start with
.........eq.2.......both sides multiply by .........eq.3
---------------------------------------------- .........eq.2 .........eq.3
-------------------------------------------------add both
........solve for ............eq.1a
go to
.........eq.1 .........eq.3......both sides multiply by
-------------------------------------------------- .........eq.1 .........eq.3
---------------------------------------------------add both
.......solve for ............eq.1b
from eq.1and eq.1b we have
.......sole for ........both sides multiply by .............eq1c
go to .........eq.3, substitute from eq.1a and from eq.1c
There are always dozens of different paths you can take to solve a system of equations like this. One path I would not take is the one the other tutor showed, which leads you through a whole bunch of ugly fractions.
In general, in my opinion, the easiest way to solve a system like this is to eliminate one variable at a time. Eliminate one variable to give a system of two equations in two unknowns; then eliminate one of the remaining variables and solve for the other. Then work back through your equations to solve for the other two variables.
Even with that general method, there are multiple paths you can take. Here is one....
The coefficients make it look as if eliminating either y or z first will be easiest; I chose to eliminate z.
If I multiply the first equation by 3 and the second by 4, the coefficients of z will be 12 and -12; when I add the two resulting equations, z will be eliminated.
Similarly if I multiply the second equation by 2 and add it to the third equation.
[1]
[2]
Now eliminate one of the variables between equations [1] and [2]. Eliminating y looks easier; multiply the first equation by 9 and the second by 2 and add.
[3]
Now work backwards to solve for the other variables.
Substitute [3] in [2]:
[4]
And now substitute [3] and [4] in one of the original equations:
