You may use elimination method, hence, if you decide to eliminate the variable x, you may multiplicate by 4 the first and you need to add the result to the second equation, such that:

`4(-x+3y+10z) = 4*8 => -4x + 12y + 40z = 32` (multiplication by 4)

`-4x + 12y + 40z + 4x - 9y - 34z= 32- 17` (addition to the second equation)

Reducing like terms yields:

`3y + 6z = 15 `

Factoring out 3 yields:

`3(y + 2z) = 15 => y + 2z = 5 => y = 5 - 2z`

You need to eliminate the variable x considering the first and the third equation, hence, you should multiplicate the first equation by 3 such that:

`3(-x+3y+10z) = 3*8 => -3x + 9y + 30z = 24`

You need to add the result to the third equation such that:

`3x + 5y - 2z - 3x + 9y + 30z= 46 + 24`

Reducing like terms yields:

`14y + 28z = 70 `

Factoring out 7 yields:

`7(2y + 4z) = 70 => 2y + 4z = 10 => y + 2z = 5`

You need to substitute `5 - 2z` for y in the given equations of system such that:

`-x+3(5-2z)+10z=8 => -x+15-6z+10z=8 => -x+4z=8-15=>-x+4z=-7`

`3x+5(5-2z)-2z=46 => 3x+25-10z-2z=46=> 3x-12z=46-25 => 3x-12z=21 => x-4z=7`

Adding the equations `x+4z=-7` and `x-4z=7` yields:

`x-4z+x+4z=7-7 => 2x = 0 => x = 0`

Substituting 0 for x in the equations of the system yields:

`{(3y+10z=8),(-9y-34z=-17),(5y-2z=46):}`

Multiplicating the first equation by 3 yields:

`9y + 30z = 24`

Add this equation to the second `-9y-34z=-17` such that:

`9y + 30z - 9y - 34z = 24 - 17`

`-4z = 7 => z = -7/4`

Substituting `-7/4` for z in equation that relates z and y yields:

`y = 5 - 2z => y = 5 + 14/4 => y = 5 + 7/2 => y = 17/2`

**Hence, evaluating the solution to the system of equations yields `x=0, y=17/2, z=-7/4.` **