Question 1060139

Since you have 2 equations and 2 unknowns, the first thing is to try to solve for one of the variables.  So long as the two equations are "linearly independent" (or not linearly-dependent) there will be a solution.
Linearly dependent means one equation can be re-arranged to exactly match the other (in essence, if they only "look" like two different equations but are in fact the same).  Ok, enough on that…
 
{{{ y=3x+2 }}}  (1)
{{{ 3x+y = 1 }}} (2)

(1) conveniently expresses y as a function of x, so let's substitute "3x+2" from (1) for "y" in (2):

{{{ 3x + (3x+2) = 1 }}}     (substituted "3x+2" for "y")
{{{  6x + 2 = 1 }}}
{{{   6x = -1 }}}
{{{  x = -1/6 }}}

If  {{{x=-1/6}}} we can plug into (1) to get  {{{ y=3(-1/6)+2 = -(1/2)+(4/2) = 3/2 }}}

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Ans:   {{{ x=-1/6 }}}  and  {{{ y=3/2 }}}
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Check:
    (1)    Does  3/2 = (3*(-1/6) + 2)?
                  3/2 = (-3/6)+(12/6)
                  3/2 = 9/6
                  3/2 = 3/2   (ok)
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   (2)   Does 3(-1/6) + (3/2) = 1 ?
                   -(1/2) + (3/2) = 1 ?
                             2/2 = 1   (ok)