-5x+y-4z=60 2x+4y+3z=-12 6x-3y-2z=-52 solving a 3 variable linear system
wow. all you have to do is add them i believe.
well my math teacher said to use elimination using two different sets of the 3 so i did but i got z as a decimal and he said all the variables were whole numbers so idk what to do
add 2 of them, so you cab get the better of the 2, then you substitute the vaiable in the last equation.
okay ill try, thanks
@kirstenande this was along question!!! i got::: x=-8 y=4 z=-4
greek, it should be y=5 not 4.
@andriod09 substitute all of my values back into one of the 3 eqns.... LHS=RHS
@kirstenande U getting it???
ohhh fail.
@andriod09 :P
i just re did it and when i got it down to the two equations with two variables i got 22x+19z=-252 -9x-14z=128
i just re did it and when i got it down to the two equations with two variables i got 22x+19z=-252 -9x-14z=128
|dw:1349115063770:dw| first matrix
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