 4y = 8\\\\\\\implies y =\dfrac 84 =2\\\\\text{Substitute y =2 in equation (ii):}\\\\x = 3(2) -2\\\\\implies x = 6-2 =4\\\\\\\text{Hence,}~ (x,y) = \left(4, 2\right)[/tex]
