Question

Solve the following system of equations.

x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1

Answer 1

x=-3,y=-1,z=-5

Answer 2

I'm really bad at these - but I will do my best, for you. Sorry if I stuff up.

x + 3y + 2z = –16
2x – y + 2z = –15
2x – 2y – z = 1

Make x the subject in the first equation.
3y-2z-16=x
Substitute this value into your second and third equation.

2(3y-2z-16) – y + 2z = –15
2(3y-2z-16) – 2y – z = 1

Distribute.

2(3y-2z-16) – y + 2z = –15
2(3y-2z-16) – 2y – z = 1

6y-4z-32-y+2z=-15
6y-4z-32-y-z=1

Gather like terms

5y-2z-32=-15
5y-5z-32=1

Plus 32 both sides.
6y-2z=17 <-- multiply by 5
5y-5z=31 <-- multiply by 2

30y-10z=17
10y-10z=62

Take away to eliminate:

20y=-45
y = -9/4

Sub this back in.
10y-10z=62
10x(-9/4)-10z=62
-22.5-10z=62
-10z = 84.5
z = -8.45

2x – 2(-9/4) – (-8.45) = 1
2x+79/20 =1
2x =-59/20
x = -2.95

x = -2.95
y = -9/4
z - 8.45

Sorry if this is wrong. I tried ):

Answer 3

This is wrong - I will leave it here though, so everyone can glory over my failure, or find where I went wrong. Sorry ):

Answer 4

are you allowed to use matrices to solve this Psssst? It would be easier with matrices.

Answer 5

Not yet. I use that in the next lesson.

Answer 6

Make x the subject in the first equation.
3y-2z-16=x
I think here is the error or am I wrong?

x=-16-3y-2z

Answer 7

More than likely - I think the procedure is correct though, I hope. Sorry again :x

Answer 8

x + 3y + 2z = –16 [1]
2x – y + 2z = –15 [2]
2x – 2y – z = 1 [3]

[2] - [3] : y + 3z = -16 [4]
[1]x2: 2x + 6y + 4z = -32 [5]
[2] - [5]: -7y -2z = 17 [6]

we now solve [4] with [6]:

[4]x7: 7y + 21z = -112 [7]
[5] + [7]: 19z = -95 hence z = -5.
From [4], y -15 = -16, hence y = -1.
From [1], x - 3 - 10 = -16 hence x = -3

Answer 9

xactxx that's fine

Answer 10

you can also solve these by method of elimination
subtract: 1 - 2:

-x + 4y = -1.............................(a)
multiply 3rd equn. by 2
4x-4y-2z = 2
add this to 2nd equn.
6x -5y = -13..............................(b)

x (a) by 6:
-6x + 24y = -6
add this to (b):
19y = -19
y = -1
substitute y = -1 into (a)
-x -4 = -1
x = -3

find z by plugging y and x in first equn.:

-3 +3*-1 + 2z = -16
2z = -16 + 6 = -10
z = -5

x=-3,y=-1,z=-5