Question 1150840

{{{Y - 4Z = 8}}}........eq.1
{{{2X - 3Y + 2Z = 1}}}........eq.2
{{{5X - 8Y + 7Z = 1}}}........eq.3

start with

{{{Y - 4Z = 8}}}........eq.1, solve for {{{Y }}}

{{{Y =4Z+ 8}}}...........eq.1a

go to

{{{2X - 3Y + 2Z = 1}}}........eq.2, substitute {{{Y }}}
{{{2X - 3(4Z+ 8) + 2Z = 1}}}
{{{2X - 12Z-24 + 2Z = 1}}}
{{{2X - 10Z = 1+24}}}
{{{2X - 10Z = 25}}}......solve for {{{X}}}
{{{2X  = 10Z+25}}}
{{{X  = 5Z+25/2}}}..........eq.2a

go to

{{{5X - 8Y + 7Z = 1}}}........eq.3, substitute {{{Y }}}

{{{5X - 8(4Z+ 8) + 7Z = 1}}}
{{{5X - 32Z- 64 + 7Z = 1}}}
{{{5X - 25Z=64+ 1}}}
{{{5X - 25Z=65}}}
{{{5X =25Z+65}}}............both sides divide by {{{5}}}
{{{X =5Z+13}}}........eq.3a

from eq.2a and eq.3a we have

{{{ 5Z+25/2=5Z+13}}}......as you can see, we have {{{5Z}}} on both side which means it will cancel each other
therefore,  {{{highlight(no)}}} {{{highlight(solutions)}}} {{{highlight(exist )}}}