EQ1: `2x+y-3z=10`

EQ2: `y+z=12`

EQ3: `z=2`

In this system of equations, the value of variable z is known. So to get the values of the variables substitute z=2 to one of the equations. It is better if it is plug-in to the second equation since it composed of two variables only.

y + z=12

y+2=12

Then, solve for y.

y=12-2

y=10

Now that the values of y and z are known, solve for x. Plug-in them to the first equation.

`2x+y-3z = 10`

`2x+10-3(2)=10`

`2x+10-6=10`

`2x+4=10`

`2x=10-4`

`2x=6`

`x=6/2`

`x=3`

**Therefore, the solution is (3,10,2).**

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