You should notice that only the second equation has all three variables, hence, you should use the first equation to write z in terms of x and the third equation to write y in terms of x such that:

`6x - 5z = 0 => -5z = -6x => z = 6x/5`

`2x + y = -41 => y = -41 - 2x`

You may substitute 6x/5 for z and `-41 - 2x` for y in the second equation such that:

`5x - y + 3z = -1`

`5x - (-41 - 2x) + 3*(6x/5) = -1`

`5x + 41 + 2x + 18x/5 = -1`

Bringing the terms to a common denominator yields:

`5*7x + 5*41 + 18x = 5*(-1)`

`35x + 18x + 205 = -5`

`53x = -205 - 5 => 53x = -210 => x = -210/53`

You need to substitute -`210/53` for `x ` in `y = -41 - 2x` such that:

`y = -41 + 420/53 => y = (-41*53 + 420)/53 => y = -1753/53`

You need to substitute -`210/53` for `x` in `z = 6x/5` such that:

`z = 6(-210/53)/5 => z = -1260/265 => z = -252/53`

**Hence, evaluating the solution to the given system of equations yields `x = -210/53 , y = -1753/53 , z = -252/53.` **

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