Question 1097690
<br>There are dozens of different paths you could take to solve this system of equations....<br>
I would be inclined to clear fractions in the first and third equations, since working with fractions always slows things down and makes it easier to make careless mistakes.  So<br>
{{{x+3y+z = 3}}}  (1)
{{{x+y+z = 5}}}  (2)
{{{2x+3y+z = 6}}}  (3)<br>
Comparing (1) and (3) immediately gives us x=3; comparing (1) and (2) yields 2y=-2, so y=-1.  Then substituting x=3 and y=-1 in (2) gives us z=3.<br>
Answer:  (x,y,z) = (3,-1,3)