5x-4y-6z = -3 X-3y+z=-1 -3x-6y+7z=1

 To solve the system of equations:

1) 5x - 4y - 6z = -3 2) x - 3y + z = -1 3) -3x - 6y + 7z = 1

We can use the method of substitution or elimination. For the sake of simplicity, let's use the elimination method:

Step A: Let's first align the equations to clearly see the coefficients:

5x - 4y - 6z = -3 x - 3y + z = -1 -3x - 6y + 7z = 1

Step B: Now let’s eliminate one variable from two pairs of equations. For example, we could eliminate x from equations 1 and 2, and from equations 1 and 3.

First, let's multiply the second equation by 5 so that we have the same coefficient for ‘x’ in the first and second equation:

(5) * (x - 3y + z = -1) => 5x - 15y + 5z = -5

Now we have:

5x - 4y - 6z = -3 ...(i) 5x - 15y + 5z = -5 ...(ii)

Let’s subtract (ii) from (i):

(5x - 4y - 6z) - (5x - 15y + 5z) = -3 - (-5) => 5x - 5x - 4y + 15y - 6z - 5z = -3 + 5 => 0x + 11y - 11z = 2 => 11y - 11z = 2 ...(iii)

Now, to eliminate x from equation 1 and equation 3, we can multiply equation 3 by 5:

(5) * (-3x - 6y + 7z = 1) => -15x - 30y + 35z = 5

Now we have:

5x - 4y - 6z = -3 ...(i) -15x - 30y + 35z = 5 ...(iv)

Let’s add (iv) and (i):

(5x - 4y - 6z) + (-15x - 30y + 35z) = -3 + 5 => 5x - 15x - 4y - 30y - 6z + 35z = 2 => -10x - 34y + 29z = 2 ...(v)

Step C: Now we have two new equations with two variables:

11y - 11z = 2 ...(iii) -10x - 34y + 29z = 2 ...(v)

The next step would be to eliminate one more variable, possibly by making y or z the subject from (iii) and substituting into (v), or continuing with the elimination method.

Step D: Let's say we make z the subject in equation (iii):

11y - 11z = 2 => z = (11y - 2) / 11 ...(vi)