1x - 1y + 0z =7
2x - 2y - 1z = 14
-3x + 1y - 1z = -7
2006-12-18 00:36:49 · 7 answers · asked by Rocstarr 2 in Science & Mathematics ➔ Mathematics
-2(x-y+0z=7)
2x-2y-z=14
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-2x+2y+0z=-14
2x-2y-z=14
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-z=0
z=0
3(2x-2y=14)
2(-3x+y=-7)
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6x-6y=42
-6x+2y=-14
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-4y=28
y=-7
x-(-7)+0(0)=7
x+7=7
x=0
(0,-7,0)
Check:
1(0)-(-7)+0(0)=7
7=7
2(0)-2(-7)-1(0)=14
-2(-7)=14
14=14
-3(0)+(-7)-1(0)=-7
-7=-7
2006-12-18 05:46:05 · answer #1 · answered by Anonymous · 4⤊ 0⤋
1x - 1y + 0z =7
2x - 2y - 1z = 14
-3x + 1y - 1z = -7
Subtract the third equation from the second equation getting:
5x -3y +0z = 21 Multiply 1st equation by -3 getting:
-3x +3y +0z = -21 Now ad these two new equations getting
2x +0y +0z = 0
So x=0
Put x=0 in 1st equation getting;
0x-1y +0z +7
-y = 7, so y= -7
Put x=0 and y=-7 in second equation getting
2*0 -2(-7) -z = 14
So 14-z =14
z=0
So x=0, y= -7, and z=0
2006-12-18 08:55:13 · answer #2 · answered by ironduke8159 7 · 2⤊ 0⤋
1x - 1y + 0z =7 (1)
2x - 2y - 1z = 14 (2)
-3x + 1y - 1z = -7 (3)
2*(1) - (2) => 2x - 2x - 2y - (-2y) - (-z) = 14 -14
z = 0
So we can ignore all reference to z
1x - 1y =7 (4)
2x - 2y = 14 (identical to (4) )
-3x + 1y = -7 (5)
(5) + (4)
-3x + x + y - y = -7 + 7
2x = 0
x = 0
So we can ignore all reference to x
-3x + 1y - 1z = -7
y = -7
So x = 0, y = -7, z = 0
2006-12-18 08:44:16 · answer #3 · answered by Tom :: Athier than Thou 6 · 2⤊ 0⤋
1x - 1y = 7
so
x = y + 7
You can now replace x in the second equation by y + 7:
2y + 14 - 2y - z = 14
<=> z = 0
Now you can replace x by y + 7 and z by 0 in the third equation:
-6y -42 + y - 1.0 = -7
<=> -5y = 35
<=> y = -7
You can now replace y by -7 and z by 0 in the first equation to find the value of x:
x + 7 + 0 = 7
<=> x = 0
So, x=0; y=-7 and z=0
2006-12-18 08:46:58 · answer #4 · answered by freekvanbaelen 2 · 2⤊ 0⤋
First of all, the x, y, and z terms are *not* equal to each other.
What you have to do is dig out your book and look in the chapter titled 'simultaneous equations' and learn all about things such as 'Gaussian Reduction', and 'Cramers Rule'. Once you understand these things, it will be fairly obvious that x = 0, y = -7, and z = 0 is the only unique, singular solution of this system.
Doug
2006-12-18 09:04:04 · answer #5 · answered by doug_donaghue 7 · 2⤊ 0⤋
Solve for variables and substitute. You can then multiply one of the equations by a number all the way through and then subtract it from the other to get variables to cancel out. You should then be able to solve the equation for another variable that should then lead you to the final answer to all. Good luck.
2006-12-18 09:03:06 · answer #6 · answered by Xbox2006 1 · 2⤊ 0⤋
x-y=7 then y=x-7(*) into
2x-2(x-7)-z=14 =>2x-2x+14-z=14 =>z=0 (**)
With (*) and (**) then we have:
-3x+(x-7)-z=-7 =.>-3x+x-7-0=-7 =>-2x=0 => x=0
y=0-7=-7 then the solution is (x,y,z)=(0,-7,0) OK???
2006-12-18 08:57:12 · answer #7 · answered by grassu a 3 · 2⤊ 0⤋