In a system of three equations, this method starts by reducing the system to one of two variables. To do this, we isolate one of the variables from one the system equations, then we substitute it in the remaining equations to eliminate said variable.
z - 2y = x & (I) y+2z=2 & (II) 3z-y-2x = 10 & (III) 1.6cm
Solve for $z$ in Eq.(I): 2cm
LHS+$2y$ = RHS+$2y$ 2cm
z = x -2y 2.7cm
Substitue $z$: 4cm
in Eq.(II) 2.75cm in Eq(III) 0.7cm
y+2(x+2y) = 2 0.85cm 3(x+2y)-y-2x=10
y +2x +4y = 2 1.1cm3x +6 y -y -2x =10
5y +2x = 2 2.7cm 5y +x = 10
We obtain a system of two equations 0.8cm
with two variables 2.25cm
5y +2x = 2 & (I) 5y +x = 10 & (II) 1.9cm
At this point we can continue by solving the new system to find the values for the variables x and y in this case. Once they are found, we can substitute their values in any of the equations of the original system to find the missing variable.
When Is It Preferred?
As the Substitution Method works by isolating a variable in one of the equations and then substituting it in the others, having a system in which we can solve for one variable easily or having it already isolated can facilitate using this method. However, preferences may vary, and we are always free to choose the method we like.
