x+y+2z=11 ... x = 11 -y -2z ... z = (11 -x -y)/2
x+y+3z=14 ... x = 14 -y -3z
x+2y -z =5 ... z = 2y +x -5
11 -y -2z = x = 14 -y -3z
11 - 14 -y +y = -3z +2z
-3 +0y = -z
z = 3
3 = 2y +x -5
8 = 2y +x =(x +y) +y
... 8 -y = (x+y)
3 = (11 -x -y)/2
6 = 11 -x -y
-5 = -x -y = -(x+y)
x+y = 5
x+y = 5 = 8 -y
y = 3
x = 2
ANS: x=2, y = z =3
2007-11-15 11:48:23
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answer #1
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answered by David F 5
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(A) x + y + 2z = 11
(B) x + y + 3z = 14
(C) x + 2y - z = 5
First, multiply Eqn A by (-1) and add that result to Eqn B:
-x - y - 2z = -11
x + y + 3z = 14
-----------------------
z = 3
Rebstitute this value of z back into either Eqn A or B to get:
(D) x + y = 5
Do the same for Eqn 3:
(E) x + 2y = 8
We have two unknowns and two equations, which is all we need to get solutions. Multiply Eqn D by (-1) and add the result to Eqn E:
-x - y = -5
x + 2y = 8
---------------
y = 3
Use any of the existing relationships to solve for x (I'll use Eqn D for the heck of it):
x = 5 - y = 5 - 3 = 2
That's it....you're done:
x =2
y = 3
z = 3
2007-11-15 12:07:25
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answer #2
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answered by The K-Factor 3
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x+y+2z=11 ...(1)
x+y+3z=14 ...(2)
x+2y-z=5 ....(3)
(2) - (1)
(x+y+3z) - (x+y+2z) =14 - 11
x+y+3z - x - y - 2z =3
z = 3
Sub z = 3 into (1)
x+y+2*3=11
x + y = 11 - 6 = 5
x + y = 5 ..... (4)
Sub z = 3 into (3)
x+2y-3=5
x+ 2y = 8 ....(5)
(4) - (5)
(x + y) - (x+ 2y) = 5 - 8
x + y - x - 2y = -3
-y = -3
y = 3
Sub y=3 into (4)
x + 3 = 5
x = 2
So x = 2, y = 3, z = 3 .
2007-11-15 11:54:16
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answer #3
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answered by Chan A 3
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Let's look at the 1st to equations... then subtract the 2nd from the 1st (remember, no matter how weird it sounds, you can subtract equations from other equations!):
[1x+1y+2z=11
[1x+1y+3z=14
1st minus 2nd:
(1-1)x+(1-1)y+(2-3)z=(11-14)
0x+0y+(-1)z=-3
-z=-3
z=3
now we'll substitute 3 back in for z in the LAST 2 equations:
x+y+3(3)=14
x+2y-(3)=5
Become
x+y+9=14 >> x+y=5
x+2y-3=5 >> x+2y=8
now we'll subtract (x+y=5) from (x+2y=8)
x+2y=8
minus
x+y=5
equals
y=3
now let's put y=3 and z=3 back into the 1st equation
x+(3)+2(3)=11
x+9=11
x=2
SO x=2,y=3,z=3
2007-11-15 11:53:08
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answer #4
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answered by SaintPretz59 4
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u must set it up, so two variables cancel when you add 2 equations, continue this process until you find all of the variables.
ex: x+y+2z=11
-x- y- 3z=-14
_____________
-z=-3
z=3
plug this in and continue the process to get the other two variables
2007-11-15 11:51:09
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answer #5
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answered by pickle 2
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You will probably have to divide each equation by the no.variable.
2007-11-15 11:45:35
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answer #6
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answered by Honorable D 2
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