Problem 1 is: Problem 2 is:
2x+y-z=10
3x+4y+z=14 6x+5y-2z=23
2y+7z=25 3x-y+z=5
-2y+3z=5
Since x is eliminated from both the 2nd and 3rd equation, just solve them:
2y+7z=25
-2y+3z=5
Add them and the y's cancel and you get
10z=30
z=3
Substitute z=3 in
2y+7z=25
2y+7(3)=25
2y+21=25
2y=4
y=2
Substitute y=2 and z=3 in
3x+4y+z=14
3x+4(2)+(3)=14
3x+8+3=14
3x+11=14
3x=3
x=1
(x,y,z) = (1,2,3)
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Problem 2 is:
2x+ y- z=10
6x+5y-2z=23
3x- y+ z= 5
Add the 1st and 3rd as they are and both the y's and will
cancel:
2x+y-z=10
3x-y+z= 5
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5x =15
x=3
Substitute x=3 in
2x+y-z=10
2(3)+y-z=10
6+y-z=10
y-z=4
Substitute x=3 in the middle original equation:
6x+5y-2z=23
6(3)+5y-2z=23
18+5y-2z=23
5y-2z=5
Now you have this system:
y- z=4
5y-2z=5
Eliminate z by multiplying the first by -2 and adding:
-2y+2z=-8
5y-2z= 5
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3y =-3
y=-1
Substitute y=-1 in
y-z=4
(-1)-z=4
-1-z=4
-z=5
z=-5
(x,y,z) = (3,-1,-5)
Edwin
