Question

1/x+2/y-4/z=1
2/x+3/y+8/z=0
-1/x+9/y+10/z=5
solve the system for x,y,z

Answer 1

\[\Large \frac1x+\frac2y-\frac4z=1\]\[\Large \frac2x+\frac3y+\frac8z=0\]\[\Large-\frac1x+\frac9y+\frac{10}z=5\]

Answer 2

This really hurts to look at ^.^
How about you do this, instead?
\[\Large x^{-1}+2y^{-1}-4z^{-1}=1\]\[\Large 2x^{-1}+3y^{-1}+8z^{-1}=0\]\[\Large -x^{-1}+9y^{-1}+10z^{-1}=5\]

Answer 3

:O

Answer 4

And now it's down to bread-and-butter systems :)
There are sooo many ways to do this, but since you said Linear Algebra, what say we use Kramer's Rule?

Think of adventure.......^.^

Answer 5

pete pan now i got idea thanks

Answer 6

That's music to my ears... even though I didn't hear a thing ^.^

Answer 7

great :) ....you can also use matrix method if you wish !

Answer 8

How about this?
\[\frac x1 + \frac y2 - \frac z4 = \frac 11\]\[\frac x2 + \frac y3 - \frac z8 = 0\]\[-\frac x1 + \frac y9 - \frac {z}{10}= \frac 15\]

1/x+2/y-4/z=1
2/x+3/y+8/z=0
-1/x+9/y+10/z=5

Answer 9

but its too complicated

Answer 10

Hey, @saifoo.khan
That really hurts :D

You're not suggesting
\[\huge a+b+c=d\] means
\[\huge a^{-1}+b^{-1}+c^{-1}=d^{-1}\]
are you...?
naughty...

Answer 11

yaa it hurts badly :/

Answer 12

For which values of a will the followingsystem have no solutions? Exactly one solution?
Infinitely many solutions?
x + 2y − 3z = 4
3x − y + 5z = 2
4x + y + (a2−14)z = a +2

Answer 13

Interesting...

Answer 14

interseting why?

Answer 15

Nothing... I'm trying to see if I'm correct :)

Answer 16

Okay, I know the answer. That was fun ^.^

Answer 17

but i dnt know the answer:(

Answer 18

This was played real nicely :)

Answer 19

So, anyway, I want to know if you have an idea, first :)

Answer 20

ya i know the idea of reduce echwelon form

Answer 21

Oh. Is that so? What happens when you add the first two equations?

Answer 22

ya its like reduce row echelon form

Answer 23

Answer me....>.>
What's the sum of the first two equations?

Answer 24

1 2 -3 4
0 1 -2 10
0 0 1 a^2-4/a^2-16
i have done till this but now the problem is that
For which values of a will the followingsystem have no solutions? Exactly one solution?
Infinitely many solutions?
how to find this

Answer 25

\[\huge \left[
\begin{array}{ccc}
1&2&-3\\
3&-1&5\\
4&1&a^2-14
\end{array}
\right|
\left.
\begin{array}{ccc}
4\\
2\\
a+2
\end{array}
\right]\]

Answer 26

yes

Answer 27

You could subtract the first and second row from the third.

Answer 28

i know that we want a solution
have you read that question

Answer 29

I did :)

Answer 30

so then

Answer 31

So then... subtract the first and the second row from the third ^.^

Answer 32

ok

Answer 33

ok? What did you get?

Answer 34

nothing

Answer 35

Well, you better get something :D

Answer 36

Or, you can forget the matrix, and just add the first two equations ^.^

Answer 37

hey you dont get my point i know all that but u dnt understand question

Answer 38

What don't I get?