The 2 main ways of solving a system of equations are elimination and substitution.
The ideal scenario for elimination is if there are additive inverses in two of the equations.. for instance: #5x-5x =0, -3y+3y =0#
We have 3 equations with 3 unknowns.
Notice that in equations B and C there are 2 pairs of additive inverses!
#color(white)(xxxxxx xx) x+""y+""z=" " 1#......................A
#color(white)(xxxxxx)color(red)(-2x)color(blue)(+2y) +3z = " "20#..................B
#color(white)(xxxxxx/.x)color(red)(2x)color(blue)(-2y) -z =-16#..................C
#B+C:color(white)(xxxxxxxxx)2z =4" "larr# only one variable!
#color(white)(xxxxxxxxxxxxxxx)color(magenta)(z=2)#
Substituting this value of 2 for #z# into equation A gives:
#color(white)(xxxxxxxxxx)x+y color(magenta)(+2) = 1#
This can be re-arranged to give #x # as the subject:
#color(white)(xxxxxxx.xx)color(lime)(x = (-y-1))#
Now replace #color(lime)(x=(-y-1)) and color(magenta)(z=2) " in " C#
#color(white)(xxxxxx/.x)2color(lime)(x)-2y color(magenta)(-z) =-16#..................C
#color(white)(xx)2color(lime)((-y-1))-2y color(magenta)(-2) =-16#..................C
#color(white)(x.xx)-2y-2-2y color(magenta)(-2) =-16#..................C #larr# only y !
#color(white)(x.xxxxxxxx.xx)-4y =-16+4#
#color(white)(x.xxxxxxxx.xx)-4y =-12#
#color(white)(x.xxxxxxxx.xxxxx)y =3#
Now that we have a value for y, we can find x.
#color(lime)(x=-y-1) = -(3)-1#
#x = -4#
