Solving Systems of Linear Equations in Three Variables -2x-3y-6z=-26 5x+5y+4z=24 3x+4y-5z=-40
hi are you familiar with matrices ?
or determinants?you can solve them using matricews or determinants
what subject are you in? algebra?
yes algebra 2 but i have to use the linear combination method
ok, so you can solve it by elimination method
not really
-2x-3y-6z=-26 eq (1) 5x+5y+4z=24 eq(2) 3x+4y-5z=-40 eq (3)
multiply eq 1 by 3, multiply eq 3 by 2
ok ill do that right now
on eq 1 you get -6x-9y-18z=-78 on eq 3 you get 6x+8y-10z=-80
then you can add them together,to get -y-28z=-158 ......eq (4)
now on the same prob multiply eq 1by 5 and eq 2 by 2
ok
on eq 1 you get -10x-15y-30z=-130 on eq 2 10x+10y+8z=48 add them together to get ------------------------- 0-5y-22z=-82
0-5y-22z=-82 eq 5 now mult eq 4 by -5 to get 5y+140z=790 eq 6 add them together to get ---------------- 118z=708 z=708/118 z=6
okay hold up a sec.
im getting a bit lost
on where?
take your time..... your getting in there now
equation 4 (-5) -1y-28z=-158 equation 5 -5y-22z=-82 5y+140z=-290 -5y-22z=-82 right?
what do i multipy equation 6 by?
how did you get this 5y+140z=790
now once you get z=6 , subs this to eq 4 to get y =-10
how did you get 790 on equation 4?
from eq 4 mult it by 5 to get 5y+140z=790
oh right right i accidentally wrote -58 instead of -158
ok, then follow the next step
so equation 4 with -1y-28(6)=-158?
yes correct
yes california so i did it with the other equation -5y -22(6)=-82 and got y=-10
yes correct
then subs them to another equations to get x
i sub them to the equations from the beginning right?
yes subs them to get the x
so i sub everything to the equation 2 and im getting a fraction 5x+5(-10)+4(6)=24 5x-50+24=24 5x-26=24 5x=50 x=10 :D i got it.... Thank you sooooooooooooo much!!!
its really nice of you taking your time to help me :)
yw.....ok gud luck now
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