`2x - 2y - 6z = -4, -3x + 2y + 6z = 1, x - y - 5z = -3` Solve the system of linear equations and check any solutions algebraically.

EQ1:   `2x-2y-6z=-4`

EQ2:   `-3x + 2y + 6z=1`

EQ3:    `x-y-5z=-3`

To solve this system of equation, let's apply elimination method. In this method, a variable or variables should be removed in order to get the value of the other variable.

Let's eliminate y. To do so, add EQ1 and EQ2.

           `2x-2y-6z=-4`

`+`      `-3x+2y+6z=1`

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

` `

                            `-x = -3`

Then, solve for x.

`(-x)/(-1)=(-3)/(-1)`

`x=3`

Isolate y again.Consider EQ1 and EQ3.

`2x-2y-6z=-4`

`x-y-5z=-3`

To be able to eliminate y, multiply EQ3 by -2. Then, add the two equations.

        `2x-2y-6z=-4`

`+` `-2x+2y+10z=6`

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

                         `4z=2`

And, solve for z.

`(4z)/4=2/4`

`z=1/2`

Now that the values of the two variables are known, let's solve for the remaining variable. Let's plug-in x=3 and z=1/2 to EQ1.

`2x - 2y -6z = -4`

`2(3)-2y-6(1/2)=-4`

`6-2y-3=-4`

`3-2y=-4`

`3-3-2y=-4-3`

`-2y=-7`

(-2y)/(-2)=(-7)/(-2)

`y=7/2`

Therefore, the solution is  `(3, 7/2, 1/2)` .