Question 1062040
You have two equations and two unknowns.  As long as they're linearly independent (i.e. not exact multiples of one another) there should be a solution:
{{{ 2x + y = 10 }}}  (1)
{{{ 2x + 3y = 6 }}}   (2)

Subtract (2) from (1), to get:
{{{ 0x - 2y = 4 }}}
{{{          y = 4/(-2) = -2 }}}
{{{  y=-2 }}} — using (1) —>  {{{ 2x - 2 = 10 }}}  or {{{ x = 6 }}}

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Ans:  x=6, y=-2
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Check (using (2), we already used (1) to find x):
             2(6) + 3(-2) = 12 - 6 = 6  (ok)