To solve this system of equation, let's apply elimination method. In elimination method, a variable/variables should be removed in order to come up with another equation.
In this system of equation, let's eliminate variable x using EQ1 and EQ2.
`2x + y + 3z=1`
`2x +6y + 8z=3`
To do so, multiply EQ2 by -1.
`-1 * (2x + 6y+8z)=3*(-1)`
`-2x - 6y-8z=-3`
And add this to EQ1.
`2x + y + 3z = 1`
`+` `-2x - 6y -8z=-3`
`-5y - 5z=-2 ` Let this be EQ4.
Eliminate x variable again to come up with another equation that has y and z only. Use EQ1 and EQ3.
So multiply EQ1 by -3.
Then, add this to EQ3.
`-6x - 3y-9z=-3`
`+` `6x + 8y+18z=5`
`5y + 9z = 2 ` Let this be EQ5.
So the two new equations are:
EQ4: `-5y-5z = -2`
EQ5: `5y+ 9z = 2`
Since there two equations still contain more than two variables, let's eliminate another variable. Let's remove the y. To do so, add these two equations.
`-5y - 5z =-2`
`+` `5y +9z=2`
Then, solve for z.
Then, plug-in this to either EQ4 or EQ5 to get the value of y.
Now that the values of y and z are known, solve for x. To do so, plug-in these two values to either EQ1, EQ2 or EQ3.
Therefore, the solution is `(3/10, 2/5,0)` .