`2x + y + 3z = 1, 2x + 6y + 8z = 3, 6x + 8y + 18z = 5` Solve the system of linear equations and check any solutions algebraically.

EQ1:  `2x+y+3z=1`

EQ2:  `2x+6y+8z=3`

EQ3: `6x+8y+18z=5`

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.

`2x+y+3z=1`

`6x+8y+18z=5`

So multiply EQ1 by -3.

`-3(2x+y+3z)=1*(-3)`

`-6x-3y-9z=-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`

`------------`

                  `4z=0`

Then, solve for z.

`4z=0`

`z=0/4`

`z=0`

Then, plug-in this to either EQ4 or EQ5 to get the value of y.

`5y+9z=2`

`5y+9(0)=2`

`5y=2`

`y=2/5`

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.

`2x+y+3z=1`

`2x+2/5+3*0=1`

`2x+2/5=1`

`2x=1-2/5`

`2x=3/5`

`x=(3/5)/2`

`x=3/10`

Therefore, the solution is `(3/10, 2/5,0)` .

-6x-3y-9z=-3