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
