Question 50099: Well I will start off saying I know I have the wrong answer somewhere and I know that because I put my answers into the original equations and they don't work. I have trouble when you throw that third variable in so if someone can tell what is right (if anything) and where I went wrong. Thanks!! Andrea
(a)
2x – y + z = 8
x + 2y + z = -3
x - 2y – z = 7
(1 & 3)
3x – 3y = 15
x – y = 5
2x = 4
x = 2
2 – y = 5
y = -3
2 + 2(-3) + z = -3
z = 1
(b)
3x - y - 2z = 11
x – 2y + 3z = 12
x + y - 2z = 5
(1 & 3) = (4)
4x – 4z = 16
(2 & 3) = (5)
x – 2y + 3z = 12
2x + 2y – 4z = 10
3x + z = 22
(4 & 5)
4x – 4z = 16
3x + z = 22
4x – 4z = 16
-12x + 4z = -88
-8x = -72
x = 9
3(9) + z = 22
27 + z = 22
z = -5
9 – y – 2(-5) = 5
9 – y + 10 = 5
19 – y = 5
-y = -14
y = 14
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! Ok, there's nothing wrong with your solution to problem a), but the solution to problem b) is another story. Let's have a look:
b)
1) 3x-y-2z = 11
2) x-2y+3z = 12
3) x+y-2z = 5
Add 1) & 3)
4) 4x-4z = 16 You are ok here.
4a) x-z = 4 You are ok here.
Now add 2) and 2 X 3)
2) x-2y+3z = 12
3a) 2x+2y-4z = 10 adding theses two, you get (or should have obtained)
5) 3x-z = 22 Here's one of your boo boo's (3z+(-4z) = -z...not +z as you had)
Now here you could have subtracted 4a) from 5) to get x.
4a) x-z = 4
5) 3x-z = 22
6) 2x = 18
6a) x = 9 ...and you got this right!
Now to find z. You can substitute x = 9 into 4a.
4b) 9-z = 4 Solve for z.
4c) z = 5 Here's another boo boo, you had z = -5
To find y, substitute x = 9 and z = 5 into any of your original equations and solve for y. Let's use equation 1)
3x-y-2z = 11
3(9)-y-2(5) = 11
27-y-10 = 11
17-y = 11
y = 6 ...and here's you third mistake. You had y = 14.
The solution:
x = 9
y = 6
z = 5
Check:
1) 3x-y-2z = 11
3(9)-6-2(5) = 11
27-6-10 = 11
27-16 = 11
11 = 11 OK here.
2) x-2y+3z = 12
9-2(6)+3(5) = 12
9-12+15 = 12
24-12 = 12
12 = 12 OK here too.
3) x+y-2z = 5
9+6-2(5) = 5
9+6-10 = 5
15-10 = 5
5 = 5 ...and alright here.
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