Question 53677: This is a question involving 3 variables, solve by elimination:
2x - 5y + 3z = -1
x + 4y - 2z = 9
x - 2y - 4z = -5
Answer by funmath(2933)   (Show Source):
You can put this solution on YOUR website! You'll like these much better once you are introduced to matrices. Until then, we eliminate.
2x-5y+3z=-1
x+4y-2z=9
x-2z-4z=-5
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(Step 1)Eliminate the x in the first 2 equations:
2x-5y+3z=-1
x+4y-2z=9
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Multiply the second equation by -2 and add the two equations together.
2x-5y+3y=-1
-2(x+4y-2z)=-2(9)
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2x-5y+3y=-1
-2x-8y+4z=-18
_____________
0x-13y+7z=-19
The result of step 1 is -13y+7y=-19
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(Step 2)Eliminate the x in the second two equations:
x+4y-2z=9
x-2y-4z=-5
----
Multiply the the last equation by -1 and add the two equations together.
x+4y-2z=9
-1(x-2y-4z)=-1(-5)
---
x+4y-2z=9
-x+2y+4z=5
__________
0x+6y+2z=14
The result of step 2 is 6y+2z=14
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(Step 3) Eliminate the z's in the two equations that are the results of steps 1 and 2 and solve for y.
-13y+7z=-19
6y+2z=14
----Multiply the result of step one by-2 and multiply the result of step 2 by 7 and add the two equations together.
-2(-13z+7z)=-2(-19)
7(6y+2z)=7(14)
----
26y-14z=38
42y+14z=98
__________
68y+0z=136
68y=136
68y/68=136/68
y=2
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(Step 4) Substitute y=2 into the result of step 2 and solve for z:
6(2)+2z=14
12+2z=14
-12+12+2z=14-12
2z=2
2z/2=2/2
z=1
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(step 5) Substitute the y=2 that we got in step 3 and the z=1 that we got from step 4 into one of the original equations (I'm picking the last one because it looks easiest) and solve for x.
x-2(2)-4(1)=-5
x-4-4=-5
x-8=-5
x-8+8=-5+8
x=3
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(Step 6) Check your answer (3,2,1) in all three equations to see if they check. There are way too many ways to mess these things up!
2(3)-5(2)+3(1)=-1
6-10+3=-1
-1=-1 Equation 1 checks, we're looking good.
(3)+4(2)-2(1)=9
3+8-2=9
9=9 Equation 2 checks, we're looking better.
(3)-2(2)-4(1)=-5
3-4-4=-5 Equation 3 checks, we're right!!!!
The answer is: (x,y,z)=(3,2,1).
Happy Calculating!!!!
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