Step-by-step explanation:
{β2x + 2y + 3z = 0
{β2x - y + z = β3 β
{2x + 3y + 3z = 5 β
2y + 4z = 2 β·
{β2x + 2y + 3z = 0 β
{β2x - y + z = β3
{2x + 3y + 3z = 5 β
5y + 6z = 5 β·
{5y + 6z = 5
{2y + 4z = 2
ββ
[5y + 6z = 5]
{β2y - 2β
z = β2 >> New Temporary Equation
{2y + 4z = 2
____________
1β
z = 0
___ _
1β
1β
[Plug this back into both equations above to get the y-term of 1, then plugging both z-term and y-term into all three equations at the very top, will give you the x-term of 1]; ;
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