Question 908468
<pre>
Substitution isn't the easiest way. Elimination is easier,
But you specified substitution, so here goes:

(1)  {{{2x+3y=-5}}}
(2)  {{{4y-5z=-32}}}
(3)  {{{3x+2z=14}}}

Pick any equation and solve for any letter:

I'll arbitrarily pick equation 2 and arbitrarily solve for y

{{{4y-5z=-32}}}
{{{4y=5z-32}}}
(4)  {{{y=(5z-32)/4}}}

Substitute for y in (1)

(1)  {{{2x+3y=-5}}}
     {{{2x+3((5z-32)/4)=-5}}}

Multiply through by 4 to clear the fraction:

     {{{8x+3(5z-32)=-20}}}
     {{{8x+15z-96=-20}}}
(5)  {{{8x+15z=76}}}

We notice that (3) contains the same letters as (5)

Solve (3) for one of its letters.

I'll arbitrarily solve (3) for z

(3)  {{{3x+2z=14}}}
     {{{2z=14-3x}}}
(6)  {{{z=(14-3x)/2}}}

Substitute in (5)

(5)   {{{8x+15z=76}}}
      {{{8x+15(14-3x)/2=76}}}

Multiply by 2 to clear the fraction:

      {{{16x+15(14-3x)=152}}}
      {{{16x+210-45x=152}}}
      {{{-29x+210=152}}}
      {{{-29x=-58}}}
      {{{x=(-58)/(-29)}}}
      {{{x=2}}}

Substitute in (6)

(6)  {{{z=(14-3x)/2}}}
     {{{z=(14-3(2))/2}}}
     {{{z=(14-6)/2}}}
     {{{z=8/2}}}
     {{{z=4}}}

(4)  {{{y=(5z-32)/4}}}  
     {{{y=(5(4)-32)/4}}}
     {{{y=(20-32)/4}}}
     {{{y=(-12)/4}}}
     {{{y=-3}}}   

(x,y,z) = (2,-3,4)

Edwin</pre>