Question 31591
if you have not done matrices then you will have to a more long-winded algebraic approach:


I shall do the second example...it looks easier:-). You can then try the first one, in the same way.


First off, we will choose one variable in one of the 3 equations and then sub this into the other 2 equations, leaving us with 2 equations with just unknowns in:


eqn1 - x+y=0
eqn2 - x+z=1
eqn3 - 2x+y+z=2


Looking at this, eqn1 already is just 2 variables...x and y. So my approach will be to get z=? in either of eqn2 or eqn3 and then sub that back into eqn3 or eqn2, depending upon which i chose.


So... i will chose to do the following:
eqn2 - x+z=1
z = 1-x


Sub this into eqn3: 2x+y+z=2
--> 2x+y+(1-x)=2
2x+y+1-x=2
x+y+1=2
x+y=1


So we have x+y=0 --eqn1
and also x+y=1


Looking at these, there can be no solution, since a number added to another number cannot simultaneously be equal to 0 and 1...so no solution.


Hopefully, the other example will lead to a solution.


jon.