Question 346673:
Algebra- solving sytems of equations in three varaiables:
( Answer is needed within 1.5 hours, by 11"30 AM Pacific time - Thanks!)
3x+5y+4z =13
5z+2y+3z=-9
6x+3y+4z=-8
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! Algebra- solving sytems of equations in three varaiables:
( Answer is needed within 1.5 hours, by 11"30 AM Pacific time - Thanks!)
3x+5y+4z =13
5z+2y+3z=-9 (--> assuming this is 5x + 2y + 3z = -9)
6x+3y+4z=-8
using method of elimination, choosing z first since I see two 4z's
use 1st equation and multiply 3rd equation by -1
3x + 5y + 4z = 13
-6x - 3y - 4z = 8
add these to get -3x + 2y = 21
use 1st equation multiplied by 3 and 2nd equation multipled by -4
9x + 15y + 12z = 39
-20x - 8y - 12z = 36
add these to get -11x + 7y = 75
we now have:
-3x + 2y = 21 (multiply this one by 7)
-11x + 7y = 75 (multiply this one by -2)
to get:
-21x + 14y = 147
22x - 14y = -150
add these to get x = -3
plug x = -3 into -3x + 2y = 21
-3(-3) + 2y = 21
9 + 2y = 21
2y = 12
y = 6
now going back to the original equations, plug x = -3 and y = 6
into 3x + 5y + 4z = 13
3(-3) + 5(6) + 4z = 13
-9 + 30 + 4z = 13
21 + 4z = 13
4z = -8
z = -2
solution is (-3,6,-2)
hope this helps you
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