Question

Help please!!!!!!!!!!!!!

Answer 1

Combine the 2nd and 3rd equations.  3x will disappear:

 2y + 5z = 7  This is true for any y and z, and is independent of x.
 
                                                            7-5z
Solving for y, 2y = 7 - 5z, so that y = ---------
                                                               2

We have to eliminate x again by using the 1st and 2nd equations:

 -2x - 6y - 2z = -2.  We want the coeff. of x in the first eqn to be -3.  

Therefore, mult. all terms of  -2x - 6y - 2z = -2  by 3/2:

(3/2)(-2x - 6y - 2z = -2) = -3x -9y  -3z =-3

Now add this version of the 1st row to the 2nd row:

  -3x -9y  -3z =-3
   3x +2y +5z = 7
----------------------
         -7y + 2z = 4
                                          7 - 5z
We found earlier that y = ----------, and can elim. y by subst. this fraction into 
                                              2

 -7y + 2z = 4.  Then     -7y + 2z = 4 becomes   -7( (7-5z)/2 ) + 2Z = 4.

Elim. the fraction by mult. all terms by 2:

-7(7-5z) + 4z = 8    which becomes    -49 + 35z + 4z = 8

so that 39z = 57, and so z = 57/39.

Subst. this back into -7y + 2z = 4 to calculate y.  Make a final subst. to calculate x.

Given the choice, I would solve this system using matrices.

Turns out that the solution, using matrices, does not exist.